“mysql_fetch_array（）期望参数1是资源”[重复]

Here is my code:

<?php

$inputuser = $_POST['username'];
$inputpass = $_POST['password'];

    $connect = mysql_connect("localhost", "root", "", "fapdatabase") or die("could not connect" . mysql_error());

    $query = mysql_query("SELECT * FROM user WHERE username = '$inputuser'");
    $querypass = mysql_query("SELECT * FROM user WHERE password = '$inputpass'");

    $result = mysql_query($query);
    $resultpass = mysql_query($querypass);

    $row = mysql_fetch_array($result);
    $rowpass = mysql_fetch_array($resultpass);

    $serveruser = $row["username"];
    $serverpass = $row["password"];

    if($serveruser&&$serverpass) {
        if(!$result) {
            die("Username or Password is INVALID");
        }

        echo "<br><center>Database Output</b></center><br><br>";
        mysql_close();

        echo $inputuser;
        echo $serverpass;

    if($inputpass == $serverpass) {
        header('Location: login/index.html');

    } else {
        echo "Sorry! Bad LOGIN!";
    }
    }

And the error is around line 14 and 15 (the one with fetch) Please help?

</div>

you should check $result and $resultpass.

if ($result) {
    $row = mysql_fetch_array($result);
}
if($resultpass){
    $rowpass = mysql_fetch_array($resultpass);
}

you can check whether your query is executing or not by using mysql_error() function if there is some error correct it

<?php

$inputuser = $_POST['username'];
$inputpass = $_POST['password'];

    $connect = mysql_connect("localhost", "root", "", "fapdatabase") or die("could not connect" . mysql_error());

    $query = mysql_query("SELECT * FROM user WHERE username = '$inputuser'") or die(mysql_error());
    $querypass = mysql_query("SELECT * FROM user WHERE password = '$inputpass'") or die(mysql_error());

    $result = mysql_query($query);
    $resultpass = mysql_query($querypass);

    $row = mysql_fetch_array($result);
    $rowpass = mysql_fetch_array($resultpass);

    $serveruser = $row["username"];
    $serverpass = $row["password"];

    if($serveruser&&$serverpass) {
        if(!$result) {
            die("Username or Password is INVALID");
        }

        echo "<br><center>Database Output</b></center><br><br>";
        mysql_close();

        echo $inputuser;
        echo $serverpass;

    if($inputpass == $serverpass) {
        header('Location: login/index.html');

    } else {
        echo "Sorry! Bad LOGIN!";
    }
    }

$conn = mysql_connect("localhost", "root", "", "fapdatabase") or die("could not connect" . mysql_error());

$inputuser = mysqli_real_escape_string($conn, $_POST['username']);
$inputpass = $_POST['password'];

$sql = 'SELECT pass FROM user WHERE username = \''.$inputuser.'\'';
$result = $conn->query($sql);

$result = mysql_query($query);
$count =$conn->affected_rows;// gives you count of returned rows

if($count==1)//since it should return only a single user
{
 $row=mysqli_fetch_row($result);
 if($row[0]==$inputpass)echo 'User Found';
}
else echo "Username or Password is INVALID";

look up mysqli_real_escape_string and why it is used. Also look up mysqli_fetch_row.

this should get you started. Note I used column names in sql. change them to the names in your mysql table. and as Jon noted, mysql is depreciated. Use mysqli. Don't worry you'll get used to it soon.