Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 9168901
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 17, 20262026-06-17T15:44:33+00:00 2026-06-17T15:44:33+00:00

I have the following PHP code: $con = mysql_connect(localhost,name,pass) or die(mysql_error()); $db = db;

  • 0

I have the following PHP code:

$con = mysql_connect("localhost","name","pass") or die(mysql_error());

$db = "db";

mysql_select_db($db,$con);

Now in my experience, $con should be true or false. When I echo $con I get:

Resource id #25

If I do the following code, the echo never fires (as to be expected after the above statement):

if($con) { echo "it worked"; }

When I run a query against this connection, everything works as expected. Is there a reason why this $con will not be true or false?

What am I doing wrong?

Thanks

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-17T15:44:34+00:00Added an answer on June 17, 2026 at 3:44 pm

    Check mysql_connect Return Values :

    Returns a MySQL link identifier on success or FALSE on failure.
    

    So to check the connection:

    if($con !== false) { echo "it worked"; }
    

    or to quit in case of an error:

    if (!$con) {
        die('Could not connect: ' . mysql_error());
    }
    

    A word of advice though, better to use the MySQLi or PDO_MySQL instead of mysql_connectsince it will soon be deprecated:

    Warning

    This 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. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:

    mysqli_connect()

    PDO::__construct()

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have the following PHP code: class SQLStatements { public function __construct($iName) { $this->name
I have the following PHP code: echo ' <td id='.$metaso['semanaventa'].'td1><form id='.$metaso['semanaventa'].'form name='.$metaso['semanaventa'].'form class=dinamic action=compromiso_funciones.php
I have the following php code : $db_host = localhost; $db_user = root; $db_pass
I have the following php code: $seasons = array(Autumn, Winter, Spring, Summer); Now I
I have the following PHP code which should load the data from a CSS
I have the following php code if (!($stream = ssh2_exec($con, 'sed -i \'s/motd=A Minecraft
I have following PHP code $val=<div id=user.$row['cid']. userid=.$row['cid']. class=innertxt><img src=images/images.jpg width=50 height=50><strong>.$uname.</strong><ul> <li>Email: .$row['cemail'].</li>
I have the following PHP code, and for the life of me I can't
I have the following php code: $k=1; for($i=0; $i < 5; $i++) { $stmt
I have the following php code: $k=1; for($i=0; $i < 5; $i++) { $stmt

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.