Bash 退出不退出
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Bash exit doesn't exit
提问by Tulains Córdova
I wonder why this script continues to run even with an explicit exit command.
我想知道为什么即使使用显式退出命令，该脚本仍会继续运行。
I have two files:
我有两个文件：
file1.txtwith the following content:
file1.txt具有以下内容：
aaaaaa bbbbbb cccccc dddddd eeeeee ffffff gggggg
file2.txtwith the following content:
file2.txt具有以下内容：
111111 aaaaaa 222222 333333 ffffff 444444
The script (test.sh) is this, two nested loops checking if any line of the first file contains any line of the second file. If it finds a match, it aborts.
脚本 ( test.sh) 是这样的，两个嵌套循环检查第一个文件的任何行是否包含第二个文件的任何行。如果找到匹配项，则中止。
#!/bin/bash
path=`dirname aaaaaa
!!!!!!!!!!
bbbbbb
cccccc
dddddd
eeeeee
ffffff
!!!!!!!!!!
gggggg
`
cat $path/file1.txt | while read line
do
echo $line
cat $RUTA/file2.txt | while read another
do
if [ ! -z "`echo $line | grep -i $another`" ]; then
echo "!!!!!!!!!!"
exit 0
fi
done
done
I get the following output even when it should exit after printing the first !!!!!!!!!!:
即使在打印第一个后应该退出，我也会得到以下输出!!!!!!!!!!：
#!/bin/bash
while read -r line
do
echo "$line"
while read -r another
do
if grep -i "$another" <<< "$line" ;then
echo "!!!!!!!!!!"
exit 0
fi
done < file2.txt
done < file1.txt
Isn't exitsupposed to end the execution of the script altogether?
不exit应该完全结束脚本的执行吗？
回答by user000001
The reason is that the pipes create sub processes. Use input redirection instead and it should work
原因是管道创建了子进程。改用输入重定向，它应该可以工作
while read -r line
do
echo "$line"
while read -r another
do
if grep -i "$another" <<< "$line" ;then
echo "!!!!!!!!!!"
exit 0
fi
done < <(command2)
done < <(command1)
In the general case, where the input comes from another program and not from a file, you can use process substitution
在一般情况下，输入来自另一个程序而不是来自文件，您可以使用进程替换
...
echo "!!!!!!!!!!"
exit 1
fi
done
[ $? == 1 ] && exit 0;
done
回答by svante
The while loops are running in their respective shells. Exiting one shell does not exit the containing ones. $? could be your friend here:
while 循环在各自的 shell 中运行。退出一个外壳不会退出包含的外壳。$? 可能是你的朋友：
##代码##