Bash：提取字符串的一部分
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Bash : extracting part of a string
提问by confusified
Say I have the string "Memory Used: 19.54M" How would I extract the 19.54 from it? The 19.54 will change frequently so i need to store it in a variable and compare it with the value on the next iteration.
假设我有字符串 "Memory Used: 19.54M" 我如何从中提取 19.54？19.54 会经常变化，所以我需要将它存储在一个变量中，并将它与下一次迭代的值进行比较。
I imagine I need some combination of grep and regex, but I never really understood regex..
我想我需要 grep 和 regex 的某种组合，但我从来没有真正理解 regex ..
回答by choroba
You probably want to extractit rather than removeit. You can use the Parameter Expansion to extract the value:
您可能想提取它而不是删除它。您可以使用参数扩展来提取值：
var="Memory Used: 19.54M"
var=${var#*: } # Remove everything up to a colon and space
var=${var%M} # Remove the M at the end
Note that bash can only compare integers, it has no floating point arithmetics support.
请注意，bash 只能比较整数，它没有浮点算术支持。
回答by Onilton Maciel
Other possible solutions:
其他可能的解决方案：
With grep:
与grep：
var="Memory Used: 19.54M"
var=`echo "$var" | grep -o "[0-9.]\+"`
With sed:
与sed：
var="Memory Used: 19.54M"
var=`echo "$var" | sed 's/.*\ \([0-9\.]\+\).*//g'`
With cut:
与cut：
var="Memory Used: 19.54M"
var=`echo "$var" | cut -d ' ' -f 3 | cut -d 'M' -f 1`
With awk:
与awk：
var="Memory Used: 19.54M"
var=`echo "$var" | awk -F'[M ]' '{print }'`
回答by Alastair
You can use bash regex support with the =~operator, as follows:
您可以将 bash regex 支持与=~运算符一起使用，如下所示：
var="Memory Used: 19.54M"
if [[ $var =~ Memory\ Used:\ (.+)M ]]; then
echo ${BASH_REMATCH[1]}
fi
This will print 19.54
这将打印 19.54
回答by Vijay
> echo "Memory Used: 19.54M" | perl -pe 's/\d+\.\d+//g'
Memory Used: M
