我在哪裏可以找到jquery.param的實現代碼？

Question

我想知道我在哪裏可以找到jquery.param的實現代碼。

Answer 1

/src/ajax.js

Answer 2

這裏是source directly from jQuery（MIT license）：

jQuery.param = function(a, traditional) { 
    var prefix, 
     s = [], 
     add = function(key, value) { 
      // If value is a function, invoke it and return its value 
      value = jQuery.isFunction(value) ? value() : (value == null ? "" : value); 
      s[ s.length ] = encodeURIComponent(key) + "=" + encodeURIComponent(value); 
     }; 

    // Set traditional to true for jQuery <= 1.3.2 behavior. 
    if (traditional === undefined) { 
     traditional = jQuery.ajaxSettings && jQuery.ajaxSettings.traditional; 
    } 

    // If an array was passed in, assume that it is an array of form elements. 
    if (jQuery.isArray(a) || (a.jquery && !jQuery.isPlainObject(a))) { 
     // Serialize the form elements 
     jQuery.each(a, function() { 
      add(this.name, this.value); 
     }); 

    } else { 
     // If traditional, encode the "old" way (the way 1.3.2 or older 
     // did it), otherwise encode params recursively. 
     for (prefix in a) { 
      buildParams(prefix, a[ prefix ], traditional, add); 
     } 
    } 

    // Return the resulting serialization 
    return s.join("&").replace(r20, "+"); 
}; 

Answer 3

萬一別人絆倒過這個，它在SRC/serialize.js現在https://github.com/jquery/jquery/blob/master/src/serialize.js