查找表的最大id并在别处输入

Question

因此，我正在尝试获取一个表的最大id，我可以使用

SELECT * FROM forsale ORDER BY StockID DESC LIMIT 0,1

然后我保存结果，以便在显示图像时可以在另一个表中使用它作为参考。唯一的问题是，当我打印结果时，它显示了MAX id，但没有将其输入到表中？有谁有建议吗？在此之前，我已经将代码输入到forsale表中，然后我将获得该记录的ID。Here's what I get in the table，这是代码：

if(isset($_POST['add'])){

include '../Login-System/db.php';

$make = mysqli_real_escape_string($conn, $_POST['Make']);
$model = mysqli_real_escape_string($conn, $_POST['Model']);
$variant = mysqli_real_escape_string($conn, $_POST['Variant']);
$year = mysqli_real_escape_string($conn, $_POST['Year']);
$mileage = mysqli_real_escape_string($conn, $_POST['Mileage']);
$fuel = mysqli_real_escape_string($conn, $_POST['Fuel']);
$doors = mysqli_real_escape_string($conn, $_POST['Doors']);
$trans = mysqli_real_escape_string($conn, $_POST['transmission']);
$enginesize = mysqli_real_escape_string($conn, $_POST['Enginesize']);
$price = mysqli_real_escape_string($conn, $_POST['Price']);
$description = mysqli_real_escape_string($conn, $_POST['description']);

$makeupper = strtoupper($make);
$modelupper =strtoupper($model);
$variantupper =strtoupper($variant);

$sqlcarinsert = "INSERT INTO forsale (make, model, variant, year, mileage, fuel, doors, trans, enginesize, price, description) VALUES ('$makeupper','$modelupper','$variantupper','$year','$mileage','$fuel','$doors','$trans','$enginesize','$price','$description');";


//Image Upload

//Find next StockID
$sql = "SELECT * FROM forsale ORDER BY StockID DESC LIMIT 0, 1";
$result = mysqli_query($conn, $sql);
$stockIDtable = mysqli_fetch_assoc($result);    
$stockID = $stockIDtable['StockID'];

if(!empty($_FILES['files']['name'][0])){

    $files = $_FILES['files'];

    //File Extensions allowed
    $allowed = array('jpg', 'jpeg', 'png');

    foreach ($files['name'] as $position => $file_name) {
        $file_tmp = $files['tmp_name'] [$position];
        $file_size = $files['size'] [$position];
        $file_error = $files['error'] [$position];

        //Order
        $orderimg = $position;

        //Get file extension
        $FileExt = explode('.', $file_name);
        $endext = end($FileExt);
        $fileActualExt = strtolower($endext);

        if (in_array($fileActualExt, $allowed)) {
        //Checks for Errors in uploading
            if ($file_error === 0) {

                //New name to remove possibilities of duplicates
                $fileNameNew = uniqid('', true).".".$fileActualExt ;
                $FileDestination = '../Photos/forsale/'.$fileNameNew;
                $SQLDestination = 'Photos/forsale/'.$fileNameNew;
                //Upload to Designated folder with name
                move_uploaded_file($file_tmp, $FileDestination);
                //Insert into forsaleimg
                $sqlimginsert = "INSERT INTO forsaleimg (id, StockID, imgOrder, FileDestination) VALUES ('NULL', '$stockID', '$orderimg', '$SQLDestination');";

                mysqli_query($conn, $sqlimginsert);
                //echo "<pre>";
                //print_r($sqlimginsert);
                //echo "</pre>";    
                $orderimg++ ;
            } else {
                header("Location: ../salelist.php?upload=error");
                exit();
            }
        } else {
            header("Location: ../salelist.php?fucked");
            exit();

        }

    }   
}

mysqli_query($conn, $sqlcarinsert);
header("Location: ../salelist.php?added=".$make);
exit();


} else {
header("Location: ../salelist.php?add=notclicked");
exit();
}

Answer 1

使用以下查询：-

$sql = "select * from forsale
        where StockID = (select max(StockID) as 'StockID'
                         from forsale)
        order by StockID" ;