如何在jquery中迭代json数据

Question

如何在jquery中迭代json数据.

[{"id":"856","name":"India"},
 {"id":"1035","name":"Chennai"},
 {"id":"1048","name":"Delhi"},
 {"id":"1113","name":"Lucknow"},
 {"id":"1114","name":"Bangalore"},
 {"id":"1115","name":"Ahmedabad"},
 {"id":"1116","name":"Cochin"},
 {"id":"1117","name":"London"},
 {"id":"1118","name":"New York"},
 {"id":"1119","name":"California"}
]

Answer 1

你可以$.each()像这样使用:

$.each(data, function(i, obj) {
  //use obj.id and obj.name here, for example:
  alert(obj.name);
});

Answer 2

你也可以使用常规的javascript,我认为会更快一些(虽然我不确定jQuery如何优化each):

var data = [{"id":"856","name":"India"},
 {"id":"1035","name":"Chennai"},
 {"id":"1048","name":"Delhi"},
 {"id":"1113","name":"Lucknow"},
 {"id":"1114","name":"Bangalore"},
 {"id":"1115","name":"Ahmedabad"},
 {"id":"1116","name":"Cochin"},
 {"id":"1117","name":"London"},
 {"id":"1118","name":"New York"},
 {"id":"1119","name":"California"}
];

var data_length = data.length;
for (var i = 0; i < data_length; i++) {
  alert(data[i]["id"] + " " + data[i]["name"]);
}

编辑以反映尼克关于表现的建议