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Grep content in xml file

 
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# 1  
Old 11-04-2013
Grep content in xml file
I have an xml file with header as below.

Code:
<Provider xmlns="http://www.xyzx.gov/xyz" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.xyzx.gov/xyz xyz.xsd" SCHEMA_VERSION="2.5" PROVIDER="5">

I want to get the schema version here that is 2.5 and put in a variable for further use in the shell script. The size of the data file is around 5 GB. can you please help how i can just grep and get the 2.5


Thanks much.
Last edited by Don Cragun; 11-04-2013 at 02:10 PM.. Reason: Change HTML tags to CODE tags.
# 2  
Old 11-04-2013
Use awk instead:
Code:
awk '/SCHEMA_VERSION/{gsub(/.*SCHEMA_VERSION="|".*/,X);print}' file.xml

# 3  
Old 11-04-2013
Code:
sed 's#.*SCHEMA_VERSION=\([^ ][^ ]*\).*#\1#' myFile

# 4  
Old 11-04-2013
If your 5MB xml file contains more than one line, I think vgersh99's script will produce more output than you want. If you are using a system where awk and sed have limited line lengths, both Yoda's script and vgersh99's script could fail if your xml file contains long lines.

The following awk script should get around these problems:
Code:
version=$(awk -F '=*"' '$1 == "SCHEMA_VERSION" { print $2; exit 0 }' RS=' ' file.xml)
printf "schema version is: %s\n" "$version"

If you want to run this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk.

With your sample input, the above script produces the output:
Code:
schema version is: 2.5

# 5  
Old 11-04-2013
You may try Grep

Code:
$ grep -Po '(?<=SCHEMA_VERSION=").*(?="[[:space:]])'  file.xml
2.5

Code:
my_variable=$(grep -Po '(?<=SCHEMA_VERSION=").*(?="[[:space:]])' file.xml)

# 6  
Old 11-05-2013
Quote:
Originally Posted by Don Cragun
If your 5MB xml file contains more than one line, I think vgersh99's script will produce more output than you want. If you are using a system where awk and sed have limited line lengths, both Yoda's script and vgersh99's script could fail if your xml file contains long lines.

The following awk script should get around these problems:
Code:
version=$(awk -F '=*"' '$1 == "SCHEMA_VERSION" { print $2; exit 0 }' RS=' ' file.xml)
printf "schema version is: %s\n" "$version"

If you want to run this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk.

With your sample input, the above script produces the output:
Code:
schema version is: 2.5

Hello Don,
Thanks for your input is there anyway to increase the performance, i am basically searching for three elements SCHEMA_VERSION, PROVIDER, UNINUM. Out of these schema version & provider are coming up fast using your line of code as they will exist only in initial lines at begining of XML file. However UNINUM exists at multiple places in 4.5 GB XML file and it is taking lot of time to retreive all of them, i modifed your code to below. Can you please help me further for faster retrieval of UNINUM's if you have any ideas. thank you.

awk -F '=*"' '$1 == "UNINUM" { print $2;}' RS=' ' file.xml

I did tried Yoda, Akshay, Vgersh99 inputs but those are not effective either.
# 7  
Old 11-05-2013
Hi,

May be this helps. Assuming like you have same pattern in all the lines in file.

Code:
$ a=`awk -F"\"" '/SCHEMA_VERSION/{print$8}' filename`
$ echo $a
2.5
$


Thanks,
R. Singh
 
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