BASH-通过subshel​​l和grep进行变量分配

Question

我要从此文件中提取匹配的正则表达式：

abc
de

{my_pattern.global} # want to extract my_pattern.global
                    # without curly brackets
123

并将其分配给shell脚本中的变量：

#!/bin/bash
l_config_file="my_file.cfg"
l_extracted_pattern=""
l_match_pattern="(?<={).+\.global(?=})"
l_my_dir=$(pwd)


echo "grep -oP '$l_match_pattern' $l_my_dir/$l_config_file"
echo "debug 1 - exit code: $?"

grep -oP '$l_match_pattern' $l_my_dir/$l_config_file
echo "debug 2 - exit code: $?"

sh -c "grep -oP '$l_match_pattern' $l_my_dir/$l_config_file"
echo "debug 3 - exit code: $?"

$l_extracted_pattern = "$(sh -c "grep -oP '$l_match_pattern' $l_my_dir/$l_config_file")"
echo "debug 4 - exit code: $?"

echo $l_extracted_pattern

输出：

grep -oP '(?<={).+\.global(?=})' /tmp/my_file.cfg
debug 1 - exit code: 0
debug 2 - exit code: 1
my_pattern.global
debug 3 - exit code: 0
./sto.sh: line 14: =: command not found.
debug 4 - exit code: 127

如您所见，grep命令运行良好（当通过sh -c执行时），但是在尝试将输出分配给具有退出代码127的变量$ l_extracted_pa​​ttern时失败。这意味着shell无法识别该命令。我怀疑正则表达式是造成麻烦的原因，但无法弄清楚具体是什么。怎么了？

Answer 1

即使我之前已经分配了它：

l_extracted_pattern=""

，然后尝试覆盖它：

$l_extracted_pattern = "$(sh -c "grep -oP '$l_match_pattern' $l_my_dir/$l_config_file")"

那是一个错误。显然，bash中的 no 变量赋值可能在变量名之前包含$-甚至早于实例化时也没有。更改为：

l_extracted_pattern = "$(sh -c "grep -oP '$l_match_pattern' $l_my_dir/$l_config_file")"