PHP mysql fetch array error

Question

  Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in C:\xampp\htdocs\arun\login.php on line 9

Code:

$db_select = mysql_select_db('my_db', $con);
$username = $_GET['username'];
$password = $_GET['password'];
$role = $_GET['role'];
$result = mysql_query("Insert into  table1(Username,Password,Role) values (Username='$username', Password='$password',Role='$role') ",$con);
$row = mysql_fetch_array($result);

if ( mysql_error() )
 {
  die ( mysql_error());
}
$row_PK = mysql_fetch_assoc($row);
mysql_close($con);
?>

[Why shouldn't I use mysql_* functions in PHP?](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) — Itay, Feb 24 '14 at 05:47
did you googling how to insert data to database in php ? this is really effortless question. put some effort from your end then come to us — Awlad Liton, Feb 24 '14 at 05:49

Shankar Narayana Damodaran · Accepted Answer · 2014-02-26T04:55:36.750

Replace

$result = mysql_query("Insert into  table1(Username,Password,Role) values (Username='$username', Password='$password',Role='$role') ",$con);

with

$result = mysql_query("Insert into  table1(`Username`,`Password`,`Role`) values ('$username', '$password','$role') ",$con);

You don't have to assign the value to the column on your INSERT query as you have done. Just map the values as shown above.

Apart from the above issues... Dissecting your code ..

The main issue with your code is you are doing an INSERT but you are expecting a resultset as you have done a SQL SELECT which is wrong.

You need to change your code like this..

The below code is just an illustration to tell you what you had done was actually wrong. Do not use this code ! You first need to migrate to the latest database API provided for you in the side note section.

$db_select = mysql_select_db('my_db', $con);
if(isset($_GET['username'],$_GET['password'],$_GET['role']))
{
$username = mysql_real_escape_string($_GET['username']);
$password = mysql_real_escape_string($_GET['password']);
$role =     mysql_real_escape_string($_GET['role']);
$result = mysql_query("Insert into  table1(`Username`,`Password`,`Role`) values ('$username', '$password','$role') ",$con);
if (!$result)
{
    die (mysql_error());
}
else
{
    echo "Record Inserted";
}
mysql_close($con);
}

SideNote

The (mysql_*) extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead,the MySQLi or PDO_MySQL extension should be used. Switching to PreparedStatements is even more better to ward off SQL Injection attacks !

Guys I have a database named 'name' and 'my_db ' by using mysql_query But as what u guys suggested i have used mysqli_query,,,during the execution it says no database exists/ mysqli_select_db($con,"name");/ — Arun Shankar, Feb 24 '14 at 06:25
@downvoters and mods, care to explain what happened with this answer, why was it downvoted twice? why was comment thread deleted? — DevZer0, Feb 26 '14 at 04:45
please improve your answer and add sufficient escaping. then i will +1 it — DevZer0, Feb 26 '14 at 04:47

score 1 · Answer 2 · answered Feb 24 '14 at 05:50

1

For a INSERT query the return value isn't a resource, it's a TRUE on success or FALSE on error.

A resource is returned for a SELECT, SHOW, DESCRIBE or EXPLAIN query.

answered Feb 24 '14 at 05:50

user210584

111
2

score 0 · Answer 3 · answered Feb 24 '14 at 05:47

0

this is your solution change your query to

$result = mysql_query("Insert into table1(Username,Password,Role) values ('$username', '$password','$role') ",$con);

answered Feb 24 '14 at 05:47

Nishant Solanki

2,119
3
19
32

score -1 · Answer 4 · answered Feb 24 '14 at 05:48

-1

your query should be like this

$result = mysql_query("Insert into  table1(Username,Password,Role) values ('$username', '$password','$role') ",$con);

NOTE: mysql_* is deprecated use mysqli_* OR PDO

answered Feb 24 '14 at 05:48

Satish Sharma

9,547
6
29
51

2

Not my downvote, but your code neglects to escape the variables properly. – Ja͢ck Feb 24 '14 at 06:38

PHP mysql fetch array error

4 Answers4

The below code is just an illustration to tell you what you had done was actually wrong. Do not use this code ! You first need to migrate to the latest database API provided for you in the side note section.