Problem in execution of query

Question

2.50/5 (2 votes)

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hi,
I have a problem here:
first, myquery not executed;
and i catch a errer like this

(Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\EasyPHP-5.3.5.0\www\jadid\logincontrol.php on line 26 )

and in addition, my database is newsdb and tables are users and news and type of username and password are varchar.

Thanks

PHP

$user = $_POST['txtuser'];
$pass = $_POST['txtpassword'];
$con=mysql_connect('localhost','root','');
if(!$con)
{
    echo "Connection is Failed!!";
    exit;
}
mysql_select_db('newsdb');
$query=""select username from users where username='$user' and password='$pass'";
$allRows=mysql_query($query);
if($allRows)
{
    echo "Query Not Executed!!";
    exit;
}
echo $allRows;
$numRows=mysql_num_rows($allRows);
if($numRows)
{
    $_SESSION['user']=$user;
    echo "<center>Welcome ".$_SESSION['user']." to your site<br>";
    echo "<a href=user.php>Please click here on this link to load your page</a></center>";

Posted 12-Jun-12 3:04am

shiiiiir

Updated 12-Jun-12 5:40am

v7

Add a Solution

3 solutions

Solution 1

hi,
you have error in SQL query It self

you did not pass the table name in SLQ query

change query to :

SQL

select username from tablename  where username='@user' and password='@pass

Posted 12-Jun-12 3:13am

vyas_pratik20

Updated 12-Jun-12 7:27am

v2

Solution 3

The mysql_query() function returns FALSE on error. See http://php.net/manual/en/function.mysql-query.php[^]

So,

PHP

if($allRows)
{
    echo "Query Not Executed!!";
    exit;
}

"Query Not Executed!!" will be printed only when the query is actually executed without an error.

Try this to locate the error in your SQL statement:

PHP

$allRows=mysql_query($query) or die('MySQL Error'.mysql_error());

Posted 12-Jun-12 6:08am

krumia

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shiiiiir · Accepted Answer · 2012-06-12T06:01:00

Thank,
I've found my answer and i explain it:
first,i use:

var_dump($query);

and second, i use:

or die(mysql_error().": $query"

it's like

$allRows=mysql_query($query)or die(mysql_error().": $query");

now,it's my right code now:

PHP

session_start();
$user = $_POST['txtuser'];
$pass = $_POST['txtpassword'];
$con=mysql_connect('localhost','root','');
if(!$con)
{
    echo "Connection is Failed!!";
    exit;
}
mysql_select_db('newsdb',$con);
$query="SELECT * FROM users WHERE username='$user' AND password='$pass'";
var_dump($query);
$allRows=mysql_query($query)or die(mysql_error().": $query");

if(!$allRows)
{
    echo "Query Not Executed!!";
    exit;
}
echo $allRows;
$numRows=mysql_num_rows($allRows);
if($numRows)
{
    $_SESSION['user']=$user;
    echo "<center>Welcome ".$_SESSION['user']." to your site<br>";
    echo "<a href=user.php>Please click here on this link to load your page</a></center>";
}

Well,i use this codes and i found my problem and then i solved it.
Thanks again.