PHP和Ajax输入验证中的If/else语句
问题描述:
我正在通过Ajax将一些数据发送到处理注册的PHP脚本。我试图从服务器得到一个响应,无论输入是否有效(实时验证)。我希望每个输入都有自己的响应,这意味着如果第一个输入仍然无效,并且某人在第二个输入字段中输入了无效的内容,那么第一个响应将保持不变,并且它会发送第二个响应。PHP和Ajax输入验证中的If/else语句
现在发生的情况是:第一个输入无效,我看到响应,但是当我进入下一个输入字段并在该字段中放置了一些无效字符时,第一个响应停留在那里(我已经用控制台的.log在Chrome)
UPDATE:来举个例子,我看到这样的:not a valid username!
,然后我把一些无效的电子邮件地址,在未来的领域,我仍然看到not a valid username!
。
这是我的PHP代码:
if(isset($_POST['username']) && isset($_POST['email']) && isset($_POST['email2']) && isset($_POST['password'])
&& isset($_POST['firstname']) && isset($_POST['surname']) && isset($_POST['gender']) && isset($_POST['day'])
&& isset($_POST['month']) && isset($_POST['year'])) {
$username = $_POST['username'];
$email = $_POST['email'];
$email2 = $_POST['email2'];
$password = $_POST['password'];
$firstname = $_POST['firstname'];
$surname = $_POST['surname'];
$gender = $_POST['gender'];
$day = $_POST['day'];
$month = $_POST['month'];
$year = $_POST['year'];
if(!preg_match("/^[a-z](?=[\w.]{3,19}$)\w*\.?\w*$/i",$username)){
echo "not a valid username.";
}
else if(filter_var($email,FILTER_VALIDATE_EMAIL)){
echo "OK!";
}
else if(!filter_var($email,FILTER_VALIDATE_EMAIL)){
echo "not a valid email address";
}
else if(strcmp($email,$email2) != 0){
echo "emails are different.";
}
else if(strcmp($email,$email2) == 0){
echo "OK!";
}
else if(!preg_match("[a-zA-Z]*",$firstname)){
echo "Not a valid firstname.";
}
else if(preg_match("[a-zA-Z]*",$firstname)){
echo "OK!";
}
else if(!preg_match("[a-zA-Z]*",$surname)){
echo "not a valid surname.";
}
else if(preg_match("[a-zA-Z]*",$surname)){
echo "OK!";
}
}
,这是JQuery的Ajax代码:
function handlePost() {
var username = $('#username').val();
var email = $('#email').val();
var email2 = $('#email2').val();
var password = $('#password').val();
var firstname = $('#firstname').val();
var surname = $('#surname').val();
var gender = $('#gender').val();
var day = $('#day').val();
var month = convertMonth($('#month').val())
var year = $('#year').val();
$.ajax({
type: "POST",
url: "handleRegister.php",
data: "username="+username+"&email="+email+"&email2="+email2+"&password="+password+"&firstname="
+firstname+"&surname="+surname+"&gender="+gender+"&day="+day+"&month="+month+"&year="+year,
success: function(resp){
// we have the response
//alert("Server said:\n '" + resp + "'");
console.log("Server said:\n '" + resp + "'")
},
error: function(e){
//alert('Error: ' + e);
console.log("Server said:\n '" + e + "'")
}
});
}
我想这是我使用的if/else来这里的路上。此外,在这种情况下,我对如何/何时使用isset($ _ POST ['submit'])有些困惑?
在此先感谢。
答
$errors = array();
if (!preg_match("/^[a-z](?=[\w.]{3,19}$)\w*\.?\w*$/i", $username))
{
$errors[] = "not a valid username.";
}
if (!filter_var($email, FILTER_VALIDATE_EMAIL))
{
$errors[] = "not a valid email address";
}
if ($email !== $email2)
{
$errors[] = "emails are different";
}
if (!preg_match("[a-zA-Z]*", $firstname))
{
$errors[] = "Not a valid firstname.";
}
if (!preg_match("[a-zA-Z]*", $surname))
{
$errors[] = "not a valid surname.";
}
if ($errors)
{
echo implode("\n", $errors);
}
else
{
echo 'OK!';
}
答
将您的else if
s中的一些更改为if
s。只要使用else if
,只要条件匹配,其余的都会被跳过。所以,当你有一个无效的用户名,你的代码从不检查电子邮件是否有效。
当只需要其中一个输出时,仅使用else if
。
答
,你会想要做的是什么,而不是别人的,只是IFS喜欢:
if(!preg_match("/^[a-z](?=[\w.]{3,19}$)\w*\.?\w*$/i",$username)){
echo "not a valid username.";
}
if(filter_var($email,FILTER_VALIDATE_EMAIL)){
echo "OK!";
}
if(!filter_var($email,FILTER_VALIDATE_EMAIL)){
echo "not a valid email address";
}
if(strcmp($email,$email2) != 0){
echo "emails are different.";
}
etc....
OK!从未实际印刷。当改变1场时,所有的错误出现?我怎样才能展示我想要的?使用数组索引? – Loolooii 2012-03-01 19:34:42
好吧,我明白了:) – Loolooii 2012-03-01 19:38:27