Question

我的源代码文件中包含以下类型的文本：

html += T.m('current value of the quality level of the security service (as established from the diagnostic phase).');
or
{header: T.m("Service to Improve"), dataIndex : 'searvicesToImprove', hideable: false},
or
Ext.getCmp('MAX_ACTION_PLANS_PAGE').setText(T.m('od') + ' ' + MAX_ACTION_PLANS_PAGE);

我想提取括号内的子串，即。从T.m（X）我想得到没有引号括号或带有它们的X，然后我会修剪它们。

所以换句话说我想要这样的东西

regex( "T.m('X')" | "T.m("X")" );
and then say:
listOfMatches.add(X);

我知道这通常是用regexp完成的，但是我对regexp并不是那么好，没有使用它，仅用于基本样本。任何帮助都非常有用。

Answer 1

try {
    Regex regexObj = new Regex(@"T\.m\([""|'](.+?)[""|']\)");
    Match matchResults = regexObj.Match(subjectString);
    while (matchResults.Success) {
        var neededvalue = matchResults.Groups[1].Value;
    } 
} catch (ArgumentException ex) {
    // Syntax error in the regular expression
}

Answer 2

var regex = new Regex(@"T\.m\((?<MyGroup>.*)\)");
var match =regex.Match(subject);
if(match.Success)
{
    var found =  match.Groups["MyGroup"].Value;
}

Answer 3

如果你使用正则表达式并希望所有部分分开，这可能会起作用

 #  @"(?s)T\.m\(((['""])((?:(?!\2).)*)\2)\)"

 (?s)                          # dot all modifier
 T \. m                        # 'T.m'
 \(                            # Open parenth
 (                             # (1 start), Quoted part
      ( ['"] )                      # (2), Open Quote
      (                             # (3 start), Body of quoted part
           (?:                           # grouping
                (?! \2 )                      # lookahead, not a quote
                .                             # any char
           )*                            # end grouping, do 0 or more times
      )                             # (3 end)
      \2                            # Close quote (group (2) backreference)
 )                             # (1 end), Quoted part
 \)                            # Close parenth

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用正则表达式提取子字符串

3 个答案: