我正在创建一个小插件,我想使用ajax提交表单数据,但它的响应为零。
这是我的电话,我暂时将所有代码添加到单个文件中
文件名:index.php
<?php wp_enqueue_script("jquery");?>
<?php
function myaddgallery(){
global $wpdb;
echo "abac";
}
add_action('wp_ajax_myaddgallery' , 'myaddgallery');
add_action('wp_ajax_nopriv_myaddgallery' , 'myaddgallery');
?>
<script type="text/javascript">
jQuery(document).ready(function(){
jQuery('#shaGalleryForm').submit(function(e){
// prevent normal submit behaviour
e.preventDefault();
var name = jQuery("#shaGalleryName").val();
alert(name);
//var postData = jQuery("#shaGalleryForm").serialize();
//console.log(postData);
var name = "shalu";
jQuery.ajax({
type: "POST",
url: "http://localhost/plugindevelop/wp-admin/admin-ajax.php",
data : {action: "myaddgallery", name : name},
success: function(response){
//console.log(postData);
console.log("added success");
alert(response);
},
error: function(data){
console.log("fail");
}
});
});
}) ;
</script>
<div class='wrap'>
<h2>Add Gallery</h2>
<p>This is where from you can create new gallery</p>
<form name="shaGalleryForm" id="shaGalleryForm" method="post">
<table class="form-table">
<tr>
<th>Gallery Name</th>
<td>
<input type="text" id="shaGalleryName" name="shaGalleryName">
<p class="description">Add gallery name also please avoid special character</p>
</td>
</tr>
</table>
<p class="submit">
<input type="submit" value="Save" id="shaSaveGallery" name="shaSaveGallery" class="button button-primary">
</p>
</form>
</div>
答案 0 :(得分:0)
尝试在回显数据后添加die()
function myaddgallery(){
global $wpdb;
echo "abac";
die();
}