【发布时间】:2018-06-20 07:37:06
【问题描述】:
header.php 文件用于发送 caegory id 并返回子类别名称
<?php
$conn = mysqli_connect("localhost", "root", "", "iraa");
?>
Category :
<select id="category_id" name="cat_id" required="" class="form-control" tabindex="2" onselect ="getSubcat()">
<option value="">-- Select --</option>
<?php
$sql = "SELECT category_id,category_name FROM categories_master";
$result = mysqli_query($conn, $sql);
while ($data=mysqli_fetch_array($result)) {
?>
<option value="<?php echo $data['category_id'] ?>"><?php echo $data['category_name'] ?></option>
<?php } ?>
</select>
Sub Category :
<select id="sub_cat_id" required name="sub_cat_id" class="form-control " tabindex="2">
<option value="">-- Select --</option>
</select>
<script src="js/jquery-1.12.4.js"></script>
<script src="//rawgithub.com/indrimuska/jquery-editable-select/master/dist/jquery-editable-select.min.js"></script>
<link href="//rawgithub.com/indrimuska/jquery-editable-select/master/dist/jquery-editable-select.min.css" rel="stylesheet">
<script type="text/javascript">
$('#category_id').editableSelect();
function getSubcat() {
var category_id=$("#category_id").val();
alert(category_id);
$.ajax({
url: 'get_subcategory.php',
type: 'POST',
data: {category_id: category_id},
})
.done(function(response) {
$('#sub_cat_id').html(response);
});
}
</script>
get_subcategory.php 文件,用于获取 ajax 响应并返回子类别
<?php
$conn = mysqli_connect("localhost", "root", "", "iraa");
$sql = "SELECT sub_category_id,sub_category_name FROM sub_categories_master WHERE category_id='$_POST[category_id]'";
$result = mysqli_query($conn, $sql);
while ($data=mysqli_fetch_array($result)) {
?>
<option value="<?php echo $data['sub_category_id'] ?>"><?php echo $data['sub_category_name'] ?></option>
【问题讨论】:
-
哥们,你检查过答案了吗?
-
我检查了答案,但它不起作用。因为我在那里使用了 editableSelect() 方法来执行脚本时将其转换
-
好的伙计,我已经在下面更新了我的答案。
-
非常感谢先生。这是工作。
标签: javascript php jquery ajax mysqli