PHP 尝试使用准备好的语句将数据插入到两个不同的表中

Question

我已经为此工作了几天，似乎找不到我哪里出错了，我想这有些愚蠢，但因为我的大学导师在他很少之前从未使用过准备好的陈述没有用。

第一个语句没有问题，第二个语句没有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，因为我确定这不是问题）并获取 PictureID 这是我图片中的主键 table 并将此作为其他信息插入我的图片价格 table.

欢迎任何帮助，我对这个网站还很陌生，所以请多多关照:)

<?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在尝试从 INSERT 语句中获取带照片的 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）。

要获得一个自动增量值，您通常会调用...

$photoID = mysqli_insert_id($conn);

PHP 尝试使用准备好的语句将数据插入到两个不同的表中

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

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prepared-statement