Bash 在字符串上使用匹配正则表达式和后视模式 [英] Bash use match regex with lookbehind pattern on string

问题描述

bash --version

GNU bash，版本 3.2.57(1)-release (x86_64-apple-darwin19)

说明:

我想从这个字符串字典中提取 WorkingDir 键值:

I want to extract the WorkingDir key value from this string dictionary:

config="""
"TerraformCommand": "terragrunt-info",
  "WorkingDir": "usr/terraform-modules/terraform-aws-codebuild/examples/.terragrunt-cache/xh9X2WgTwVjjRHiPBjKUl0Lr86w/SGN8gG45haGoXT7IhOh9_iuKkbc"
}
"""

在这种情况下，预期的输出是:

In this case, the expected output would be:

"usr/terraform-modules/terraform-aws-codebuild/examples/.terragrunt-cache/xh9X2WgTwVjjRHiPBjKUl0Lr86w/SGN8gG45haGoXT7IhOh9_iuKkbc"

尝试:

到目前为止，我已经尝试过使用这种正向后视/前瞻模式的不同方法:'(?<=("WorkingDir":s")).+(?=")'1.

So far I've tried using different methods using this positive lookbehind/lookahead pattern: '(?<=("WorkingDir":s")).+(?=")' 1.

echo `expr "$config" : '(?<=("WorkingDir":s")).+(?=")'`

输出:0

2.

pat='(?<=("WorkingDir":s")).+(?=")'
[[ $config =~ $pat ]]
echo "${BASH_REMATCH[0]}"
echo "${BASH_REMATCH[1]}"

输出:

echo $config | grep -o '(?<=("WorkingDir":s")).+(?=")'

输出:

推荐答案

正如评论中所指出的，expr 只使用basic"；(又名过时")正则表达式，并且 bash 的 =~ 运算符使用的正则表达式引擎不支持前瞻或后视(或 s 空格的简写)).但是你不需要环顾四周，只需匹配一切并使用捕获组来挑选你想要的部分(并将模式存储在一个变量中以避免可能的解析不一致):

As was pointed out in the comments, expr only uses "basic" (aka "obsolete") regular expressions, and the regex engine that bash's =~ operator uses doesn't support lookahead or lookbehind (or the s shorthand for space either). But you don't need lookaround, just match everything and use a capture group to pick out the part you want (and store the pattern in a variable to avoid possible parsing inconsistencies):

WorkingDirPattern='"WorkingDir":[[:space:]]"([^"]+)"'
if [[ "$config" =~ $WorkingDirPattern ]]; then
    WorkingDir="${BASH_REMATCH[1]}"    # get the contents of the first capture group
    echo "WorkingDir is $WorkingDir"
else
    echo "No WorkingDir found" >&2
fi

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