如何替換數組？

Question

下面的代碼（從stackedit - 減價編輯器）將取代

\begin{thm} 
blabla... 
\end{thm} 

與

<div class="latex_thm"> 
blabla... 
</div> 

的代碼是：

userCustom.onEditorConfigure = function(editor) { 
    var converter = editor.getConverter(); 
    converter.hooks.chain("preConversion", function(text) { 
     return text.replace(/\\begin{thm}([\s\S]*?)\\end{thm}/g, function(wholeMatch, m1) { 
      return '<thm>' + m1 + '</thm>'; 
     }); 
    }); 
    converter.hooks.chain("preBlockGamut", function(text, blockGamutHookCallback) { 
     return text.replace(/<thm>([\s\S]*?)<\/thm>/g, function(wholeMatch, m1) { 
      return '<div class="latex_thm">' + blockGamutHookCallback(m1) + '</div>'; 
     }); 
    }); 
}; 

但是如果我想不僅可以取代thm與latex_thm，但lem與latex_lem，如何做到這一點？我想也許用array解決吧？但它同樣不適合我：

userCustom.onEditorConfigure = function(editor) { 
    var converter = editor.getConverter(); 
    converter.hooks.chain("preConversion", function(text) { 
     var array = { 
      "thm": "thm", 
      "lem": "lem" 
     }; 
     for (var val in array) { 
      return text.replace(/\\begin{array[val]}([\s\S]*?)\\end{array[val]}/g, function(wholeMatch, m1) { 
       return '<div class="latex_"' + array[val] + '>' + m1 + '</div>'; 
      }); 
     }; 
    }); 
}; 
}; 

你能幫我嗎？

Answer 1

像（未經測試）：

converter.hooks.chain("preConversion", function(text) { 
     return text.replace(/\\begin{(thm|lem)}([\s\S]*?)\\end{\1}/g, function(wholeMatch, m1, m2) { 
      return '<' + m1 + '>' + m2 + '</' + m1 + '>'; 
     }); 
    }); 
    converter.hooks.chain("preBlockGamut", function(text, blockGamutHookCallback) { 
     return text.replace(/<(thm|lem)>([\s\S]*?)<\/\1>/g, function(wholeMatch, m1, m2) { 
      return '<div class="latex_' + m1 + '">' + blockGamutHookCallback(m2) + '</div>'; 
     }); 
    });