Php fetch mysql result as variable

<?php
$result = mysql_query("SELECT id,email FROM people WHERE id = '42'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
?>

Source: stackoverflow.com

Similar codes

2 votes321 views1 answer
PhpMyAdmin is not able to cache templates
PHP
6 votes324 views1 answer
Laravel scaffolding
PHP
0 vote401 views1 answer
Laravel run php server by ipv4
PHP Laravel
35 votes328 views6 answers
Php current time
PHP
0 vote382 views1 answer
Login with email or username codeigniter 4
PHP CodeIgniter
1 vote448 views1 answer
How login with phone in laravel
PHP Laravel
18 votes308 views5 answers
If statement in laravel blade
PHP
4 votes455 views6 answers
How to remove index.php in codeigniter
PHP CodeIgniter
0 vote361 views1 answer
In Connection.php line 664: could not find driver (SQL: select * from information_schema.tables where table_schema = news and table_name = migrations) In Connector.php line 67: could not find driver
PHP
2 votes360 views1 answer
Wordpress if is in category
PHP Wordpress