Question

我正在尝试使用图片上传数据库中的图片。它一直给我这个错误：

Array ( [0] => images (1).jpg 
        [1] => images (1).jpg 
        [2] => images (1).jpg [3] => 
        images (1).jpg )

注意：未定义的偏移量：第82行的C：\ xamppv2 \ htdocs \ fabriek \ invullen.php中的4

致命错误：未捕获异常'PDOException'，消息'SQLSTATE [HY093]：参数号无效：绑定变量数与C中的令牌数不匹配'：\ xamppv2 \ htdocs \ fabriek \ invullen.php：82 Stack trace：＃0 C：\ xamppv2 \ htdocs \ fabriek \ invullen.php（82）：在第82行的C：\ xamppv2 \ htdocs \ fabriek \ invullen.php中抛出的PDOStatement-＆gt;执行（数组）＃1 {main}

因此，我们的想法是创建一个管理员可以上传产品的网上商店。现在唯一不起作用的是将多个图片上传到数据库中。我在互联网上搜索了这个错误，但是所有的错误都是用不同的代码编写的，而且他们大多都在谈论“错误。有谁知道这可能是什么问题？

希望任何人都可以提供帮助，

托马斯

代码：

$db_server = "localhost";
$db_username = "root";
$db_password = "";
$db_database = "meubelfabriek";
$conn = new PDO("mysql:host=$db_server;dbname=$db_database", $db_username, $db_password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
if (isset($_POST['submit']))
{
    $tipe_file1 = $_FILES['image1']['type'];
    $tipe_file2 = $_FILES['image2']['type'];
    $tipe_file3 = $_FILES['image3']['type'];
    $tipe_file4 = $_FILES['image4']['type'];
    $prijs = $_POST['prijs'];
    $naam = $_POST['fname'];
    $beschrijving = $_POST['desc'];
    $maat1 = $_POST['maat1'];
    $maat2 = $_POST['maat2'];
    $maat3 = $_POST['maat3'];
    $maat4 = $_POST['maat4'];
    if ($tipe_file1 == "image/jpeg" || $tipe_file1 == "image/jpg"){
        $aMyUploads = array();
        foreach ($_FILES as $aFile) {
            if(0 === $aFile['error']){
                $newLocation = ''.$aFile['name'];
                if(0 === $aFile['error'] && (false !== move_uploaded_file($aFile['tmp_name'], $newLocation))){
                    $aMyUploads[] = $newLocation;
                } else {
                    $aMyUploads[] = '';
                }
            }
        }

        print_r($aMyUploads);
        $stmt = $conn->prepare("INSERT INTO `producten` (naam, beschrijving, prijs, maat1, maat2, maat3, maat4, image1, image2, image3, image4) VALUES
    (:naam, :beschrijving, :prijs,  :maat1, :maat2, :maat3, :maat4,'".$aMyUploads[0]."','".$aMyUploads[1]."','".$aMyUploads[2]."','".$aMyUploads[3]."',)");
        $stmt->execute(array(":naam"=>$naam, ":beschrijving"=>$beschrijving, ":prijs"=>$prijs, ":maat1"=>$maat1, ":maat2"=>$maat2, ":maat3"=>$maat3, ":maat4"=>$maat4,":image1"=>$aMyUploads[1], ":image2"=>$aMyUploads[2], ":image3"=>$aMyUploads[3], ":image4"=>$aMyUploads[4]));
    }else{
        echo "<script>alert('Failed to add data! Images must be inputed at last one and filetype JPG/JPEG')</script>
            <script>setTimeout(\"self.history.back();\",0000);</script>";
    }
}

Answer 1

您可以通过字符串连接添加一些列，但将值绑定为Parameter。那样不行。因此，对所有列使用参数：

  $stmt = $conn->prepare("INSERT INTO `producten` (naam, beschrijving, prijs, maat1, maat2, maat3, maat4, image1, image2, image3, image4) VALUES
(:naam, :beschrijving, :prijs,  :maat1, :maat2, :maat3, :maat4,:image1,: image2, :image3, :image4)");
    $stmt->execute(array(":naam"=>$naam, ":beschrijving"=>$beschrijving, ":prijs"=>$prijs, ":maat1"=>$maat1, ":maat2"=>$maat2, ":maat3"=>$maat3, ":maat4"=>$maat4,":image1"=>$aMyUploads[1], ":image2"=>$aMyUploads[2], ":image3"=>$aMyUploads[3], ":image4"=>$aMyUploads[4]));

这个PHP错误是什么？

1 个答案: