我有以下 AJAX 脚本,但由于某种原因,var ok 它没有返回 true 或 false,因此表单可以继续:
function ajax_call(email,title,url){
var email = document.getElementById("email").value;
var title = document.getElementById("title").value;
var url = document.getElementById("url").value;
var parametros = {"emaail":email, "tiitle":title, "uurl":url};
var ok = true;
$.ajax({
data: parametros,
url: 'validate.php',
type: 'post',
error: function () {
alert("An error has occurred! Try Again!");
},
success: function (response) {
if(response == 'bien') { ok = true; } else { $("#ajax_cal").html(response); ok = false; }
}
});
return ok;
}
HTML:
<form onsubmit="return ajax_call();">
...
</form>
PHP:
<?php
//////....
if(!empty($errors)) {
foreach($errors as $error) {
echo '<li>'.$error.'</li>';
}
} else { echo 'bien'; }
?>
一切都很好,除了返回值。提前致谢。