sed extract part of string from a file

user_9733887 0
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I've ben trying to extract only part of string from a file looking like this:

str1=USER_NAME
str2=justAstring
str3=https://product.org/v-4.5-bin.zip
str4=USER_HOME

I need to extract ONLY the version - in this case: 4.5

I did it by grep and then sed but now the output is 4.5-bin.zip

-> grep str3 file.txt
str3=https://product.org/v-4.5-bin.zip
-> echo str3=https://product.org/v-4.5-bin.zip | sed -n "s/^.*v-\(\S*\)/\1/p"
4.5-bin.zip

What should I do in order to remove also the -bin.zip at the end? Thanks.

sed grep bash
2 Answers

1st solution: With your shown samples, please try following sed code.

sed -n '/^str3=/s/.*-\([^-]*\)-.*/\1/p' Input_file

Explanation: Using sed's -n option which will STOP printing of values by default, to only print matched part. In main program checking condition if line starts from str3= then perform substitution there. In substitution catching everything between 1st - and next - in a capturing group and substituting whole line with it by using \1 and printing the matched portion only by using p option.

2nd solution: Using GNU grep you could try following grep program.

grep -oP '^str3=.*?-\K([^-]*)' Input_file

3rd solution: Using awk program for getting expected output as per shown smaples.

awk -F'-' '/^str3=/{print $2}'  Input_file

4th solution: Using awk's match function to get expected results with help of using RSTART and RLENGTH variables which get set once a TRUE match is found by match function.

awk 'match($0,/^str3=.*-/){split(substr($0,RSTART,RLENGTH),arr,"-");print arr[2]}'  Input_file
user_58665800

If you know the version contains just digits and dots, replace \S by [0-9.]. Also, match the remaining characters outside of the capture group to get it removed.

sed -n 's/^.*v-\([0-9.]*\).*/\1/p'
user_10306750
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