I attempt to replace the Carriage Return with Space. In
this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first carriage
return but not all. Is there any suggestion ?
Thanks
try removing the line feed character as well - char(10)
regards,
Mark Baekdal
http://www.dbghost.com
http://www.innovartis.co.uk
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
> I attempt to replace the Carriage Return with Space. In
> this way, in the SQL Statement, I add this script:
> replace(t1.address, char(13), char(32))
> However, it seems that it only remove the first carriage
> return but not all. Is there any suggestion ?
> Thanks
>
|||Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
[vbcol=seagreen]
>--Original Message--
>try removing the line feed character as well - char(10)
>
>regards,
>Mark Baekdal
>http://www.dbghost.com
>http://www.innovartis.co.uk
>+44 (0)208 241 1762
>Database change management for SQL Server
>
>"Peter" wrote:
In[vbcol=seagreen]
carriage
>.
>
|||Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @.tmp varchar(100)
set @.tmp = 'line1
line2
line3
line4
line5'
select @.tmp
select replace(@.tmp, char(13)+char(10), char(32))
*mike hodgson* |/ database administrator/ | mallesons stephen jaques
*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907
*E* mailto:mike.hodgson@.mallesons.nospam.com |* W* http://www.mallesons.com
Peter wrote:
[vbcol=seagreen]
>Does it mean that I have to run like this
>replace(replace(t1.address, char(13), char(32)), char(10),
>char(32)) ?
>Thanks
>
>In
>
>carriage
>
|||Thank you for your advice.
When I use char(13)+char(10), most of the problems are
fixed.
However, for unknown reason, there is record that works
fine with char(13) only becomes like this:
Aboukir Street
MONGKOK ABN 3355
Do you have any suggestion ?
Besides, when I use Notepad, is it possible for me to see
whether it is only CR or CRLF ?
Thanks
>--Original Message--
>Have you tried
> replace(t1.address, char(13)+char(10), char(32))
>as Mark suggested? It works for me (see example below).
>declare @.tmp varchar(100)
>set @.tmp = 'line1
>line2
>line3
>line4
>line5'
>select @.tmp
>select replace(@.tmp, char(13)+char(10), char(32))
>--
>*mike hodgson* |/ database administrator/ | mallesons
stephen jaques
>*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61
(408) 675 907
>*E* mailto:mike.hodgson@.mallesons.nospam.com |* W*
http://www.mallesons.com[vbcol=seagreen]
>
>Peter wrote:
(10),
>
Showing posts with label space. Show all posts
Showing posts with label space. Show all posts
Wednesday, March 28, 2012
Remove Carriage Return
I attempt to replace the Carriage Return with Space. In
this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first carriage
return but not all. Is there any suggestion ?
Thankstry removing the line feed character as well - char(10)
regards,
Mark Baekdal
http://www.dbghost.com
http://www.innovartis.co.uk
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
> I attempt to replace the Carriage Return with Space. In
> this way, in the SQL Statement, I add this script:
> replace(t1.address, char(13), char(32))
> However, it seems that it only remove the first carriage
> return but not all. Is there any suggestion ?
> Thanks
>|||Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
>--Original Message--
>try removing the line feed character as well - char(10)
>
>regards,
>Mark Baekdal
>http://www.dbghost.com
>http://www.innovartis.co.uk
>+44 (0)208 241 1762
>Database change management for SQL Server
>
>"Peter" wrote:
>
In[vbcol=seagreen]
carriage[vbcol=seagreen]
>.
>|||Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @.tmp varchar(100)
set @.tmp = 'line1
line2
line3
line4
line5'
select @.tmp
select replace(@.tmp, char(13)+char(10), char(32))
*mike hodgson* |/ database administrator/ | mallesons stephen jaques
*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907
*E* mailto:mike.hodgson@.mallesons.nospam.com |* W* http://www.mallesons.com
Peter wrote:
[vbcol=seagreen]
>Does it mean that I have to run like this
>replace(replace(t1.address, char(13), char(32)), char(10),
>char(32)) ?
>Thanks
>
>
>In
>
>carriage
>|||Thank you for your advice.
When I use char(13)+char(10), most of the problems are
fixed.
However, for unknown reason, there is record that works
fine with char(13) only becomes like this:
Aboukir Street
MONGKOK ABN 3355
Do you have any suggestion ?
Besides, when I use Notepad, is it possible for me to see
whether it is only CR or CRLF ?
Thanks
>--Original Message--
>Have you tried
> replace(t1.address, char(13)+char(10), char(32))
>as Mark suggested? It works for me (see example below).
>declare @.tmp varchar(100)
>set @.tmp = 'line1
>line2
>line3
>line4
>line5'
>select @.tmp
>select replace(@.tmp, char(13)+char(10), char(32))
>--
>*mike hodgson* |/ database administrator/ | mallesons
stephen jaques
>*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61
(408) 675 907
>*E* mailto:mike.hodgson@.mallesons.nospam.com |* W*
http://www.mallesons.com
>
>Peter wrote:
>
(10),[vbcol=seagreen]
>
this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first carriage
return but not all. Is there any suggestion ?
Thankstry removing the line feed character as well - char(10)
regards,
Mark Baekdal
http://www.dbghost.com
http://www.innovartis.co.uk
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
> I attempt to replace the Carriage Return with Space. In
> this way, in the SQL Statement, I add this script:
> replace(t1.address, char(13), char(32))
> However, it seems that it only remove the first carriage
> return but not all. Is there any suggestion ?
> Thanks
>|||Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
>--Original Message--
>try removing the line feed character as well - char(10)
>
>regards,
>Mark Baekdal
>http://www.dbghost.com
>http://www.innovartis.co.uk
>+44 (0)208 241 1762
>Database change management for SQL Server
>
>"Peter" wrote:
>
In[vbcol=seagreen]
carriage[vbcol=seagreen]
>.
>|||Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @.tmp varchar(100)
set @.tmp = 'line1
line2
line3
line4
line5'
select @.tmp
select replace(@.tmp, char(13)+char(10), char(32))
*mike hodgson* |/ database administrator/ | mallesons stephen jaques
*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907
*E* mailto:mike.hodgson@.mallesons.nospam.com |* W* http://www.mallesons.com
Peter wrote:
[vbcol=seagreen]
>Does it mean that I have to run like this
>replace(replace(t1.address, char(13), char(32)), char(10),
>char(32)) ?
>Thanks
>
>
>In
>
>carriage
>|||Thank you for your advice.
When I use char(13)+char(10), most of the problems are
fixed.
However, for unknown reason, there is record that works
fine with char(13) only becomes like this:
Aboukir Street
MONGKOK ABN 3355
Do you have any suggestion ?
Besides, when I use Notepad, is it possible for me to see
whether it is only CR or CRLF ?
Thanks
>--Original Message--
>Have you tried
> replace(t1.address, char(13)+char(10), char(32))
>as Mark suggested? It works for me (see example below).
>declare @.tmp varchar(100)
>set @.tmp = 'line1
>line2
>line3
>line4
>line5'
>select @.tmp
>select replace(@.tmp, char(13)+char(10), char(32))
>--
>*mike hodgson* |/ database administrator/ | mallesons
stephen jaques
>*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61
(408) 675 907
>*E* mailto:mike.hodgson@.mallesons.nospam.com |* W*
http://www.mallesons.com
>
>Peter wrote:
>
(10),[vbcol=seagreen]
>
Remove Carriage Return
I attempt to replace the Carriage Return with Space. In
this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first carriage
return but not all. Is there any suggestion ?
Thankstry removing the line feed character as well - char(10)
regards,
Mark Baekdal
http://www.dbghost.com
http://www.innovartis.co.uk
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
> I attempt to replace the Carriage Return with Space. In
> this way, in the SQL Statement, I add this script:
> replace(t1.address, char(13), char(32))
> However, it seems that it only remove the first carriage
> return but not all. Is there any suggestion ?
> Thanks
>|||Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
>--Original Message--
>try removing the line feed character as well - char(10)
>
>regards,
>Mark Baekdal
>http://www.dbghost.com
>http://www.innovartis.co.uk
>+44 (0)208 241 1762
>Database change management for SQL Server
>
>"Peter" wrote:
>> I attempt to replace the Carriage Return with Space.
In
>> this way, in the SQL Statement, I add this script:
>> replace(t1.address, char(13), char(32))
>> However, it seems that it only remove the first
carriage
>> return but not all. Is there any suggestion ?
