php code works wrong

Question

Well, something going wrong here. I type username and password which are exist in my DB. It should echo this string in such situations

need to activate

but it echo this

You need to reg

init.php

<?php
    //error_reporting(0);
    session_start();

    require 'dbconnect.php';//this works okay so i wouldn't post this file code
    require 'users.php';

    $errors = array();
?>

users.php

        <?php

        function user_exists($username){
            $username = mysql_real_escape_string($username);

            $query = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE 'username' = '$username'");

            if (!$query) {
                die('Could not query:' . mysql_error());
            }

            return (mysql_result($query, 0) == 1) ? true : false;

        }

        function user_active($username){

            $username = mysql_real_escape_string($username);

            $querytoo = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE 'username' = '$username' AND 'active' = 1");

            if (!$querytoo) {
                die('Could not query:' . mysql_error());
            }

            return (mysql_result($querytoo , 0) == 1) ? true : false;

        }
    ?>

login.php

<?php
include 'init.php';


if(empty($_POST) === false){
    $username = $_POST['username'];
    $password = $_POST['password'];

    if(empty($username) === true || empty($password) === true){
        $errors[]='You need to enter a username and password';
    } 
    elseif(user_exists($username) === false){
        $errors[]='You need to reg';
    } 
    elseif(user_active($username) === false){
        $errors[]='need to activate';
    } 
    else {
    //
    }

    print_r($errors);
}



?>

part of html

<form action="login.php" method = "post">
                    <ul id="login">
                        <li>
                            username:<br>
                            <input type="text" name="username" size="30" value=""/></li>
                        <li>password:<br>
                            <input type="password" name="password" size="30" value=""/></li>
                        <li><input type = "submit" value ="Log in"></li>
                        <li>
                            <a href="register.php"> Register</a>
                        </li>
                    </ul>

P.S. text of querys is works fine, i checked it in mysql. in php code when I type `` instead of '' here

SELECT COUNT('user_id') FROM users WHERE 'username' = '$username' AND 'active' = 1

appears

'Could not query:'

thing

and i tried elseif and else if things so i don't think that problems is there

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Robik
– Robik, Commented Oct 13, 2012 at 17:07

Bart Friederichs · Accepted Answer · 2012-10-13 17:08:21Z

6

Your SQL is wrong. You are comparing a static strings to your value, not a column name:

    SELECT COUNT('user_id') FROM users WHERE 'username' = '$username' AND 'active' = 1

should be

SELECT COUNT(user_id) FROM users WHERE username = '$username' AND active = 1

Note that I removed the ' from the column names. It is also valid to use backticks:

SELECT COUNT(`user_id`) FROM users WHERE `username` = '$username' AND `active` = 1

answered Oct 13, 2012 at 17:08

Bart Friederichs

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5 Comments

DanilGholtsman Over a year ago

It says >Could not query:Unknown column 'user_id' in 'field list'

DanilGholtsman Over a year ago

but column user_id is exist i checked it

Bart Friederichs Over a year ago

could you run SHOW CREATE TABLE users?

DanilGholtsman Over a year ago

weel, if i understood you right i typed echo mysql_query("SHOW CREATE TABLE users"); inside function user_exists and now it's says Resource id #6 Could not query:Unknown column 'user_id' in 'field list'

DanilGholtsman Over a year ago

oh damn guys it was smth with sql i just reebot it and now it works fine

Marc B · Accepted Answer · 2012-10-13 17:08:51Z

1

using ' around table/field names in a query turns them into strings, and they will not longer be treated as field/table names. That's why there's backticks, for escaping such field/table names that happen to be keywords. The ONLY time you HAVE to escape field/table names is if they're reserved words.

Which means that...

SELECT COUNT('user_id')
             ^-      ^-
FROM `users`
WHERE 'username' = '$username' AND 'active' = 1");
      ^--      ^--                 ^-     ^-

is completely wrong. You're counting a fixed string, you're comparing the provided username against a string whose value is username, and ditto for active.

answered Oct 13, 2012 at 17:08

Marc B

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Comments

Yanick Rochon · Accepted Answer · 2012-10-13 17:09:37Z

1

Your query to check the existence of a user is probably one of your problem :

$query = mysql_query("SELECT COUNT('user_id') FROM `users` WHERE 'username' = '$username'");

In that query, you are using single quotes on your column name. Which results in comparing two strings, and, unless you have a user name called username, it will always return 0. Escape your columns with an ` char.

$query = mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username'");

or don't use anything

$query = mysql_query("SELECT COUNT(user_id) FROM `users` WHERE username = '$username'");

answered Oct 13, 2012 at 17:09

Yanick Rochon

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3 Comments

DanilGholtsman Over a year ago

same here Could not query:Unknown column 'user_id' in 'field list'

Yanick Rochon Over a year ago

I just used what you wrote (user_id). However, I'm glad it worked now that you rebooted.

DanilGholtsman Over a year ago

thanks! and now is apache crushes gyazo.com/a3e65d30d251d5dd94608893c73d3c59 i think i should reinstall Denwer. again. oh.

imp · Accepted Answer · 2012-10-13 17:10:19Z

1

Have you remembered to register the username into the session, as well as get it afterwards?

$username = $_SESSION['username'];

answered Oct 13, 2012 at 17:10

imp

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1 Comment

imp Over a year ago

Well where are the $username variable coming from?

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