[英]Bash quotes disable escaping
I want to run some command, let's name it "test" from my bash script and put there some of params from bash variable. 我想运行一些命令,让我们从我的bash脚本命名为“test”,然后从bash变量中添加一些params。
My script: 我的剧本:
#!/bin/bash -x
PARAMS="-A 'Foo' -B 'Bar'"
./test $PARAMS
I've got: 我有:
+ PARAMS='-A '\''Foo'\'' -B '\''Bar'\'''
+ ./test -A ''\''Foo'\''' -B ''\''Bar'\'''
It's wrong! 这是不对的!
Another one case: 另一个案例:
#!/bin/bash -x
PARAMS='-A '"'"'Foo'"'"' -B '"'"'Bar'"'"
./test $PARAMS
Result is sad too: 结果是伤心过:
+ PARAMS='-A '\''Foo'\'' -B '\''Bar'\'''
+ ./test -A ''\''Foo'\''' -B ''\''Bar'\'''
So, question is – how can I use bash variable as command line arguments for some command. 所以,问题是 - 如何将bash变量用作某些命令的命令行参数。 Variable is something like "-A 'Foo' -B 'Bar'" (exactly with single-quotes) And result must be calling of program "./test" with arguments "-A 'Foo' -B 'Bar'" like this:
变量类似于“-A'Foo'-B'Bar'”(完全用单引号)并且结果必须调用带有参数“-A'Foo'-B'Bar'”的程序“./test”这个:
./test -A 'Foo' -B 'Bar'
Thanks! 谢谢!
It is safer to use BASH arrays for storing full or partial command lines like this: 使用BASH数组存储完整或部分命令行更安全,如下所示:
params=(-A 'Foo' -B 'Bar')
then call it as: 然后将其称为:
./test "${params[@]}"
which will be same as: 这将是:
./test -A 'Foo' -B 'Bar'
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