Goo*_*tes 1 javascript php ajax json
是否可以使用PHP中的AJAX请求响应?我不是一个真正的JS开发人员,所以我用这个来调查我的头发.
我有点一起攻击这个:
var base_url = 'http://dev.local/westview/public';
$('select.child_id').change(function() {
var child_id = $('#child_id');
var dataString = 'child_id=' + child_id;
$.ajax({
type: "POST",
url: base_url + "/finance/payment-history",
data: dataString,
dataType: 'html',
success: function(html) {
alert(html);
},
});
return false;
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});
该功能似乎工作正常,它给我一个正确数据的警报.
{"payments":[{"id":"19","child_id":"21","club":"Breakfast Club","term":"Half Term 3","amount":"15.00","pdate":"2015-02-25","notes":"","created_at":"2015-02-11 12:16:32","updated_at":"2015-02-11 12:16:32","starting_debt":"0","debt_start_date":"2015-01-05"},{"id":"20","child_id":"21","club":"After School Club","term":"Half Term 3","amount":"11.50","pdate":"2015-02-25","notes":"","created_at":"2015-02-11 12:16:49","updated_at":"2015-02-11 12:16:49","starting_debt":"0","debt_start_date":"2015-01-05"}]}
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我需要能够将其输出给用户,以便它可读.我找到的很多指南都描述了替换数据,但是直到选择了child_id才会有数据.然后我希望它以可读的方式显示上述数据.
我不知道如何开始使用我的视图文件(php)中的数据.
谢谢
[编辑]更新了工作代码:
var base_url =' http://dev.local/westview/public ';
$('select.child_id').change(function() {
var response = "";
var child_id = $('#child_id').val();
var dataString = 'child_id=' + child_id;
$.ajax({
type: "POST",
url: base_url + "/finance/payment-history",
data: dataString,
success: function(response) {
var json_obj = $.parseJSON(response);
var output = "<ul>";
for (i=0; i < json_obj.payments.length; i++)
{
var payment = json_obj.payments[i];
var date = moment(payment.pdate).format('Do MMM YYYY');
output += "<li>£" + payment.amount + " - " + date + " (" + payment.club + ")</li>";
}
output += "</ul>";
$('.history-section').html(output);
},
dataType: "html"
});
});
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这样做.
var data = $.parseJSON("your_json");
var output= "<ul>";
for (i=0; i < data.payments.length; i++){
output += "<li>" + data.payments[i].id + ", " + data.payments[i].child_id + "</li>";
}
output += "</ul>";
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