Question

我正在尝试让jQuery从变量处理JSON字符串而不进行AJAX调用，因为这个JSON字符串将从JSON2HTML生成，但jQuery.getJSON需要进行AJAX调用，这是浪费资源。有没有人有这个问题的解决方案或解决方法？

从JSON2HTML生成的JSON字符串。

[{"tag":"div","id":"0","class":"sgph","children":[
      {"tag":"div","data-scoretype":"","class":"sgi","id":"1","children":[
          {"tag":"button","data-btnrole":"satis_group","class":"up","html":"up"},
          {"tag":"button","data-btnrole":"satis_group","class":"down","html":"down"},
          {"tag":"button","data-btnrole":"satis_group","class":"edit","html":"change"},
          {"tag":"button","data-btnrole":"satis_group","class":"del","html":"delete"}
        ]},
      {"tag":"div","class":"scph","children":[
          {"tag":"button","data-btnrole":"score_type","class":"add","html":"score type"}
        ]},
      {"tag":"div","class":"stph","children":[
          {"tag":"div","class":"sti","id":"1","children":[
              {"tag":"button","data-btnrole":"satis_topics","class":"up","html":"up"},
              {"tag":"button","data-btnrole":"satis_topics","class":"down","html":"down"},
              {"tag":"button","data-btnrole":"satis_topics","class":"edit","html":"change"},
              {"tag":"button","data-btnrole":"satis_topics","class":"del","html":"delete"}
            ]},
          {"tag":"div","class":"sti","id":"3","children":[
              {"tag":"button","data-btnrole":"satis_topics","class":"up","html":"up"},
              {"tag":"button","data-btnrole":"satis_topics","class":"down","html":"down"},
              {"tag":"button","data-btnrole":"satis_topics","class":"edit","html":"change"},
              {"tag":"button","data-btnrole":"satis_topics","class":"del","html":"delete"}
            ]},
          {"tag":"button","data-btnrole":"satis_topics","class":"add","html":"add topic"}
        ]}
    ]},
      {"tag":"div","id":"1","class":"sgph","children":[
      {"tag":"button","data-btnrole":"satis_group","class":"add","html":"add group"}
    ]}
]

我做的jQuery代码。

//json_str will store JSON string which had been generate from JSON2HTML.
var json_str=$.parseJSON(FormatJSON(toTransform($('div.satis_form').children()))); 
$.getJSON(json_str, function(data){
    $.each(data,function(i,val){
        if(val.tag == 'button'){
            delete val[i];
        }
    })
})

Answer 1

代码应该是这样的，因为json_str是JSON对象。

var json_str=$.parseJSON(FormatJSON(toTransform($('div.satis_form').children()))); 
    $.each(data,function(i,val){
        if(val.tag == 'button'){
            delete val[i];
    }
})

jQuery从变量处理JSON字符串而不进行AJAX调用

1 个答案: