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

您可以像这样使用$.each()：

$.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，我认为会更快一点（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"]);
}

已编辑以反映尼克关于表现的建议

Answer 3

您可以使用.each()功能：

$(yourjsondata).each(function(index, element) {
    alert('id: ' + element.id + ', name: ' + element.name);
});

Answer 4

使用$ .each函数迭代所有对象的属性。在每次迭代中，您将获得名称/密钥和属性的值：

$.each(data, function(key, val) {
  alert(key+ " *** " + val);
});

如何在jquery中迭代json数据

4 个答案: