php - 无法正确获取 JSON 数据

Question

我通过这样做得到了 json 的输出

$.get('getDataPartySize.php', function(data) {
      alert(data);
});

输出是:

[object Object],[object Object],[object Object],[object Object],[object Object],
[object Object],[object Object],[object Object],[object Object],[object Object],
[object Object],[object Object],[object Object],[object Object],[object Object],
[object Object],[object Object],[object Object],[object Object],[object Object],
[object Object],[object Object],[object Object],[object Object],[object Object],
[object Object],[object Object]

我尝试了 getDataPartySize.php 的输出，如下所示，并且 getDataPartySize.php 也有这个 header("Content-type: text/json");

[{"name":"waqar","size":"3"},{"name":"waqas","size":"8"},{"name":"jahangir","size":"6"},{"name":"wawar","size":"3"},
 {"name":"basit","size":"5"},{"name":"basit","size":"5"},{"name":"Ali","size":"4"},
 {"name":"basit","size":"4"},{"name":"Malik","size":"5"},{"name":"basit","size":"5"},
 {"name":"Kashif","size":"7"},{"name":"test","size":"5"},{"name":"Ch","size":"7"},
 {"name":"first","size":"10"},{"name":"2nd","size":"6"},{"name":"qasim","size":"10"},
 {"name":"saleem","size":"14"},{"name":"testing","size":"16"},
 {"name":"kashif","size":"10"},{"name":"Ch","size":"14"},{"name":"saleem","size":"6"},
 {"name":"new","size":"1"},{"name":"abc","size":"2"},{"name":"lahori","size":"3"},
 {"name":"newpak","size":"1"},{"name":"abc","size":"2"},{"name":"tabaq","size":"6"}
]

我正在使用 jQuery。我不知道问题出在哪里。

编辑:也尝试过这个，得到相同的输出。

$.getJSON("getDataPartySize.php",function(result){
  alert(result);
}

Answer 1

指定数据类型

$.get('getDataPartySize.php', function(data) {
      alert(data);
},'json');

或者手动解析json

$.get('getDataPartySize.php', function(data) {
var json = $.parseJSON(data);
      alert(json);
//i would prefer 
//console.log(json);
//because alert tries to convert the objects to strings as mentioned by @Rocket in comments
});

这是DEMO