How to update a row in sql with loop? - php

Question

I'm trying to figure how to call a query once.
I have 6 different variables for images, title and desc.
In this code, I need to know how to loop for id from 0 to 6.
$date = new DateTime("NOW");
$image1 = 'SSSS';
$title1 = 'AAAA';
$desc1 = 'BBBB';
$image2 = 'RRRR';
$title2 = 'GGGG';
$desc2 = 'VVVV';
/// 4 vars later....
$id = 6;
$get = $this->db->queryRow("UPDATE `featured` SET `image` = '{$image.$id}', `title` = '{$title.$id}', `desc` = '{$desc.$id}', `date` = '{$date->format('Y-m-d H:i:s')}' WHERE id = '{$id}'");
return(object) $get;

Answer 1

To build a collection of Querys use the multi_query function.
Loop to build your Query string to pass to the db and concatenated by a semicolon.
<?php
for($i=0;$i <= $maxquerys;$i++){
$query = "UPDATE `featured` SET `image` = '".$image.$id."', `title` = ".$title.$id."', `desc` = '".$desc.$id."', `date` = '".$date->format('Y-m-d H:i:s')."' WHERE id = '".$id."';"
}
/* execute multi query */
if ($mysqli->multi_query($query)) {
while ($mysqli->next_result());
}
/* close connection */
$mysqli->close();
you may check also the result by
echo $mysqli->affected_rows;
?>

Answer 2

I have tried to use a simple $query model and it works fine.
Create a valid Query string to pass to the db
<?php
$query = "UPDATE `featured` SET `image` = '".$image.$id."', `title` = ".$title.$id."', `desc` = '".$desc.$id}."', `date` = '".$date->format('Y-m-d H:i:s')."' WHERE id = '".$id."';"
$result=$mysqli->query($query);
// Verify results
if(!$result) {
$ErrMessage = "ErrSqlQuery:" . $mysqli->error . "\n";
$mysqli->close();
die($ErrMessage);
}
you can check also the result by
echo $mysqli->affected_rows;
?>

Answer 3

$query_build = "";
foreach($arr as $$image){
$query_build .= "UPDATE `featured` SET `image` = '{$image.$id}', `title` = '{$title.$id}', `desc` = '{$desc.$id}', `date` = '{$date->format('Y-m-d H:i:s')}' WHERE id = '{$id}';";
}
$get = $this->db->queryRow($query_build);
Accumulate all the queries and execute all at once.