Replace String in XML file

Question

I try to replace string in XML file with sed, but since its regular expression doesn't support non-greedy expressions, I encounter a problem.

XML Example:

<opt>
  <Node active="yes" file="/home/user/random_filename" last_time="17/07/14-00:02:07" time_in_min="5" />
</opt>

I want to find the file attribute, which is random string, and replace it with another string.

This command replaces the string, BUT removes the trailing data.

sed 's/file=".*"/file="new_file_name"/' file.xml

Output:

<opt>
   <Node active="yes" file="new_file_name" />
</opt>

How should I handle it?

possible duplicate of regex match non greedy

whereswalden
– whereswalden

2014-07-17 02:20:54 +00:00
Commented Jul 17, 2014 at 2:20 — whereswalden
– whereswalden, Commented Jul 17, 2014 at 2:20

OnlineCop · Accepted Answer · 2014-07-16 21:37:16Z

2

Use "[^"]*" instead of ".*":

Input:

<opt>
  <Node active="yes" file="/home/user/random_filename" last_time="17/07/14-00:02:07" time_in_min="5" />
</opt>

Command:

sed 's/file="[^"]*"/file="new_file_name"/'

Output:

<opt>
  <Node active="yes" file="new_file_name" last_time="17/07/14-00:02:07" time_in_min="5" />
</opt>

answered Jul 16, 2014 at 21:37

OnlineCop

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Comments

Cole Tierney · Accepted Answer · 2014-07-17 12:57:21Z

1

This doesn't use sed but when manipulating xml documents, you might want to consider xslt. It may be an over the top solution to your problem, but it's a great fit for working with xml.

filter.xsl:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" omit-xml-declaration="yes"/>

<!-- The replacement string is passed as a parameter.
     You can have as many param elements as you need. -->
<xsl:param name="NEW_NAME"/>

<!-- Copy all attributes and nodes... -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<!-- ...but when you encounter an element with a file attribute,
     replace it with the value passed in the parameter named $NEW_NAME. -->
<xsl:template match="@file">
    <xsl:attribute name="file">
        <xsl:value-of select="$NEW_NAME"/>
    </xsl:attribute>
</xsl:template>

</xsl:stylesheet>

A libxslt example (add as many --stringparam name/value pairs that you need):

xsltproc --stringparam NEW_NAME new_file_name filter.xsl file.xml

Result:

<opt>
    <Node active="yes" file="new_file_name" last_time="17/07/14-00:02:07" time_in_min="5"/>
</opt>

edited Jul 17, 2014 at 12:57

answered Jul 16, 2014 at 22:56

Cole Tierney

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2 Comments

Alex Over a year ago

Nice one! I will consider it. I need to change more than one argument and it looks like most convenient way. Thanks.

Cole Tierney Over a year ago

The match attribute xpath expression can be made more specific if needed. For example to only change file attributes in Node elements, change match="@file" to match="Node/@file". You can also match based on the contents of the file attribute. To change only those that start with '/home/user/' use match="Node/@file[starts-with(., '/home/user/')]".

Ed Morton · Accepted Answer · 2014-07-17 00:10:46Z

0

$ sed 's/\(.*file="\)[^"]*/\1new_file_name/' file
<opt>
  <Node active="yes" file="new_file_name" last_time="17/07/14-00:02:07" time_in_min="5" />
</opt>

answered Jul 17, 2014 at 0:10

Ed Morton

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