表单提交使用jQuery阿贾克斯 [英] Form submission using jQuery ajax

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本文介绍了表单提交使用jQuery阿贾克斯的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

表没有提交使用ajax.form提交关于点击华里。给我一些解决方案

Form is not submit using ajax.form submit on click li. Give me some solution

我的js code是这里

My js code is here

$(document).ready(function(){

$('#sortable li').click(function() {

$("#frmgallery").submit(function(event) {
    event.preventDefault();
    var formdata = $(this).serialize();
    alert(formdata);
    $.ajax({
        type: "POST",
        url: "gallery.php",
        data: formdata,
        success: function(){alert('success');}
    });
});

});

HTML是这里

HTML is here

 <form method="post" enctype="multipart/form-data" id="frmgallery" name="gallery" action="<?php get_template_directory();?>admin.php?page=gallery/gallery.php">

<ul id="sortable">

Query

<li  class="ui-state-default" name='liindex' id="<?php echo $row['glryRecordId'];?>" >
<span style="display:none"><?php echo $row['glryRecordId'];?></span>
<img class='thumbnail' alt=''  src='<?php echo get_site_url();?>/wp-content/themes/townsley/upload/<?php echo $row['glryName']; ?>' width='80' height='60' 
 style="border: 1px solid #D3D3D3;padding:2px;"/><input type="hidden" value="<?php echo $row['glryRecordId'];?>" name="recordId[]" />
    <a href="<?php get_template_directory();?>admin.php?page=gallery/gallery.php&delid=<?php echo $row['glryRecordId'];?>" style="display:block;text-align:center"  title="DELETE this image from the record" class="arial-red">Remove</a>
</li>


</ul>
</form> 

请帮我 谢谢

AJAX的jQuery JavaScript的

ajax jquery javascript

推荐答案

您应该为您的HTML过你的问题,但据我所看到的,你有事件事件回调有其实也没什么启动提交事件。所以基本上你应该考虑是这样的:

You should provide your HTML too in your question, but as far as i can see, you have event in event callbacks with actually nothing to initiate the submit event. So basically you should consider something like this :

$(document).ready(function(){

    $('#sortable li').click(function() {
        event.preventDefault();
        var formdata = $("#frmgallery").serialize();
        alert(formdata);
        $.ajax({
            type: "POST",
            url: "gallery.php",
            data: formdata,
            success: function(){alert('success');}
        });
    });

});

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