Ajax返回不起作用

时间:2015-12-29 18:08:22

标签: php jquery ajax

<form class="form-horizontal" method="post" id="sponsor" name="sponsor" onsubmit="return ajaxSubmit();" >   
    <input type="text" class="form-control"  id="sponsor_name" name="sponsor_name" placeholder="Enter Name"/>   
    <input type="submit" class="btn btn-default" name="sponsor_form" value="Register" />                
</form>
function ajaxSubmit(){
    // fetch where we want to submit the form to
    var url = url;
    // fetch the data for the form
    var email = $('#sponsor #email').val();
    // setup the ajax request

    $.ajax({
        url: url,
        method:'POST',
        data: {email:email},
        success: function(data) {
            if(data!=0){
                alert('ifenter');
                $('#msg_section').html('The email has already registered with '+data+' plan <br/> <button type="button" class="btn btn-default form_sponsor" id="form_sponsor" name="form_sponsor" onclick="submitform()">Click Here to Continue</button>');

                $('#msg_section').css('display','block');
                $('#form_sponsor').css('display','block');
                return false;
            }else{
                alert('enter');
                $('#msg_section').css('display','none');
                $('.form_sponsor').css('display','none');
                $('#msg_section').html('');
                return true;
            }
        }
    });
    return false;
}

返回true它假设进入isset($ _ POST [&#39; sponsor_form&#39;])。但它没有任何帮助被接受。这是ajax无法正常工作。

0 个答案:

没有答案