Question

我是jquery和JSON的新手。我有以下JSON结构。

{
   "address":{
      "suburb":[
         "HACKHAM WEST",
         "HUNTFIELD HEIGHTS",
         "ONKAPARINGA HILLS",
         "m"
      ],
      "state":"SA"
   }
}

所以基本上就是上面的反应：

$.ajax({
    type:'POST',
    url: 'getAddress.php',
    data:postCode='+postCode',
    success: function(response) {
        alert(response)
    }
});

所以，我想要的是一个包含状态的变量，以及一个包含郊区的数组。

Answer 1

检查您是否有有效的Ajax请求：

$.ajax({
    type: "POST",
    url: "getAddress.php",
    data: {
        postCode : postCode          // data as object is preferrable
    },
    dataType: "json",                // to parse response as JSON automatically
    success: function(response) {
        var state = response.address.state;
        var suburbs = response.address.suburb;
    }
});

Answer 2

这应该可以解决问题

$.ajax({type:'POST', 
    url: 'getAddress.php', 
    dataType: 'json',
    data:'postCode='+postCode, 
    success: function(response) {
        var state = response.address.state;
        var suburbs = response.address.suburb;   
    }
});

添加了dataType:'json'并修复了data参数。

Answer 3

您需要解析所获得的JSON。 $.ajax可以为您做到这一点。

$.ajax({
    type:'POST',
    url: 'getAddress.php',
    data: 'postCode='+postCode, // make sure this line is correct
    dataType: 'json', // this tells jQuery to parse it for you
    success: function(response) {
        var state = response.address.state;
        var suburbs = response.address.suburb;
    }
});

尝试使用jquery迭代JSON结构

3 个答案: