Question

我在提交表单后没有从输入文件中获取任何数据，但是当我删除了try和catch语句时，其他输入能够插入到db中..我在这里做错了什么..

$author = $_SESSION['name'];
$product_image =addslashes(file_get_contents($_FILES['product_image']['tmp_name']));
$imageProperties = getimageSize($_FILES['product_image']['tmp_name']);

try
    {
    $mysqli->begin_Transaction();

    $insert_image = $mysqli->query
    (" 
        INSERT INTO product_image
        (
            author,
            regist_date,
            fileType,
            imageData
        )
        VALUES
        (
            '{$author}',
            now(),
            '{$imageProperties['mime']}',
            '{$product_image}'
        )"
    );


    $mysqli->commit();

}

catch (Exception $e) 
{
    $mysqli->rollback();
}

和我的表格

<form role="form" id="upload_form" method="post" enctype="multipart/form-data">
....
  <input name="product_image" type="file" class="inputFile" required >`
....
  <input type="submit" value="Submit" class="btnSubmit" />
</form>

Answer 1

理论上，如果你想将数据提交到同一个php页面，你的代码是有效的。但尝试通过向表单添加操作属性来执行此操作

<form role="form" id="upload_form" action="PageNameHere.php" method="post" enctype="multipart/form-data">

无法从输入文件图像类型中获取Post值

1 个答案: