所以我用PHP为一个网站构建了一个简单的登录脚本。它运行良好，但我最近对它进行了一些更改，似乎阻止了它正常运行。
基本上，当我将表放入数组时，我会使用变量$y来跟踪正在登录的用户的"类型"。然而，当登录成功时，在回显$y和$type时，它们都会返回0。用户可以是类型0或类型1，但当找到用户时，似乎由于某种原因没有分配$y。
为了确认，登录语句等确实有效，如果用户名和密码正确，则会显示正确的用户名和相关详细信息。目前，出于某种原因，它似乎不想给$y赋值。
// If statement that seems to be giving me trouble
global $arrayofdata;
$arrayofdata = array();
$n = 0;
$y = 0;
// Put tables into an array
while ($row = mysql_fetch_array($resource)) {
// If statement to find position of username in array
if($arrayofdata[$n]['username'] == $username){
$y = $n;}
$arrayofdata[$n] = $row;
$n++;
}
// FULL CODE BENEATH HERE
<?php
session_start(); ?>
<html>
<head>
<title>:: clubb3r ::</title>
</head>
<body>
<?php
loginscript::login();
class loginscript {
// Login function..
static function login() {
$host = "gcdsrv.com";
global $username;
if(isset($_SESSION['username'])){
$username = $_SESSION['username'];}
else{
$username = $_POST[uname];
$_SESSION['username'] = $username;} // Store username for later
if(isset($_SESSION['password'])){
$password = $_SESSION['password'];}
else{
$password = $_POST[pword];
$_SESSION['password'] = $password;} // Store password for later
$connect = mysql_connect("gcdsrv.com", "", "");
if(!$connect) {
echo "<h1>500 Server Error</h1>";
}
$db_select = mysql_select_db("c2h5oh_database", $connect);
$resource = mysql_query("SELECT username, password, type, picture, rating FROM accounts;");
global $arrayofdata;
$arrayofdata = array();
$n = 0;
$y = 0;
// Put tables into an array
while ($row = mysql_fetch_array($resource)) {
// If statement to find position of username in array
if($arrayofdata[$n]['username'] == $username){
$y = $n;}
$arrayofdata[$n] = $row;
$n++;
}
$n = 0;
// Set user type (normal user or bar/club, 0 for user and 1 for bar/club)
if(isset($_SESSION['type'])){
$type = $_SESSION['type'];}
else{
$type = $arrayofdata[$y]['type'];
$_SESSION['type'] = $type;
}
// Counts entries
$count = count($arrayofdata);
global $count2;
// Login check loop, searches array for username and password in POST, also stores balance of that user for later
for($x = 0; $x < $count; $x++) {
if($username == $arrayofdata[$x]['username'] && $password == $arrayofdata[$x]['password'] && $username != "" && $password != "") {
$z = 1;
}
}
// Fail
if($z != 1) {
echo "<h1>Bad Username or Password</h1><br />";
echo "<h1><a href='logout.php'>Try Again</a></h1>";
}
// Success
// If for user success
if($z == 1 && $type == 0) {
echo "<h1>Login Successful!</h1><br />";
echo "<h1><a href='mainuser.html'>Proceed</a></h1>";
echo $type;
echo $y;
}
//Success
//If for bar/club success
if($z == 1 && $type == 1){
echo "<h1>Login Successful!</h1><br />";
echo "<h1><a href='mainbar.html'>Proceed</a></h1>";
echo $type;
}
}
}
?>
</body>
</html>
就我所见，您有两行，它们设置了$y。
`$y = 0;`
和
// Put tables into an array
while ($row = mysql_fetch_array($resource)) {
// If statement to find position of username in array
if($arrayofdata[$n]['username'] == $username){
$y = $n;
}
$arrayofdata[$n] = $row;
$n++;
}
再往前看，你会看到这三条线：
$arrayofdata = array(); //create EMPTY array
[...]
if($arrayofdata[$n]['username'] == $username){ //check index $n
[...]
$arrayofdata[$n] = $row; //set index $n
你在设置$arrayofdata[$n]之前访问它。我猜你的if语句会抛出PHP警告，控制你的日志；）如果if-语句为true，则在下面的代码中设置$y = $n。由于$arrayofdata[$n]不存在，if子句if始终为false，$y保持为0。