Thanks for the reply it fix my problem.
Still i was wondering is there a way that we can remove the spaces from
the character ~ to linefeed/eod like the following
LINE1: ISA*00* *00* *ZZ*7654321 *ZZ*1234567
LINE2:*020503*1705*^*00401*000010232*0*P*:~
LINE3:GS*HC*7654321 *1234567*20020503*1705*20213*X*004010X096A1~
LINE4:ST*837 *0001~
LINE5:BHT*0019*00*123B*20010329*1310*CH~REF*87*004 010X096A1~
I am looking for some thing like
LINE1:ISA*00* *00* *ZZ*7654321 *ZZ*1234567
LINE2:*020503*1705*^*00401*000010232*0*P*:~GS*HC*7654321
*1234567*20020503*1705*20213*X*004010X096A1~ST*837
*0001~BHT*0019*00*123B*20010329*1310*CH~REF*87*004 010X096A1~
Dont think about LINE1:,LINE2:,LINE3: and LINE4, for understanding i
have written edi format like that.
Basically i need to go charater ~ to line feed and remove the spaces
b/w them even linefeed and club first line and second line.
Does any one how ideas how to fix this.
--- On Thu, 11/11/10, Abel Braaksma <abel(_dot_)online(_at_)xs4all(_dot_)nl>
wrote:
From: Abel Braaksma <abel(_dot_)online(_at_)xs4all(_dot_)nl>
Subject: Re: [xsl] removing crlf character with out white spaces
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Date: Thursday, 11 November, 2010, 1:25 PM
I dont want to
remove these whitespaces but want to remove the cr's n lf's
See Brandon Ibach's reply, which does exactly that. Make
sure to _not_
use normalize-space anymore, just translate(.,
'&_#10;&_#13;', '')
(remove underscore).
Kind regards,
Abel Braaksma
On 11-11-2010 5:52, ram wrote:
what exactly i am looking is my EDI format is going to
be some thing like this. I want to remove the carriage
returns and line spaces from this text. There might be white
spaces b/w ISA*00*
*00* *zz*7654321
I am not how many white spaces are going to come.
I dont want to remove these whitespaces but want to
remove the cr's n lf's
ISA*00* *00*
*ZZ*7654321
*ZZ*1234567
*020503*1705*^*00401*000010232*0*P*:~GS*HC*7654321
*1234567*20020503*1705*20213*X*004010X096A1~ST*837
*0001~BHT*0019*00*123B*20010329*1310*CH~REF*87*004
010X096A1~NM1*41*2*SUBMITTER
ORGANIZATION*****46*ETIN123~PER*IC*SUBMITTER CONTACT
NAME*TE*SUBMITTER COMM
NUMBER-EMAIL~NM1*40*2*RECEIVER
ORGANIZATION*****46*ETIN123~HL*1**20*1~PRV*BI*ZZ*2
82N00000N~NM1*85*2*BILLINGPROVIDER
ORGANIZATION*****24*EID123456~N3*123
BILLINGPROVIDERADDRESS
LINE~N4*BILLINGPROVIDER
CITY*MD*34567~REF*1C*MEDICAREPN123~HL*2*1*22*1~SBR
*P*18*******MA~NM1*IL*1*SUBSCRIBER-LASTNAME*SUBSCRIBERFIRSTNAME*SUBSCRIBERMI***MI*MEM
BERIDNUMBER123456789~N3*123
SUBSCRIBERADDRESS LINE~N4*SUBSCRIBER
--- On Thu, 11/11/10, Brandon
Ibach<brandon(_dot_)ibach(_at_)single-sourcing(_dot_)com>
wrote:
From: Brandon Ibach<brandon(_dot_)ibach(_at_)single-sourcing(_dot_)com>
Subject: Re: [xsl] removing crlf character with
out white spaces
To: "xsl-list"<xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Date: Thursday, 11 November, 2010, 9:57 AM
Perhaps something like:
<xsl:template
match="text()">
<xsl:value-of select="translate(.,
'
', '')" />
</xsl:template>
-Brandon :)
On Wed, Nov 10, 2010 at 11:09 PM,
ram<ram_kurra(_at_)yahoo(_dot_)co(_dot_)in>
wrote:
Hi,
I am trying
to write an xsl which will
take read text which is in EDI format and crlf
characters.
My code is
<?xml version="1.0" encoding="UTF-8"?>
version="1.0"
xmlns:xalan="http://xml.apache.org/xslt">
<xsl:template
match="text()">
<xsl:value-of
select="normalize-space()"
/>
</xsl:template>
<xsl:template
match="*">
<xsl:copy>
<xsl:copy-of select="@*" />
<xsl:apply-templates />
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
The normalize-space() is removing
all the crlf characters and also trimming white
spaces
trimming into single space which i dont want to do
it.
i need only cr and line feeds needs to be
removed. How
can i do that.
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