所以即时通讯使用简单的登录和用户区域来创建此网站。使用会话变量在SQL中创建查询
我试图执行一个SQL行,但它显示一个错误。
这是我设置会话变量的checklogin.php页面。
// Check if session is not registered , redirect back to main page.
// Put this code in first line of web page.
<?php
session_start();
$tbl_name= "members";
$con=mysqli_connect("localhost","root","","bst_planner");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysqli_query($con,$sql)or die(mysqli_error($con));
// Mysql_num_row is counting table row
$count=mysqli_num_rows($result);
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$row = mysql_fetch_object($result);
$_SESSION["userID"] = $row->id;
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
echo $myusername;?>
<html>
<body>
</body>
</html>
这是出现的错误消息:
这是两个码我在我的user_area.php
<?php
session_start();
$tbl_members= "members";
$tbl_directorio= "subject_directorio";
$con=mysqli_connect("localhost","root","","bst_planner");
if(isset($_SESSION['myusername'])){
echo "You're logged in";
}
else {
header("location:index.php");
}
?>
这
<?php
$userID = $_SESSION["userID"];
$sql="SELECT * FROM $tbl_directorio WHERE Teacher_id='$userID'";
$result=mysqli_query($con,$sql)or die(mysql_error());
// Mysql_num_row is counting table row
$count=mysqli_num_rows($result);
if($count>=1){
echo "Your current planners";
}
else {
echo "You haven't created a planner yet";
}
?>
Catchable fatal error: Object of class mysqli_result could not be converted to string in C:\wamp\www\user_area.php on line 69
这是69行
$sql="SELECT * FROM $tbl_directorio WHERE Teacher_id='$userID'";
我真的很感谢您提供的任何帮助。
'session_start();'loaded?另外,将'或者死(mysql_error())'改为'或死(mysqli_error($ con))'''$ con'是你的数据库连接变量。你不能混合使用MySQL API。 – 2014-09-05 16:27:30
您还需要向我们展示更多代码。你发布的代码和错误消息一样,没有人能够给你一个明确的答案。投票结束为*“这个问题是由于无法再生产的问题引起的......”* – 2014-09-05 16:34:37
对于缺乏信息感到抱歉。我现在已经添加了更多的PHP代码。 – fmonper1 2014-09-05 22:41:46