mysql_num_rows() returns zero

Question

<form action = "index.php" method = "post">
username : <input type = "text" name = "uname" /><br>
password : <input type = "text" name = "pass" /><br>
submit : <input type = "submit" name = "submit" value = "submit" />
</form>

<?php
if(isset($_SESSION['id'])){echo $_SESSION['id'];}
if(isset($_POST['submit'])){
if ($_POST['submit'] == 'submit'){

$uname = $_POST['uname'];
$pass = $_POST['pass'];
$db = "davidedwardcakes";
$connect = mysql_connect('localhost', 'root', 'wtfiwwu');
$db_connect = mysql_selectdb($db, $connect);
if(!$db_connect){echo 'no';}

$query = "SELECT * FROM `users` WHERE uname ='$uname' AND pass = '$pass'";
$result = mysql_query($query, $connect);
if(mysql_num_rows($result) > 0){//echo 'index failed'; var_dump($result);}
while($row = mysql_fetch_array($result)){echo $row['uname']
 . "<br>";

session_start();
echo '<a href = "test.php">peruse</a>';
$_SESSION['id'] = $row['id'];}}

else{echo 'lol'; var_dump($query);}}

Whenever I want to login, i get the error: string 'SELECT * FROM users WHERE uname ='brown' AND pass = 'kenji'' (length=61)

meaning that theres a problem with my $query. If I remove the $pass query from $query it works fine but doesn't when it is included. Can anybody help please.

Huh? Why does that mean there's a problem with your query? That's your var_dump code. — BenM
– BenM, Commented May 14, 2014 at 10:51
1. Don't store passwords as plain text. 2. Don't use mysql_* functions; they're deprecated. 3. Your code is wide open to SQL injection... — BenM
– BenM, Commented May 14, 2014 at 10:51
If you really had a SQL syntax error (as the question title states) the query would not even run. — Álvaro González
– Álvaro González, Commented May 14, 2014 at 10:52

BenM · Accepted Answer · 2014-05-14 10:57:04Z

1

There's nothing wrong with your query, but there is something wrong with your error reporting. For example, the following line:

if(mysql_num_rows($result) > 0){//echo 'index failed'; var_dump($result);}

Will dump the $result variable when the SQL query returns at least 1 row. Perhaps you meant to use:

if(mysql_num_rows($result) < 1) { echo 'index failed'; var_dump($result); }

Also, some notes of caution:

Don't store passwords as plain text. http://alias.io/2010/01/store-passwords-safely-with-php-and-mysql/
Don't use mysql_* functions; they're deprecated. Look at PDO or MySQLi.
Your code is wide open to SQL injection.

answered May 14, 2014 at 10:57

BenM

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Comments

wild · Accepted Answer · 2014-05-14 10:52:09Z

1

change it

mysql_selectdb($db, $connect);

as

mysql_select_db($db, $connect);

answered May 14, 2014 at 10:52

wild

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3 Comments

wild Over a year ago

according your code connection not create so no query was run

BenM Over a year ago

Why? If his DB connection does not work, he'd get no: if(!$db_connect){echo 'no';}

wild Over a year ago

@BenM : You will try i think you will get no : statement. according to my knowlage

sathishkumar · Accepted Answer · 2014-05-14 11:02:14Z

0

There is a error in the syntax

$result = mysql_select($query, $connect);

Instead of the above code replace with this code

$result = mysql_select_db($query, $connect);

edited May 14, 2014 at 11:02

answered May 14, 2014 at 10:56

sathishkumar

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Comments

Alok Jha · Accepted Answer · 2014-05-14 11:07:09Z

0

If there is some error in your Syntax , Try to use

mysql_num_rows($result) or die('Could not Show result: ' . mysql_error());

answered May 14, 2014 at 11:07

Alok Jha

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