PHP Trying to insert data into two different tables using prepared statment - php

Question

I've been working on this for a few days now and can not seem to find where i am going wrong, I imagine its something silly but as my university tutor has never used prepared statements before he has been of little to no use.
The first statement works a treat with no problems, the second doesn't input any of my data into my database. My goal is to take the information passed through the form (which i can include didn't want to bombard with information as i'm sure that is not the problem)and take the PictureID which is the primary key in my pictures table and insert this aswel as the other inforamtion into my pictureprice table.
any help would be welcomed, I'm fairly new to the site so be gentle please:)
<?php
include_once "dbh.php";
if (empty($imageTitle) || empty($imageDesc)) {
header("Location:changes.php?upload=empty");
exit();
} else {
$sql = "SELECT * FROM pictures;";
$sqltwo = "SELECT * FROM pictureprice;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
header("Location: changes.php?sqlerror=failed");
exit();
} else { //Gallery order//
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$rowCount = mysqli_num_rows($result);
$setImageOrder = $rowCount + 1;
$sql = "INSERT INTO pictures (PhotographerID, PictureFolderPath,
imageDesc, imgFullNameGallery, orderGallery) VALUES (?, ?, ?, ?,
?);";
if (!mysqli_stmt_prepare($stmt, $sql)) {
header("Location: changes.php?sqlerror=failedtoinputdata");
exit();
} else {
mysqli_stmt_bind_param($stmt, "issss", $_SESSION['PhotographerID'], $fileDestination, $imageDesc, $imageFullName, $setImageOrder);
mysqli_stmt_execute($stmt);
move_uploaded_file($fileTempName, $fileDestination);
$result = mysqli_stmt_get_result($stmt);
$row = mysqli_fetch_assoc($result);
$photoID = $row["PictureID"]; //new
header("Location:changes.php?upload=success11");
}
$sqltwo = "INSERT INTO pictureprice
(PictureID, PictureSize, PictureSize2, PictureSize3, PictureSize4,
PicturePrice, PicturePrice2, PicturePrice3, PicturePrice4) VALUES (?,
?, ?, ?, ?, ?, ?, ?, ?);";
if (!mysqli_stmt_prepare($stmt, $sqltwo)) {
header("Location: changes.php?
sqlerror=failedtoinputdatapictureprice");
exit();
} else {
mysqli_stmt_bind_param($stmt, "issssiiii", $photoID, $picturesize1, $picturesize2, $picturesize3, $picturesize4, $price1, $price2, $price3, $price4);
mysqli_stmt_execute($stmt);
header("Location:changes.php?upload=success");
}

Answer 1

I think the problem is that the you are trying to get the photo ID from an INSERT statement...
$result = mysqli_stmt_get_result($stmt);
$row = mysqli_fetch_assoc($result);
$photoID = $row["PictureID"]; //new
This probably won't fetching anything meaningful (as far as I can tell).
To get an auto increment value you would normally call...
$photoID = mysqli_insert_id($conn);