PHP replace multiple MYSQL table value - php

Question

In case i need to replace table value with 500 rows contains random numbers. the problem is how to looping to get the ID 1 - 500, it is possible ?
<?php
$first = sprintf("%06d", mt_rand(1, 999999));
$second = sprintf("%06d", mt_rand(1, 999999));
$third = sprintf("%06d", mt_rand(1, 999999));
for($x = 1; $x <= 500; $x++) {
$sql = "UPDATE `random_num` SET
`rand_1` = '$first',
`rand_2` = '$second',
`rand_3` = '$third'
WHERE id = '$x';";
}
?>

Answer 1

whats wrong with your code?
just move the random number generation inside the for.. loop and u are done
<?php
function s_random()
{
return sprintf("%06d", mt_rand(1, 999999));
}
for($x = 1; $x <= 500; $x++) {
$sql = "UPDATE `random_num` SET
`rand_1` = '".s_random()."',
`rand_2` = '".s_random()."',
`rand_3` = '".s_random()."'
WHERE id = '".$x."'";
}
if you do it this way, that will be 500 calls to the DB and each row will have different random number
if you want the same random number in each row, just use the in keyword
"UPDATE `random_num` SET
`rand_1` = '".s_random()."',
`rand_2` = '".s_random()."',
`rand_3` = '".s_random()."'
WHERE id IN ('".implode(',',range(1,500))."'";
you can also create the query in one go
<?php
function s_random()
{
return sprintf("%06d", mt_rand(1, 999999));
}
$sql = "INSERT into random_num (id,rand_1,rand_2,rand_3) VALUES";
$val = [];
$i = 1; //begin ID
do{
$val[] = "($i, ".s_random().",".s_random().",".s_random().")";
$i++;
}while ($i < 500); //end ID
echo '<pre>';
$sql .= implode(',',$val);
$sql .= ' ON DUPLICATE KEY UPDATE rand_1 = VALUES(rand_1), rand_2 = VALUES(rand_2), rand_3 = VALUES(rand_3)';
echo ($sql);