请附上我的代码:
<?PHP
session_start();
if (isset($_POST["usernameform"])){
$username = $_POST['usernameform'];
$password1 = $_POST['passwordform'];
$user_name = "XXXX";
$password = "XXXXX";
$database = "XXX";
$server = "XXXX";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
$SQL = "SELECT * FROM login WHERE Username = '$username' AND Password = '$password1' "; //grab all the records from table
$result = mysql_query($SQL)
or die("Error:" . mysql_error());
if (mysql_num_rows($result) > 0){ //if username and password match return number of rows is always 1
$_SESSION['login'] = $username; //by placing this in session it will remember this variable on the page it directs too
while ( $row = mysql_fetch_assoc($result) ){ //lays out array in $result
$_SESSION['ID'] = $row['ID']; //selects from list of array ID
$_SESSION['firstname'] = $row['First_Name'];
echo'<script> window.location="page1.php"; </script> ';
}
} else {
$_SESSION['login'] = '';
print('
<script type="text/javascript"> //place html script for alert. Use single comma for print command here.
alert("Sorry, your username or password could not be recognized")
</script>
');
session_destroy();
}
}
?>
虽然代码在我的localhost(wampserver)上完美运行,但它在我的主机上无效并且我收到了错误:
警告:session_start()[function.session-start]:无法发送会话 cookie - 已经发送的标题(输出开始于 /.../vhindex.php:1)在第2行的/.../vhindex.php
警告:session_start()[function.session-start]:无法发送会话 缓存限制器 - 已发送的标头(输出始于 /.../vhindex.php:1)在第2行的/.../vhindex.php
session_start();
代码已放置在PHP块的顶部,在进行任何HTML输出之前,我完全被难倒了。
有什么想法吗?
答案 0 :(得分:9)
<?PHP
<?PHP
之前没有像“”或制表符或换行符这样的空白字符。
Encoding
,然后单击UTF-8 without BOM
将文件转换为不带BOM的UTF-8,然后保存。 ob_start();
之前添加session_start();
以确保安全。 答案 1 :(得分:-1)
如果从vhindex.php加载此-.php文件。确保以正确的顺序完成此操作。 例如。 require_once&#39;处理用户自的login.php&#39 ;; callToYourFunction();
如果您的会话已经发送,则会收到此警告。 (是的,就像错误代码所说的那样)。