cut the string using string as delimiter in Shell

Question

Need to print libcom.so by stripping out 1.0.0 when it prints first line.

#!/bin/sh
data=$(sed '0,/^__VERSION__$/d' "$0")
for d in $data
do
     echo with version:
     echo $d
     echo "with no version" 
     d1="$(cut -d\.so -f1 <<< $d)"
     #Need to save it in a variable for further processing
     echo $d1
done

exit

__VERSION__
libcom.so.1.0.0
libclock.so.1.0.0
libstdc++.so.6.0.20
libxml.so.2.7.8
libmi.so.1.0.0
librt.so.6.6.5
libcsapi.so.1.0.0

error out as cut: the delimiter must be a single character, whats the best way to do this?

Answer 1

If you just want the string up until the first .so, bash will happily handle it for you:

d1="${d%%.so*}.so"

The ${d%%.so*} deletes the longest string from the right side %% of the variable d that matches .so*. This effectively removes everything starting with .so until the end of the string. We then manually re-add the .so we removed because %% removes the entire matched portion, not just the wildcard portion.

See this guide on parameter expansion for more detail.

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Alternately, you could use sed even though it would be considerably slower:

d1=$(sed 's/\(\.so\).*$/\1/' <<< $d)