问题
My file looks like this:
record=123
date=2012.01.20 10:22
In the bash file I do cat myfile.ini, and then I need to use something like explode, because I need ONLY to grep the 123 numeric record data, and nothing else.
How it can be done in the bash ?
回答1:
awk -F'=| ' '/record/ {print $2}'
Substitute "date" for "record" in the command above to get the date (the time would be in $3).
This keys on the names of the variables rather than depending on a regex match of the value. It uses spaces or equal signs as field separators.
回答2:
If you have control over the ini file, rewrite it as:
record=123
date="2012.01.20 10:22"
Then in your "bash file" you do
. myfile.ini
echo $record
This is the typical approach. (if you have control over the ini file)
回答3:
You can use:
#!/bin/sh
VALUE="`grep '^record=' myfile.ini |sed 's/[^0-9]//g'`"
回答4:
You can do something like this:
awk '/=[0-9]+$/' file.txt
to print only the line with numeric content after = sign on stdout.
However if you just want to capture 123 into a variable then you can use:
val=$(awk -F"=" '/=[0-9]+$/{print $2}' file.txt)
来源：https://stackoverflow.com/questions/8943132/bash-string-explode