bash 在 shell 脚本变量中查找子字符串
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Find substring in shell script variable
提问by batty
I have a string
我有一个字符串
$VAR="I-UAT";
in my shell script code. I need a conditional statement to check if "UAT"is present in that string.
在我的 shell 脚本代码中。我需要一个条件语句来检查"UAT"该字符串中是否存在。
What command should I use to get either true or false boolean as output? Or is there any other way of checking it?
我应该使用什么命令来获取 true 或 false 布尔值作为输出？或者有没有其他的检查方法？
采纳答案by Andrew Clark
What shell? Using bash:
什么壳？使用 bash：
if [[ "$VAR" =~ "UAT" ]]; then
echo "matched"
else
echo "didn't match"
fi
回答by Diego Sevilla
You can do it this way:
你可以这样做：
case "$VAR" in
*UAT*)
# code when var has UAT
;;
esac
回答by Jonathan Leffler
The classic way, if you know ahead of time what string you're looking for, is a casestatement:
如果您提前知道要查找的字符串，那么经典的方法是case声明：
case "$VAR" in
*UAT*) : OK;;
*) : Oops;;
esac
You can use an appropriate command in place of the :command. This will work with Bourne and Korn shells too, not just with Bash.
您可以使用适当的命令代替:命令。这也适用于 Bourne 和 Korn shell，而不仅仅是 Bash。
回答by jman
found=`echo $VAR | grep -c UAT`
Then test for $found non-zero.
然后测试 $found 非零。
回答by Neil
In bashscript you could use
在bash脚本中你可以使用
if [ "$VAR" != "${VAR/UAT/}" ]; then
# UAT present in $VAR
fi
回答by marcelog
try with grep:
尝试使用 grep：
$ echo I\-UAT | grep UAT
$ echo $?
0
$ echo I\-UAT | grep UAX
$ echo $?
1
so testing
所以测试
if [ $? -ne 0 ]; then
# not found
else
# found
fi
