在json中循環 - jquery

Question

{ 
"info": { 
    "limit": "0", 
    "xdata": { 
     "hospital": { 
      "name": "qwewe", 
      "street": "cxcxc" 
     }, 
     "factory": { 
      "name": "wrwr", 
      "street": "xzcc" 
     }, 
     "industry": { 
      "name": "lll", 
      "street": "sdfdsfdf" 
     } 
    } 
    } 

如何打印xdata的內容？即我想要醫院，工廠的數據&工業 我正在使用$.each jQuery循環，但無法得到醫院的細節。

Answer 1

嘗試下一

xdata = your_data.info.xdata; 

for (xd in xdata) { 
    console.log('Name: ' + xdata[xd].name); 
    console.log('Street: ' + xdata[xd].street); 
} 

Answer 2

var data = { 
"info": { 
    "limit": "0", 
    "xdata": { 
     "hospital": { 
      "name": "qwewe", 
      "street": "cxcxc" 
     }, 
     "factory": { 
      "name": "wrwr", 
      "street": "xzcc" 
     }, 
     "industry": { 
      "name": "lll", 
      "street": "sdfdsfdf" 
     } 
    } 
}}; 

$.each(data.info.xdata, function(key, value) { 
    var type = key; // e.g. hospital/factory/industry 
    var name = value.name; 
    var street = value.street; 

    // do something with the values 
    console.log(type, name, street); 
}); 

喜歡這個？說實話，JSON並沒有真正格式化。 XDATA最好包含一組對象，但是meh。

欲瞭解更多信息，請參閱此jsfiddle：http://jsfiddle.net/fZBYG/1/請務必打開您的控制檯。

Answer 3

希望這會幫助你 - http://jsfiddle.net/ZE47n/7/

Answer 4

$.each(info.xdata, function(key, value) { 
    // key is equal to hospital, factory, industry 
    // valus is equal to { "name": "qwewe", "street": "cxcxc" }, ... 
    // this in the scope is the same as arguments[1] - value; 
    // this === value 
}); 

Answer 5

var json = $.parseJSON(j); 
//console.log(json.info.xdata); 
$.each(json.info.xdata,function(k,v){ 
console.log(v.name+" -- "+ v.street); 
}); 

DEMO