>> Thanks
>.
>|||This is a multi-part message in MIME format.
--060207020402090904070407
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit
Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @.tmp varchar(100)
set @.tmp = 'line1
line2
line3
line4
line5'
select @.tmp
select replace(@.tmp, char(13)+char(10), char(32))
--
*mike hodgson* |/ database administrator/ | mallesons stephen jaques
*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907
*E* mailto:mike.hodgson@.mallesons.nospam.com |* W* http://www.mallesons.com
Peter wrote:
>Does it mean that I have to run like this
>replace(replace(t1.address, char(13), char(32)), char(10),
>char(32)) ?
>Thanks
>
>>--Original Message--
>>try removing the line feed character as well - char(10)
>>
>>regards,
>>Mark Baekdal
>>http://www.dbghost.com
>>http://www.innovartis.co.uk
>>+44 (0)208 241 1762
>>Database change management for SQL Server
>>
>>"Peter" wrote:
>>
>>I attempt to replace the Carriage Return with Space.
>>
>In
>
>>this way, in the SQL Statement, I add this script:
>>replace(t1.address, char(13), char(32))
>>However, it seems that it only remove the first
>>
>carriage
>
>>return but not all. Is there any suggestion ?
>>Thanks
>>
>>.
>>
--060207020402090904070407
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta content="text/html;charset=ISO-8859-1" http-equiv="Content-Type">
</head>
<body bgcolor="#ffffff" text="#000000">
<tt>Have you tried<br>
<br>
replace(t1.address, char(13)+char(10), char(32))<br>
<br>
as Mark suggested? It works for me (see example below).<br>
<br>
declare @.tmp varchar(100)<br>
set @.tmp = 'line1<br>
line2<br>
line3<br>
line4<br>
line5'<br>
<br>
select @.tmp<br>
select replace(@.tmp, char(13)+char(10), char(32))<br>
</tt>
<div class="moz-signature">
<title></title>
<meta http-equiv="Content-Type" content="text/html; ">
<p><span lang="en-au"><font face="Tahoma" size="2">--<br>
</font> </span><b><span lang="en-au"><font face="Tahoma" size="2">mike
hodgson</font></span></b><span lang="en-au"> <font face="Tahoma"
size="2">|</font><i><font face="Tahoma"> </font><font face="Tahoma"
size="2"> database administrator</font></i><font face="Tahoma" size="2">
| mallesons</font><font face="Tahoma"> </font><font face="Tahoma"
size="2">stephen</font><font face="Tahoma"> </font><font face="Tahoma"
size="2"> jaques</font><font face="Tahoma"><br>
</font><b><font face="Tahoma" size="2">T</font></b><font face="Tahoma"
size="2"> +61 (2) 9296 3668 |</font><b><font face="Tahoma"> </font><font
face="Tahoma" size="2"> F</font></b><font face="Tahoma" size="2"> +61
(2) 9296 3885 |</font><b><font face="Tahoma"> </font><font
face="Tahoma" size="2">M</font></b><font face="Tahoma" size="2"> +61
(408) 675 907</font><br>
<b><font face="Tahoma" size="2">E</font></b><font face="Tahoma" size="2">
<a href="http://links.10026.com/?link=mailto:mike.hodgson@.mallesons.nospam.com">
mailto:mike.hodgson@.mallesons.nospam.com</a> |</font><b><font
face="Tahoma"> </font><font face="Tahoma" size="2">W</font></b><font
face="Tahoma" size="2"> <a href="http://links.10026.com/?link=/">http://www.mallesons.com">
http://www.mallesons.com</a></font></span> </p>
</div>
<br>
<br>
Peter wrote:
<blockquote cite="mid0ea101c518a6$4f165ab0$a401280a@.phx.gbl" type="cite">
<pre wrap="">Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
</pre>
<blockquote type="cite">
<pre wrap="">--Original Message--
try removing the line feed character as well - char(10)
regards,
Mark Baekdal
<a class="moz-txt-link-freetext" href="http://links.10026.com/?link=http://www.dbghost.com</a>">http://www.dbghost.com">http://www.dbghost.com</a>
<a class="moz-txt-link-freetext" href="http://links.10026.com/?link=http://www.innovartis.co.uk</a>">http://www.innovartis.co.uk">http://www.innovartis.co.uk</a>
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
</pre>
<blockquote type="cite">
<pre wrap="">I attempt to replace the Carriage Return with Space.
</pre>
</blockquote>
</blockquote>
<pre wrap=""><!-->In
</pre>
<blockquote type="cite">
<blockquote type="cite">
<pre wrap="">this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first
</pre>
</blockquote>
</blockquote>
<pre wrap=""><!-->carriage
</pre>
<blockquote type="cite">
<blockquote type="cite">
<pre wrap="">return but not all. Is there any suggestion ?
Thanks
</pre>
</blockquote>
<pre wrap="">.
</pre>
</blockquote>
</blockquote>
</body>
</html>
--060207020402090904070407--|||Thank you for your advice.
When I use char(13)+char(10), most of the problems are
fixed.
However, for unknown reason, there is record that works
fine with char(13) only becomes like this:
Aboukir Street
MONGKOK ABN 3355
Do you have any suggestion ?
Besides, when I use Notepad, is it possible for me to see
whether it is only CR or CRLF ?
Thanks
>--Original Message--
>Have you tried
> replace(t1.address, char(13)+char(10), char(32))
>as Mark suggested? It works for me (see example below).
>declare @.tmp varchar(100)
>set @.tmp = 'line1
>line2
>line3
>line4
>line5'
>select @.tmp
>select replace(@.tmp, char(13)+char(10), char(32))
>--
>*mike hodgson* |/ database administrator/ | mallesons
stephen jaques
>*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61
(408) 675 907
>*E* mailto:mike.hodgson@.mallesons.nospam.com |* W*
http://www.mallesons.com
>
>Peter wrote:
>>Does it mean that I have to run like this
>>replace(replace(t1.address, char(13), char(32)), char
(10),
>>char(32)) ?
>>Thanks
>>
>>--Original Message--
>>try removing the line feed character as well - char(10)
>>
>>regards,
>>Mark Baekdal
>>http://www.dbghost.com
>>http://www.innovartis.co.uk
>>+44 (0)208 241 1762
>>Database change management for SQL Server
>>
>>"Peter" wrote:
>>
>>I attempt to replace the Carriage Return with Space.
>>
>>In
>>
>>this way, in the SQL Statement, I add this script:
>>replace(t1.address, char(13), char(32))
>>However, it seems that it only remove the first
>>
>>carriage
>>
>>return but not all. Is there any suggestion ?
>>Thanks
>>
>>.
>>
>sql
this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first carriage
return but not all. Is there any suggestion ?
Thankstry removing the line feed character as well - char(10)
regards,
Mark Baekdal
http://www.dbghost.com
http://www.innovartis.co.uk
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
> I attempt to replace the Carriage Return with Space. In
> this way, in the SQL Statement, I add this script:
> replace(t1.address, char(13), char(32))
> However, it seems that it only remove the first carriage
> return but not all. Is there any suggestion ?
> Thanks
>|||Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
>--Original Message--
>try removing the line feed character as well - char(10)
>
>regards,
>Mark Baekdal
>http://www.dbghost.com
>http://www.innovartis.co.uk
>+44 (0)208 241 1762
>Database change management for SQL Server
>
>"Peter" wrote:
>> I attempt to replace the Carriage Return with Space.
In
>> this way, in the SQL Statement, I add this script:
>> replace(t1.address, char(13), char(32))
>> However, it seems that it only remove the first
carriage
>> return but not all. Is there any suggestion ?
>> Thanks
>.
>|||This is a multi-part message in MIME format.
--060207020402090904070407
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit
Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @.tmp varchar(100)
set @.tmp = 'line1
line2
line3
line4
line5'
select @.tmp
select replace(@.tmp, char(13)+char(10), char(32))
--
*mike hodgson* |/ database administrator/ | mallesons stephen jaques
*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907
*E* mailto:mike.hodgson@.mallesons.nospam.com |* W* http://www.mallesons.com
Peter wrote:
>Does it mean that I have to run like this
>replace(replace(t1.address, char(13), char(32)), char(10),
>char(32)) ?
>Thanks
>
>>--Original Message--
>>try removing the line feed character as well - char(10)
>>
>>regards,
>>Mark Baekdal
>>http://www.dbghost.com
>>http://www.innovartis.co.uk
>>+44 (0)208 241 1762
>>Database change management for SQL Server
>>
>>"Peter" wrote:
>>
>>I attempt to replace the Carriage Return with Space.
