How do I update multiple tables using prepared statements with mySQLi? - php

Question

I have a form with two fields with the name attribute of 'photo_title' and 'photographer_name', and a hidden field named 'photo_id'. When the user pushes the submit button, i want it to update two separate tables in the database. I can get it to update a single table, but as soon as I try to leftjoin the second table, it doesn't like it.
I think there may be something wrong with my query string or the binding. How can I update two separate values in two separate tables in my Mysql database while still using prepared statements?
Here's the PHP:
if (array_key_exists('update', $_POST)) {
$sql = 'UPDATE photos SET photos.photo_title = ?, photographers.photographer_name = ?
LEFT JOIN photographers ON photos.photographer_id = photographers.photographer_id
WHERE photo_id = ?';
$stmt = $conn->stmt_init();
if ($stmt->prepare($sql)) {
$stmt->bind_param('ssi', $_POST['photo_title'], $_POST['photographer_name'], $_POST['photo_id']);
$done = $stmt->execute();
}
}
Here's the form:
<form id="form1" name="form1" method="post" action="">
<input name="photo_title" type="text" value=""/>
<textarea name="photographer_name"></textarea>
<input type="submit" name="update" value="Update entry" />
<input name="photo_id" type="hidden" value="<?php echo $photo_id ?>"/>
</form>

Answer 1

Here's an answer so folks who read this question see it, instead of finding it in your comment above. I'll mark this CW so I don't get any points for it.
UPDATE photos LEFT JOIN photographers
ON photos.photographer_id = photographers.photographer_id
SET photos.photo_title = ?, photographers.photographer_name = ?
WHERE photos.photo_id = ?
FWIW, the documentation for MySQL's UPDATE syntax is illustrative.

Answer 2

I was working on something similar. Here is a few things that I did. Hope it helps
if (isset($_POST['update'])) {
$id=intval($_GET['photo_id']);
$photo_title=$_POST['photo_title'];
$photographer_name=$_POST['photographer_name'];
$sql = "update photos p, photographers pg set p.photo_title=:photo_title, pg.photographer_name=:photographer_name where p.photographer_id=$id and pg.photographer_id=$id";
$query = $dbh->prepare($sql);
$query->bindParam(':photo_title',$photo_title,PDO::PARAM_STR);
$query->bindParam(':photographer_name',$photographer_name,PDO::PARAM_STR);
$query->execute();
}
You can even add your photo_id to the form action in case you want to use the id on a different page.
<form id="form1" name="form1" method="post" action="index.php?id=<?php echo $photo_id;?>">
<input name="photo_title" type="text" value=""/>
<textarea name="photographer_name"></textarea>
<input type="submit" name="update" value="Update entry" />
</form>
I have a file I created that connects to the database that I named config that has the following codes. Include it with this code where your form is at the top so you don't get errors executing the code above.
<?php
// DB credentials.
define('DB_HOST','localhost');
define('DB_USER','root');
define('DB_PASS','');
define('DB_NAME','photographer');
// Establish database connection.
try{
$dbh = new PDO("mysql:host=".DB_HOST.";dbname=".DB_NAME,DB_USER,
DB_PASS,array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES 'utf8'"));
}
catch (PDOException $e){
exit("Error: " . $e->getMessage());}
?>