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I am using bash script for testing purpose.During my testing I have to find the line number of first occurrence of a string in a file. I have tried "awk" and "grep" both, but non of them return the value.
Awk example
#/!bin/bash .... VAR=searchstring ... cpLines=$(awk '/$VAR/{print NR}' $MYDIR/Configuration.xml this does not expand $VAR. If I use the value of VAR it works, but I want to use VAR
Grep example
#/!bin/bash ... VAR=searchstring ... cpLines=grep -n -m 1 $VAR $MYDIR/Configuration.xml |cut -f1 -d: this gives error line 20: -n: command not found
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Show Line NumbersThe -n ( or --line-number ) option tells grep to show the line number of the lines containing a string that matches a pattern. When this option is used, grep prints the matches to standard output prefixed with the line number.
To Display Line Numbers with grep MatchesAppend the -n operator to any grep command to show the line numbers.
By default grep prints the lines matching a pattern, so if the pattern appears one or more times into a line, grep will print that whole line. Adding the flag -m 7 will tell grep to print only the first 7 lines where the pattern appears.
sed ( sed 1q or sed -n 1p ) and head ( head -n 1 ) are the more appropriate tools for printing the first line from a file.
grep -n -m 1 SEARCH_TERM FILE_PATH |sed 's/\([0-9]*\).*/\1/' grep switches
-n = include line number
-m 1 = match one
sed options (stream editor):
's/X/Y/' - replace X with Y
\([0-9]*\) - regular expression to match digits zero or multiple times occurred, escaped parentheses, the string matched with regex in parentheses will be the \1 argument in the Y (replacement string)
\([0-9]*\).* - .* will match any character occurring zero or multiple times.
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You need $() for variable substitution in grep
cpLines=$(grep -n -m 1 $VAR $MYDIR/Configuration.xml |cut -f1 -d: ) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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