>>
>In
>
>>this way, in the SQL Statement, I add this script:
>>replace(t1.address, char(13), char(32))
>>However, it seems that it only remove the first
>>
>carriage
>
>>return but not all. Is there any suggestion ?
>>Thanks
>>
>>.
>>
--060207020402090904070407
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta content="text/html;charset=ISO-8859-1" http-equiv="Content-Type">
</head>
<body bgcolor="#ffffff" text="#000000">
<tt>Have you tried<br>
<br>
replace(t1.address, char(13)+char(10), char(32))<br>
<br>
as Mark suggested? It works for me (see example below).<br>
<br>
declare @.tmp varchar(100)<br>
set @.tmp = 'line1<br>
line2<br>
line3<br>
line4<br>
line5'<br>
<br>
select @.tmp<br>
select replace(@.tmp, char(13)+char(10), char(32))<br>
</tt>
<div class="moz-signature">
<title></title>
<meta http-equiv="Content-Type" content="text/html; ">
<p><span lang="en-au"><font face="Tahoma" size="2">--<br>
</font> </span><b><span lang="en-au"><font face="Tahoma" size="2">mike
hodgson</font></span></b><span lang="en-au"> <font face="Tahoma"
size="2">|</font><i><font face="Tahoma"> </font><font face="Tahoma"
size="2"> database administrator</font></i><font face="Tahoma" size="2">
| mallesons</font><font face="Tahoma"> </font><font face="Tahoma"
size="2">stephen</font><font face="Tahoma"> </font><font face="Tahoma"
size="2"> jaques</font><font face="Tahoma"><br>
</font><b><font face="Tahoma" size="2">T</font></b><font face="Tahoma"
size="2"> +61 (2) 9296 3668 |</font><b><font face="Tahoma"> </font><font
face="Tahoma" size="2"> F</font></b><font face="Tahoma" size="2"> +61
(2) 9296 3885 |</font><b><font face="Tahoma"> </font><font
face="Tahoma" size="2">M</font></b><font face="Tahoma" size="2"> +61
(408) 675 907</font><br>
<b><font face="Tahoma" size="2">E</font></b><font face="Tahoma" size="2">
<a href="http://links.10026.com/?link=mailto:mike.hodgson@.mallesons.nospam.com">
mailto:mike.hodgson@.mallesons.nospam.com</a> |</font><b><font
face="Tahoma"> </font><font face="Tahoma" size="2">W</font></b><font
face="Tahoma" size="2"> <a href="http://links.10026.com/?link=/">http://www.mallesons.com">
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</div>
<br>
<br>
Peter wrote:
<blockquote cite="mid0ea101c518a6$4f165ab0$a401280a@.phx.gbl" type="cite">
<pre wrap="">Does it mean that I have to run like this
replace(replace(t1.address, char(13), char(32)), char(10),
char(32)) ?
Thanks
</pre>
<blockquote type="cite">
<pre wrap="">--Original Message--
try removing the line feed character as well - char(10)
regards,
Mark Baekdal
<a class="moz-txt-link-freetext" href="http://links.10026.com/?link=http://www.dbghost.com</a>">http://www.dbghost.com">http://www.dbghost.com</a>
<a class="moz-txt-link-freetext" href="http://links.10026.com/?link=http://www.innovartis.co.uk</a>">http://www.innovartis.co.uk">http://www.innovartis.co.uk</a>
+44 (0)208 241 1762
Database change management for SQL Server
"Peter" wrote:
</pre>
<blockquote type="cite">
<pre wrap="">I attempt to replace the Carriage Return with Space.
</pre>
</blockquote>
</blockquote>
<pre wrap=""><!-->In
</pre>
<blockquote type="cite">
<blockquote type="cite">
<pre wrap="">this way, in the SQL Statement, I add this script:
replace(t1.address, char(13), char(32))
However, it seems that it only remove the first
</pre>
</blockquote>
</blockquote>
<pre wrap=""><!-->carriage
</pre>
<blockquote type="cite">
<blockquote type="cite">
<pre wrap="">return but not all. Is there any suggestion ?
Thanks
</pre>
</blockquote>
<pre wrap="">.
