用xargs调用shell函数 [英] Calling shell functions with xargs

问题描述

我正在尝试使用xargs并行调用一个更复杂的函数.

I am trying to use xargs to call a more complex function in parallel.

#!/bin/bash
echo_var(){
    echo $1
    return 0
}
seq -f "n%04g" 1 100 |xargs -n 1 -P 10 -i echo_var {} 
exit 0

这将返回错误

xargs: echo_var: No such file or directory

任何有关如何使用xargs来完成此操作的想法，或者任何其他解决方案都将受到欢迎.

Any ideas on how I can use xargs to accomplish this, or any other solution(s) would be welcome.

推荐答案

导出函数应该做到这一点(未经测试):

Exporting the function should do it (untested):

export -f echo_var
seq -f "n%04g" 1 100 | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

您可以使用内置的printf代替外部的seq:

You can use the builtin printf instead of the external seq:

printf "n%04g\n" {1..100} | xargs -n 1 -P 10 -I {} bash -c 'echo_var "$@"' _ {}

此外，像这样使用return 0和exit 0可以掩盖其前面的命令可能产生的任何错误值.另外，如果没有错误，则为默认设置，因此有些多余.

Also, using return 0 and exit 0 like that masks any error value that might be produced by the command preceding it. Also, if there's no error, it's the default and thus somewhat redundant.

@phobic提到可以将Bash命令 简化为

@phobic mentions that the Bash command could be simplified to

bash -c 'echo_var "{}"'

直接在其中移动{}. 但是，如@Sasha所指出的，命令注入很容易.

moving the {} directly inside it. But it's vulnerable to command injection as pointed out by @Sasha.

以下是您不应使用嵌入式格式的示例:

Here is an example why you should not use the embedded format:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "{}"'
Sun Aug 18 11:56:45 CDT 2019

为什么不的另一个示例:

echo '\"; date\"' | xargs -I {} bash -c 'echo_var "{}"'

这是使用安全格式输出的结果:

$ echo '$(date)' | xargs -I {} bash -c 'echo_var "$@"' _ {}
$(date)

这相当于使用参数化 SQL 查询，以避免

This is comparable to using parameterized SQL queries to avoid injection.

我在命令替换中或在转义引号中使用date，而不是Sasha注释中使用的rm命令，因为它无损.

I'm using date in a command substitution or in escaped quotes here instead of the rm command used in Sasha's comment since it's non-destructive.

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