php - 条件语句中的 !preg_match() 函数始终为 false

Question

请看下面的代码。我认为这是正确的。当我使用其他角色时包含字母和数字或单独显示时，我应该会看到仅字母和数字消息。但即使当我只使用字母和数字时，我也会再次看到仅字母和数字。

<?php

if($_SERVER['REQUEST_METHOD'] == "POST") {
        $username = $_POST['username'];
$login = $_POST['pass'];
$repass = $_POST['repass'];
if(empty($username) || empty($login) || empty($repass)) {

echo "one or more fields is empty";
}

 else if(!preg_match("[A-Za-z0-9]",$username) || !preg_match("[A-Za-z0-9]", $login) || !preg_match("[A-Za-z0-9]",$repass)){

echo "only the letters and numbers";
}

        else if($login != $repass){

echo "the password is not match with the repassword";
}
        else {

        $connection = mysqli_connect("127.0.0.1", "root", "", "hello");
if(!$connection) { echo "error in connecting to the database";}
        else {


$query = mysqli_query($connection,"SELECT * FROM test WHERE username='$username'");

$rows = mysqli_num_rows($query);
if($rows == 1 ) {

echo "there is this username currently";
}
        else {

        $sql = "INSERT INTO test (username, login, repass) VALUES ('$username', '$login', '$repass')";

if(mysqli_query($connection,$sql)) {



echo "you are register";

} else{
        echo "please again";
 mysqli_close($connection);}
        } } } }
else {
echo "invalid request";
}
 ?>

Answer 1

只需添加分隔符

else if(preg_match("#[^a-z0-9]+#i",$username))