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);
?>

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
– Awlad Liton, Commented Feb 24, 2014 at 5:49

Shankar Narayana Damodaran · Accepted Answer · 2014-02-26 04:55:36Z

1

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 !

edited Feb 26, 2014 at 4:55

answered Feb 24, 2014 at 5:47

Shankar Narayana Damodaran

68.6k43 gold badges102 silver badges129 bronze badges

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5 Comments

Arun Shankar Over a year ago

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");/

DevZer0 Over a year ago

@downvoters and mods, care to explain what happened with this answer, why was it downvoted twice? why was comment thread deleted?

Shankar Narayana Damodaran Over a year ago

@DevZer0, That is because a lot drama happened ;)

DevZer0 Over a year ago

please improve your answer and add sufficient escaping. then i will +1 it

Shankar Narayana Damodaran Over a year ago

@DevZer0, Yeah did it.

user210584 · Accepted Answer · 2014-02-24 05:50:38Z

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, 2014 at 5:50

user210584

1112 bronze badges

Comments

Nishant Solanki · Accepted Answer · 2014-02-24 05:47:48Z

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, 2014 at 5:47

Nishant Solanki

2,1283 gold badges19 silver badges32 bronze badges

Comments

Satish Sharma · Accepted Answer · 2014-02-24 05:48:09Z

-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, 2014 at 5:48

Satish Sharma

9,6356 gold badges31 silver badges52 bronze badges

1 Comment

Ja͢ck Over a year ago

Not my downvote, but your code neglects to escape the variables properly.

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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.

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