我的ajax函数将数据类型设置为JSON后会出错。那是代码:
ajax脚本:
$('#da').on("change",function() {
$.ajax({
url: "callAjaxIndex.php",
type: "POST",
dataType: "json",
data: {
method: 1,
id: $('#da').val(),
},
success: function() {
alert('test');
},
error: function() {
alert('error');
}
});
});
callajaxindex.php
<?PHP
require('includes/core.php');
if ( isset($_POST['method']) ) {
$sql = "SELECT tratte.nome as 'nome_arrivo', tratte.id as 'id_arrivo' FROM tariffe, tratte WHERE id_arrivo = tratte.id AND id_partenza = '".$_POST['id']."'";
$query = $conn->query($sql);
while ( $tariffe = $query->fetch_array() ) {
$result[] = array(
'id' => $tariffe['id_arrivo'],
'nome' => $tariffe['nome_arrivo']
);
}
echo json_encode($result);
}
?>
怎么了?谢谢
您可以尝试此
$(document).on('change', '#da', function(){
$.post("callAjaxIndex.php", {'method': 1, 'id': $(this).val()}, function(data){
var d = $.parseJSON(data); //here is the data parsed as JSON
//data is that returned from callAjaxIndex.php file
});
});
<?php
require('includes/core.php');
if ( isset($_POST['method']) ) {
$sql = "SELECT tratte.nome as nome_arrivo, tratte.id as id_arrivo FROM tariffe INNER JOIN tratte ON id_arrivo = tratte.id WHERE id_partenza = '".$_POST['id']."'";
$query = $conn->query($sql);
while ( $tariffe = $query->fetch_array() ) {
$result[] = array(
'id' => $tariffe['id_arrivo'],
'nome' => $tariffe['nome_arrivo']
);
}
echo json_encode($result);
}
您可以通过更改函数来发现错误:
//other code
error: function(data)
{
console.log(data.responseText)
}
//other code
这将告诉您为什么失败,可能是通用的东西,但比"错误"
更好还注意:
- 这是通过电话完成的,因此请原谅任何错误
- 我宁愿将其视为评论,直到我可以使用机器来帮助更多:)