警告:mysql_fetch_assoc()期望参数1为资源,给定对象
问题描述:
Possible Duplicate:
PHP Error: mysql_fetch_array() expects parameter 1 to be resource, boolean given
我似乎无法弄清楚我做错了什么.因此,当我提交表单时,会收到警告错误和
I cant seem to figure out what I'am doing wrong. So when I submit my form I get Warning error and the
通知:未定义的变量:/Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php在第30行上的dbusername
Notice: Undefined variable: dbusername in /Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php on line 30
$username = $_POST['username'];
$password = $_POST['password'];
if($username&&$password)
{
require 'conn.php';
$query = "SELECT * FROM users WHERE username='$username'";
$result = $mysql->query($query) or die(mysqli_error($mysql));
$numrows = $result->num_rows;
if ($numrows!=0)
{
while($row = mysql_fetch_assoc($result))
{
$dbusername = $row['username'];
$dbpassword = $row['password'];
}
//check to see if they match!
if($username==$dbusername&&$password==$dbpassword)
{
echo "youre In!";
}
else
echo "incorrect password!";
}
else
die("that user is dead");
//echo $numrows;
}
else
echo ("Please Enter Username")
我可能做错了什么?
答
更改
while($row = mysql_fetch_assoc($result))
到
while($row = $result->fetch_assoc())
您错过了函数名称的i
,并且混合了OO和过程样式代码,因此您正在混合mysql_*
结果资源和mysqli_result
对象.
You missed an i
off the function name, and mixed up OO and procedural style code, so you are mixing up mysql_*
result resources and mysqli_result
objects.