PHP尝试使用准备好的语句将数据插入两个不同的表

Question

我已经为此工作了几天，似乎找不到哪里出错了，我想这很愚蠢，但由于我的大学导师在他几乎没有甚至没有之前从未使用过准备好的陈述采用。

第一个语句可以正常工作，第二个不会将我的任何数据输入到我的数据库中。我的目标是获取通过表单传递的信息（我可以包括不想用信息轰炸，因为我确信这不是问题）并获取作为我图片表中的主键的 PictureID 并插入这也是我的图片价格表中的其他信息。

欢迎任何帮助，我是该网站的新手，所以请温柔：)

<?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

我认为问题在于您正试图从 INSERT 语句中获取照片 ID...

    $result = mysqli_stmt_get_result($stmt);
    $row = mysqli_fetch_assoc($result);
    $photoID = $row["PictureID"];     //new

这可能不会获取任何有意义的东西（据我所知）。

要获得自动增量值，您通常会调用...

$photoID = mysqli_insert_id($conn);