我今天一整天都在试图弄清楚问题出在哪里,我在这里阅读了很多类似的问题,但这似乎是一个特殊的问题。
我想要一个 AJAX 调用来填充一个基于之前从 MySQL 数据库中选择的下拉列表,但是它不起作用,因为 POST 没有向 PHP $id = $_POST['id']
文件,我将它发送到。但是,如果我为它分配一个常量值(例如 $id = 1
),它就可以正常工作并从数据库返回预期的数据。
这是 AJAX 代码:
<script type="text/javascript">
$(document).ready(function()
{
$("#state").change(function()
{
var menuId = $(this).val();
var request = $.ajax({
url: "schools.php",
type: "POST",
data: { id : menuId },
dataType: "html"
});
request.done(function( msg ) {
$("#schools").html( msg );
});
});
});
</script>
PHP 代码:
<?php
include('../connection.php');
if(isset($_POST['id']) && $_POST['id'] != '')
{
$id = $_POST['id'];
$SQ = "SELECT c_name FROM courses WHERE fac_id = ?";
$q = $conn->prepare($SQ) or die("ERROR: " . implode(":", $conn->errorInfo()));
$q->bindParam(1, $id);
$q->execute();
$total = $q->fetchAll();
$i=1;
foreach ($total as $sch){
echo "<option class = 'option' value = '$i'>". $sch[0].</option>";
$i++;
}
}
?>
编辑:
这是 HTML 代码:
<form class="form-horizontal">
<!-- Form Name -->
<!-- Select Basic -->
<div class="control-group">
<label class="control-label" for="selectbasic">State where your school is located</label>
<div class="controls">
<select id="state" name="state" class="form-control input-xlarge">
<?php
$i=1;
$SQ = "SELECT * FROM states";
$schools = $conn->prepare($SQ);
$schools->execute();
$total = $schools->fetchAll();
foreach ($total as $sch){
echo "<option class = 'option' value = '$i'>". $sch[1]. "</option>";
$i++;
}
?>
</select>
</div>
</div>
<!-- Select Basic -->
<div class="control-group">
<label class="control-label" for="selectbasic">School</label>
<div class="controls">
<select name="school" id = 'schools' class="form-control input-xlarge">
<?php
$i=1;
$SQ = "SELECT * FROM schools";
$schools = $conn->prepare($SQ);
$schools->execute();
$total = $schools->fetchAll();
foreach ($total as $sch){
echo "<option class = 'option' value = '$i'>". $sch[1]. "</option>";
$i++;
}
?>
</select>
</div>
</div>
</form>
最佳答案
我会尝试直接使用 $.ajax 而不是将其设置为变量。
$.ajax({
url: "schools.php",
type: "POST",
data: { id : menuId },
dataType: "html",
success: function(data){
append data stuff here.
},
error: function(xhr, status, error){
console.log(error + " error from here");
}
});
如果返回任何错误,您可以使用错误函数打印出您的错误。它可能会帮助您调试问题。
关于php - JQuery AJAX POST 没有将数据发送到 PHP 文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20078072/