当用户输入已经使用的用户名时,我遇到了出现错误的问题。
在下面的代码中,数据库在成功输入后会更新。但是,如果使用重复的用户名创建条目,则该条目将放在数据库中,并且不会显示任何错误消息。我已经看过网并尝试了一些方法,这就是我到目前为止所做的。谢谢你看看:))
<?php
// Create connection
$con = mysqli_connect('172.16.254.111', "user", "password", "database"); //(connection location , username to sql, password to sql, name of db)
// Check connection
if (mysqli_connect_errno($con)) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} //sql syntax below, first line is collumn titles on the db and second line is values from the html document
//$_post is a form of sending information in php
{
$username = strip_tags($_POST['username']);
$password = md5(strip_tags($_POST['pass']));
$password2 = md5(strip_tags($_POST['pass2']));
$fullname = strip_tags($_POST['fullname']);
$email = strip_tags($_POST['email']);
$department = strip_tags($_POST['department']);
if ($password != $password2) //password doestn equal same as password 2 then the message below is displayed (working)
{
echo "<H2>password doesn't match</H2>";
}
$usercheck = "SELECT * FROM Users WHERE username=$username";
$usercheck2 = mysql_query($usercheck);
if (mysql_fetch_assoc($usercheck2)) {
echo "<H2>This username already exists, please pick another</H2>";
} else {
$sql = "INSERT INTO Users(username, password, password2, email, fullname, department)
VALUES('$username','$password','$password2','$email','$fullname','$department')";
if (!mysqli_query($con, $sql)) {
die('Error: ' . mysqli_error($con));
}
echo "<H2>Registration was successful, please use the access console above</H2>";
}
}
?>
请原谅代码中的任何评论;我是PHP和编码的初学者。
答案 0 :(得分:1)
你错过了一些引用。试试这个:
$usercheck = "SELECT * FROM Users WHERE username = '$username'";
// ----------------------------------was missing---^---------^
$usercheck2 = mysql_query($usercheck);
if (mysql_num_rows($usercheck2)) {
echo 'user exists';
}
此外,您不应该使用mysql_*
功能。请查看使用PDO
答案 1 :(得分:0)
更改$ usercheck =“SELECT * FROM Users WHERE username = $ username”; 为此$ usercheck =“SELECT * FROM Users WHERE username ='$ username'”;告诉我它是否有效。