我有一個login.html
其中form
定義如下:VAR轉儲不工作的一個特定的PHP文件
<form method="post" action= "do_authorize.php" name="lform">
<span class="style1">First Initial Plus Last Name :</span>
<input type="text" name="user" size="25">
<input type="submit" value="login">
</form>
我do_authorize
如下:
<?
session_start();
require('../databaseConnectionFileFolder/dbconnection.php');
$user = $_POST["user"];
var_dump($user);
$_SESSION['username']=$user;
var_dump($user);
$sql="SELECT * FROM $table_name_users WHERE username = \"$user\"";
var_dump($sql);
[email protected]_query($sql,$connection) or die("couldn't execute query");
$num=mysql_numrows($result);
if ($num != 0) {
/*$cookie_name="$user";
$cookie_value="ok";
$cookie_expire=time()+86400;
$cookie_domain=".columbia.edu";
setcookie($cookie_name, $cookie_value, $cookie_expire, "/", $cookie_domain, 0);
*/
print "<script>";
print "self.location='somethingelse.php';";
print "</script>";
} else {
echo "<p>you're not authorized";
}
?>
我已經包括var_dump($user);
和var_dump($sql);
在適當的地方,它似乎並不打印用戶值和SQL。所有它打印在瀏覽器每次如下:
"; print "self.location='somethingelse.php';"; print ""; } else { echo "
you're not authorized"; } ?>
我指的是this post,我做了全面的檢查,通過Dan Nissenbaum
在下面的評論中提到,它爲我的作品:
For sanity checking, does the same problem occur in an independent .php file in the same webroot - something simple like <?php $x = array(1,2,3); var_dump($x); ?>?
嘗試使用'<?php'而不是'<?'。 http://php.net/manual/en/ini.core.php#ini.short-open-tag – AbraCadaver
工作,謝謝。 – Tan