如何在jquery中迭代json数据
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"} ] 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); });
你也可以使用普通的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"]); }
编辑以反映尼克关于表现的建议
您可以使用.each()函数：
$(yourjsondata).each(function(index, element) { alert('id: ' + element.id + ', name: ' + element.name); });
使用$ .each函数迭代所有对象的属性。 在每次迭代中，您将获得名称/键和属性的值：
$.each(data, function(key, val) { alert(key+ " *** " + val); });