Shell Programming and Scripting

Hi Gurus,
I'm a total newbie to Perl and Awk scripting. Let me explain the scenario, there is a DB2 table with 5 columns and one of the column is a CLOB datatype containing XML. We need all the 4 columns but only a portion of string from the XML column.
We decided to export DB2 table to a .del file and process it using Perl or Awk script. I need a script to process the .del file so that I have column1, column2, column3 and in column 4 which is XML, we just need the string which is in between <text> and </text> (there may be multiple occurrence of this so they can be seperated by number) plus the column 5.
I know it will be piece of cake for the experts.

Thanks,

try this

Code:

I=0
while read LINE
do
    TEXT[$I]=$(echo $LINE | grep -o '<text>.*</text>' | sed -e 's/<text>//' -e 's/<\/text>//'
    (( I ++ ))
done < $FILENAME

That creates an array where each item is the text contained between <text> and </text>.
Hope this helps

Thanks Frans for your prompt reply but I've a question which may sound stupid to you but still.
1) What is the extension of the file I should save? Is it .awk or .ksh?
2) Where do I replace the input and output file name in the code?

Thanks,

It's bash scripting so on the first line of the script you write #!/bin/bash If the path of your shell is /bin/bash, of course.
The full code :

Code:

#!/bin/bash
while read LINE
do
    echo $LINE | grep -o '<text>.*</text>' | sed -e 's/<text>//' -e 's/<\/text>//'
    (( I ++ ))
done < INPUTFILE > OUTPUTFILE

You could use variables for the input and output files if they have to be re-used later, else just replace 'INPUTFILE' and 'OUTPUTFILE' by your own file names.

P.S. no matter the extension ! make it executable with chmod +x and go.

Thank you very much Frans, I'll test and let you know.

---------- Post updated at 12:46 PM ---------- Previous update was at 12:05 PM ----------

Hi Frans,
There is no directory #!/bin/bash in the unix box, but there is directory for #/usr/bin/bsh. Are these 2 same? I tried to run the script but I get an error as below

grep: Not a recognized flag: o
Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwy] [-p[parasep]] -e pattern_list...
[-f pattern_file...] [file...]
Please clarify.

Thanks.

bsh is Bourne SHell, bash is Bourne Again SHell, so the options of the grep command doesn't seem to be the same.
the -o option tells grep to output only the matching part of the line, it's helpful.
Tell me what's the output when you use grep without option.
I believe that there's only one occurence of the <text>....</text> in each line.
To go faster in trying, don't redirect to the output file ( > OUTPUTFILE ) so you directly see what happens.
a possibility is to use the extract like

Code:

LINE="lbla bla bla jcjfd<text>what i want</text>flh%(j blablabla" # for testing
LINE=$(echo $LINE | grep <text>*</text>) # Returns every line containing the match.
echo $LINE
LINE=${LINE##*<text>} # Deletes the matching from the beginning
echo $LINE
LINE=${LINE%%</text>*} # Deletes the matching from the end
echo $LINE

and see what happens

No there are multiple occurence of this pattern in the record