I have a bash script where I get the disk usage, for example 60%
. How can I remove the %
symbol? Using grep or awk?
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5 Answers
12
There is no need for external tools like tr
or even sed
as bash can do it on its own since forever.
percentage="60%"
number=${percentage%\%}
This statement removes the shortest matching substring (in this case an escaped %
) from the end of the variable. There are other string manipulating facilities built into bash. It even supports regular expressions. Generally, most of the stuff you normally see people using tr
, sed
, awk
or grep
for can be done using a bash builtin. It's just almost noody knows about that and brings the big guns...
See http://tldp.org/LDP/abs/html/parameter-substitution.html#PSOREX1 for more information.
Holger Just
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6
This should do it:
sed 's/%//'
Pipe your string through it for the best results.
womble
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2There's a "g" missing because otherwise it would only match the first % in each line. So make it 's/%//g'. – joechip Aug 23 '11 at 17:14
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1@joechip: If there's only one `%`, as per the question, you don't need `g`. – womble Aug 23 '11 at 22:10
2
If the disk usage is in a variable, bash can do the removal as part of a variable substitution:
diskusagepct="60%"
echo "disk usage: ${diskusagepct%\%} percent" # prints disk usage: 60 percent
diskusagenum="${diskusagepct%\%}" # sets diskusagenum to 60
Gordon Davisson
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