Question

我从2天开始就遇到了这个问题，我需要使用javascript从以下网址获取数据。

http://www.apexweb.co.in/Apex_Quote/Download_Data.asp?un=9393910169

我解析了静态数据，但我需要解析网址中的数据。帮帮我

var employees = [{"ContactName":"AksharmaDelhi","ContactNumber":"9313146690","MainCategory":"NONLOCAL","SubCategory":"LIG"},
{"ContactName":"AbhijeetPune","ContactNumber":"9271263359","MainCategory":"NONLOCAL","SubCategory":"LIG"},
{"ContactName":"Abhinandan","ContactNumber":"9954534007","MainCategory":"NONLOCAL","SubCategory":"LIG"},
{"ContactName":"Abhishek","ContactNumber":"9302172932","MainCategory":"NONLOCAL","SubCategory":"LIG"},
{"ContactName":"AbhishekPuri","ContactNumber":"9303928233","MainCategory":"NONLOCAL","SubCategory":"LIG"},
{"ContactName":"AbhishekZhakkas","ContactNumber":"9863027167","MainCategory":"NONLOCAL","SubCategory":"MIG"},
{"ContactName":"Abin","ContactNumber":"8891886340","MainCategory":"NONLOCAL","SubCategory":"MIG"},
{"ContactName":"AccountantSrinivas","ContactNumber":"9949861074","MainCategory":"LOCAL","SubCategory":"MIG"},];

alert(employees.length);   //get length 

for(var i = 0;i<employees.length;i++){

alert(employees[i].ContactName + employees[i].ContactNumber + employees[i].MainCategory + employees[i].SubCategory);

}
</script>

Answer 1

有很多方法可以做到这一点，我使用$.ajax方式：

$.ajax({
            url : "http://www.apexweb.co.in/Apex_Quote/Download_Data.asp?un=9393910169",
            type : "get",
            dataType : 'text',
            success : function(data) {                                  
                var employees = $.parseJSON(data);
                for(var i = 0;i<employees.length;i++){
                alert(employees[i].ContactName + employees[i].ContactNumber + employees[i].MainCategory + employees[i].SubCategory);
                }                       
            },
            error : function() {
                //console.log("your call failed");
                }                                   
        });

从您的网址中提取的数据将作为data回调中的success参数传递。反对$.getJson，这使我能够更好地控制我的回归。

在这种方法中，我将dataType设置为text。意思是jquery不会尝试用我的结果做任何“聪明”的事情。如果我将其设置为json，则data参数将被解析并准备使用。您不需要手动解析它：

var employees = $.parseJSON(data);

Answer 2

对于跨域ajax，我建议使用$ .getJSON（）

$.getJSON("http://my-site-return-json.com/index.php?", function(data){
  // Load Finish
});

在服务器端

记得返回回调

$callback = $_GET['callback'];
$arr = array("a" => "1", "b" => "2");
echo $callback."(".json_encode($arr).")";

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2 个答案: