Bash: Call shell script but do not wait for return code

bashjob-control

Is it possible in Bash to call a shell script from another shell script but not have the original script wait for the sub-script to complete?

Best Answer

Just fork it with a &. As in, sh /path/to/script/script.sh &

This will print messages from the subscript, but you can replace the & with >/dev/null & and suppress the output.

Related Solutions

Linux Bash – How to Sort du -h Output by Size

As of GNU coreutils 7.5 released in August 2009, sort allows a -h parameter, which allows numeric suffixes of the kind produced by du -h:

du -hs * | sort -h

If you are using a sort that does not support -h, you can install GNU Coreutils. E.g. on an older Mac OS X:

brew install coreutils
du -hs * | gsort -h

From sort manual:

-h, --human-numeric-sort compare human readable numbers (e.g., 2K 1G)

Bash – Clean way to write complex multi-line string to a variable

This will put your text into your variable without needing to escape the quotes. It will also handle unbalanced quotes (apostrophes, i.e. '). Putting quotes around the sentinel (EOF) prevents the text from undergoing parameter expansion. The -d'' causes it to read multiple lines (ignore newlines). read is a Bash built-in so it doesn't require calling an external command such as cat.

IFS='' read -r -d '' String <<"EOF"
<?xml version="1.0" encoding='UTF-8'?>
 <painting>
   <img src="madonna.jpg" alt='Foligno Madonna, by Raphael'/>
   <caption>This is Raphael's "Foligno" Madonna, painted in
   <date>1511</date>-<date>1512</date>.</caption>
 </painting>
EOF

Best Answer

Related Solutions

Linux Bash – How to Sort du -h Output by Size

Bash – Clean way to write complex multi-line string to a variable

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