</pre>
</blockquote>
</blockquote>
</body>
</html>
--060207020402090904070407--|||Thank you for your advice.
When I use char(13)+char(10), most of the problems are
fixed.
However, for unknown reason, there is record that works
fine with char(13) only becomes like this:
Aboukir Street
MONGKOK ABN 3355
Do you have any suggestion ?
Besides, when I use Notepad, is it possible for me to see
whether it is only CR or CRLF ?
Thanks
>--Original Message--
>Have you tried
> replace(t1.address, char(13)+char(10), char(32))
>as Mark suggested? It works for me (see example below).
>declare @.tmp varchar(100)
>set @.tmp = 'line1
>line2
>line3
>line4
>line5'
>select @.tmp
>select replace(@.tmp, char(13)+char(10), char(32))
>--
>*mike hodgson* |/ database administrator/ | mallesons
stephen jaques
>*T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61
(408) 675 907
>*E* mailto:mike.hodgson@.mallesons.nospam.com |* W*
http://www.mallesons.com
>
>Peter wrote:
>>Does it mean that I have to run like this
>>replace(replace(t1.address, char(13), char(32)), char
(10),
>>char(32)) ?
>>Thanks
>>
>>--Original Message--
>>try removing the line feed character as well - char(10)
>>
>>regards,
>>Mark Baekdal
>>http://www.dbghost.com
>>http://www.innovartis.co.uk
>>+44 (0)208 241 1762
>>Database change management for SQL Server
>>
>>"Peter" wrote:
>>
>>I attempt to replace the Carriage Return with Space.
>>
>>In
>>
>>this way, in the SQL Statement, I add this script:
>>replace(t1.address, char(13), char(32))
>>However, it seems that it only remove the first
>>
>>carriage
>>
>>return but not all. Is there any suggestion ?
>>Thanks
>>
>>.
>>
>sql
Monday, March 26, 2012
Remove "blank" space?
I have a report that contains a table with two rows. Each of these rows
contains a table with a header row, detail row and footer row. When the
report loads, these inner tables expand to 5-20 rows of data. All well and
good. However, sometimes the top inner table has no data to render. But when
the report renders, this space is blank (the height of the row of the outer
table). So there's a big gap of whitespace at the top of the report.
Is there any way to have this space shrink to nothing (or nearly nothing)?
Or at least very small? I can't seem to find any way to make this happen.
Thanks,
Ripleywrite an expression on the visibilty expression
"Ripley" <ripley@.ihatespam.com> wrote in message
news:epyIiUg0GHA.4228@.TK2MSFTNGP06.phx.gbl...
>I have a report that contains a table with two rows. Each of these rows
>contains a table with a header row, detail row and footer row. When the
>report loads, these inner tables expand to 5-20 rows of data. All well and
>good. However, sometimes the top inner table has no data to render. But
>when the report renders, this space is blank (the height of the row of the
>outer table). So there's a big gap of whitespace at the top of the report.
> Is there any way to have this space shrink to nothing (or nearly nothing)?
> Or at least very small? I can't seem to find any way to make this happen.
> Thanks,
> Ripley
>sql
contains a table with a header row, detail row and footer row. When the
report loads, these inner tables expand to 5-20 rows of data. All well and
good. However, sometimes the top inner table has no data to render. But when
the report renders, this space is blank (the height of the row of the outer
table). So there's a big gap of whitespace at the top of the report.
Is there any way to have this space shrink to nothing (or nearly nothing)?
Or at least very small? I can't seem to find any way to make this happen.
Thanks,
Ripleywrite an expression on the visibilty expression
"Ripley" <ripley@.ihatespam.com> wrote in message
news:epyIiUg0GHA.4228@.TK2MSFTNGP06.phx.gbl...
>I have a report that contains a table with two rows. Each of these rows
>contains a table with a header row, detail row and footer row. When the
>report loads, these inner tables expand to 5-20 rows of data. All well and
>good. However, sometimes the top inner table has no data to render. But
>when the report renders, this space is blank (the height of the row of the
>outer table). So there's a big gap of whitespace at the top of the report.
> Is there any way to have this space shrink to nothing (or nearly nothing)?
> Or at least very small? I can't seem to find any way to make this happen.
> Thanks,
> Ripley
>sql
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