PHP Resource ID error

Question

I want to retrieve or output data in the database but I kept on getting the error called "Resource ID".

Here is my code:

<?php 

$host="localhost";
$username="root";
$password ="123192";
$db_name = "customers";

//Connecting to your Host
mysql_connect("$host","$username","$password") or die("Failed To Connect The server");
//Selecting your Database
mysql_select_db("$db_name") or die("Failed To Select The DB");

$name = $_REQUEST['customerName'];

echo 'WELCOME! <b>'.$name.'</b> We hope that you\'ll Enjoy your stay ';

$sql="SELECT Name FROM `people` WHERE id =2 && Name = 'Kyel'";
$rs=mysql_query($sql);
echo "$rs";
?>

If I need improvement regarding my code kindly tell me.

alex · Accepted Answer · 2011-08-22 03:02:30Z

4

mysql_query() returns a resource. The to string (implicitly triggered by using echo to output it) of that is Resource ID # followed by the id.

A resource in PHP is only supposed to be used with other PHP functions. This includes but is not limited to file, curl, ftp handles, etc.

I could tell you to..

(a) use mysql_fetch_array() (or similar) or

(b) use PDO.

The latter is by far much better advice.

answered Aug 22, 2011 at 3:02

alex

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Comments

Len · Accepted Answer · 2011-08-22 03:06:09Z

0

Try this instead of the echo statement:

$array = mysql_fetch_assoc($rs);
var_dump ($array);

answered Aug 22, 2011 at 3:06

Len

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4 Comments

user759630 Over a year ago

it gives me an array, I only want the data itself so that I could compare it with the userinput . I'll get data from the database and then Compare it with the user input. is that even possible? I wanted to retrieve data so that I can compare it with user input.

Phil Over a year ago

@user So you're saying you want to get the data from the database and compare it with user input?

user759630 Over a year ago

Yes Precisely. that's the reason why I wanted to retrieve and get data from the database so that I can compare it with the user input

Len Over a year ago

while ($row = mysql_fetch_assoc($result)) { echo $row["Name"]; }

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