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

0 vote120 views1 answer
Wp_delete_attachment unlink
PHP
6 votes126 views5 answers
Jetstream laravel
PHP
53 votes272 views6 answers
Laravel where in
PHP Laravel
9 votes154 views4 answers
Pretty json php
PHP
0 vote132 views1 answer
Laravel override eloquent all function
PHP Laravel
5 votes129 views2 answers
Laravel collection chunk
PHP
4 votes134 views2 answers
Laravel eloquent get data without duplicates
PHP Laravel
0 vote125 views2 answers
PHP $argv echo with number of words
PHP
15 votes127 views6 answers
Laravel custom validation message
PHP Laravel
43 votes144 views2 answers
Php can't upload gif
PHP