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"}
]
发布于 2010-11-20 23:10:01
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);
});
发布于 2010-11-20 23:09:10
您可以使用.each()
函数:
$(yourjsondata).each(function(index, element) {
alert('id: ' + element.id + ', name: ' + element.name);
});
发布于 2017-10-25 21:26:15
使用$.each函数迭代对象的所有属性。在每次迭代中,您将获得名称/键和属性的值:
$.each(data, function(key, val) {
alert(key+ " *** " + val);
});
https://stackoverflow.com/questions/4233354
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