我正在尝试使用
将表单输入转换为Json对象(function($){
$.fn.serializeObject = function(){
var self = this,
json = {},
push_counters = {},
patterns = {
"validate": /^[a-zA-Z][a-zA-Z0-9_]*(?:\[(?:\d*|[a-zA-Z0-9_]+)\])*$/,
"key": /[a-zA-Z0-9_]+|(?=\[\])/g,
"push": /^$/,
"fixed": /^\d+$/,
"named": /^[a-zA-Z0-9_]+$/
};
this.build = function(base, key, value){
base[key] = value;
return base;
};
this.push_counter = function(key){
if(push_counters[key] === undefined){
push_counters[key] = 0;
}
return push_counters[key]++;
};
$.each($(this).serializeArray(), function(){
// skip invalid keys
if(!patterns.validate.test(this.name)){
return;
}
var k,
keys = this.name.match(patterns.key),
merge = this.value,
reverse_key = this.name;
while((k = keys.pop()) !== undefined){
// adjust reverse_key
reverse_key = reverse_key.replace(new RegExp("\\[" + k + "\\]$"), '');
// push
if(k.match(patterns.push)){
merge = self.build([], self.push_counter(reverse_key), merge);
}
// fixed
else if(k.match(patterns.fixed)){
merge = self.build([], k, merge);
}
// named
else if(k.match(patterns.named)){
merge = self.build({}, k, merge);
}
}
json = $.extend(true, json, merge);
});
return json;
};
})(jQuery);
我可以将表单输入转换为JSON,如
$('#my-form').serializeObject();
但它也返回null对象。如果我没有在文本框中填充值,它会给我输出 “keyWord”:“”但我希望它不应该包含在Json中(如果文本框值没有被用户填充,那么它不应该包含在json中的键和值)。我在哪里缺少?