嗨,我想创建一个非常简单的日志功能使用PHP然而,通过代码去后,我的状态似乎能够正确,只有当我输入正确的用户名执行的功能和设置密码,如果我不,它并不在这里履行职责,是我的,我已经简化代码:登录找不到错误
的login.php
<form name="userlogin" action="phpprocess/loginprocess.php" method="POST">
<p>Username : <input type="text" id="username" name="username"></p>
<p>Password : <input type="password" id="password" name="password"></p>
<p><input type="submit" id="loginbtn" value="login" ></p>
</form>
loginproccess.php
include "mysqli.connect.php";
$username = $mysqli->real_escape_string($_REQUEST["username"]);
$password = $mysqli->real_escape_string($_REQUEST["password"]);
echo "$username";
echo "$password";
$sql = "select * from users.userlogin where username ='".$username."' and
password = '".$password."'";
$result = $mysqli->query($sql);
if($result == null){
echo"null";
}
if($mysqli -> errno){
error_log($mysqli -> error);
echo $mysqli -> error;
echo " hello";
exit();
}else{
while(list($index, $user1, $pass1) = $result -> fetch_array()){
if($user1 != null && $pass1 != null){
echo "$index $user1, $pass1";
}
}
}
$mysqli->close();
mysqli.c onnect.php
$host="localhost";
$user="root";
$password="";
$database="users";
$mysqli = new mysqli($host, $user, $password, $database);
if ($mysqli->errno) {
echo "Unable to connect to the database: <br />".$mysqli->error;
exit();
}
发生的事情是,如果我输入了错误的用户名和密码,我需要的网页呼应你好,如果我输入正确我需要它呼应了正确的用户,并通过但当输入错误, $结果看起来不是null,因为null和hello没有被打印出来。我的错误日志不显示任何内容。希望听到您的建议!先谢谢你 !
哎barmar!感谢您的支持!它真的很有用,因为我自己学习PHP!谢谢你,生病了:) –