Question

我有一个json对象构造如下：

<?
$t['w'][$id]['marchin']=$machin;
$t['w'][$id]['c'][$id_com]['machin']=$machin;    
echo json_encode($t);
?>

我像这样浏览对象

// here i access to all the $t['w'][$id]

$.each(data.w, function(k,v){

              var that=this;
              doSomething();

              // now i want to access to all the $t['w'][$id]['c']
               $.each(that.c, function(k1,v1){
                     doSomething();       
               });

});

但是在第二个每个jquery发生错误.. 如何访问所有$ t ['w'] [$ id] ['c']？！

谢谢

好的，我试过了：

              $.each(data.w, function(k,v){
                  var that = $.parseJSON(this);
                        doSomething();

                    $.each(that[k]['c'], function(k1,v1){
                        doSomething();

                 });

       });

但它不再起作用，

这是我的json的一个例子，

{"w":
   {"3":
      {"test":"test","c":
        {"15":
           {"test2":"test2"}
        }
      }
    }
}

Answer 1

数据......

var data = {"w":
   {"3": {
       "test":"test",
       "c": {
          "15": {"test2":"test2"}
       }
      }
    }
};

循环......

$.each(data.w, function(key, value){
    // you are now lopping over 2nd dimension
    // On each loop, 'key' will be equal to another [$id] value
    // since you know you'd like to crawl 'c' sub entry, you can reference it
    // and loop over it
    $('body').append('<h3>'+key+'</h3>');
    if( this['c'] )
    $.each(this['c'], function(k,v){
        // Here you have access to ['c'][...]
        $('body').append('<span>'+k+'</span>');
    });
});

Answer 2

你可以在没有.each的情况下做到这一点：

var hsh = $.parseJSON(data.w);

for (var i in hsh) {
  var that = hsh[i];
  doSomething();

  for (var j in hsh[i].c) {
    doSomething();
  }
}

使用jquery浏览JSON对象

2 个答案: