我无法传递ajax响应数据进行查看。视图不会传递数据。没有为registernumberP显示任何内容。请帮我。谢谢
AJAX响应:
[
{
"id": "1",
"studentId": "16",
}
]
AJAX致电:
$.ajax({
type: "POST",
dataType: "json",
url: "<?php echo base_url(); ?>" + "index.php/Payments/GetStudentPaymentDetails",
data: {
studentId: studentId
},
success: function (data) {
$('#registernumberP').val(data['studentId']);
},
error: function () {
alert('Some error occurred!');
},
});
查看:
<?php
$data = array(
'id' => 'registernumberP',
'name' => 'registernumberP'
);
echo form_input($data);
?>
</div>
答案 0 :(得分:0)
$('#registernumberP').val(data['studentId']);
应该是
$('#registernumberP').val(data[0]studentId);
可以移除data
:
data: {
studentId: studentId
},