这是我的代码。
我认为我的道路有误
这是我第一次将AJAX放入域名中......
$.ajax({
url: 'http://www.komatpillar.com/ajax/check_email.php',
type: 'POST',
data: {pass_email_address : email_address},
dataType: 'json',
success: function(result) {
var res = result.user_id;
if(res > 0){
$('#message_email').focus();
window.scrollTo(0, 540);
$('#danger_container_email').attr('class', 'alert alert-danger');
$('#message_email').html('This email address is already existing!');
$('#register').attr('disabled','disabled');
$('#register').removeAttr('name');
$('#message').css('color', 'red');
$('#success_email').attr('class','form-group has-error has-feedback');
$('#email_glyp').attr('class', 'glyphicon glyphicon-remove form-control-feedback');
}else{
$('#danger_container_email').removeAttr('class');
$('#message_email').html('');
$('#success_email').attr('class','row form-group has-success has-feedback');
$('#email_glyp').attr('class', 'glyphicon glyphicon-ok form-control-feedback');
$('#register').removeAttr('disabled');
$('#register').attr('name','cmd_add_new_user');
}
以下是其余部分。
答案 0 :(得分:-1)
改变这个:
data: {pass_email_address : email_address},
到此:
data: "{pass_email_address:'" + email_address + "'}",