Question

我使用下面的代码从mysql数据库上传和检索图像。图像已成功上载并移动到单独的文件夹，但图像路径未存储在数据库中，因此我无法从数据库中检索图像。请检查问题在哪里

if(isset($_POST["submit"])){
              $check = getimagesize($_FILES["image"]["tmp_name"]);
            if($check !== false){
              $target="images/".basename($_FILES["image"]["tmp_name"]);
              $image = $_FILES['image']['tmp_name'];
              $imgContent = addslashes(file_get_contents($image));
              $uploadfilename=$_FILES['image']['name'];
        /*
         * Insert image data into database
         */

       $db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);
        $dataTime = date("Y-m-d H:i:s");

        //Insert image content into database
          $insert = $db->query("INSERT into images (image,created)   VALUES('$imgContent', '$dataTime')");
      //move uploaded file to the folder images//


      if($uploadfilename!=''){
        move_uploaded_file($_FILES["image"]["tmp_name"],$target)
          $query="INSERT INTO images SET imgepath='$uploadfilename' ";
        mysqli_query($query);
        if($query){

           { 
            echo "File uploaded  successfully.";
           }
        }else{
            echo "File upload failed, please try again.";
        } 
    }else{
        echo "Please select an image file to upload.";
    }
    }
    }
   }
    ?>

Answer 1

您应该放置图片的文件夹路径或Break when this exception type is user-unhandled变量而不是$target。

$imgContent

这也很容易进行SQL注入，使用预处理语句：

"INSERT into images (image,created) VALUES('$target', '$dataTime')"

Answer 2

将图像名称保存在数据库中并在检索时将图像上传到服务器中，您可以使用文件夹名称后跟图像名称。例如，如果上传图像的文件夹名称是上传

＆＃13;

<img src="uploads/<?= $v1['image']?>" alt="product">

＆＃13;

Answer 3

感谢您的帮助。下面的代码正在运行。我创建了一个名为＆＃34; images＆＃34;使用colms＆＃34; id＆＃34;，＆＃34; name＆＃34;（我保存了图片的名称）和＆＃34; url＆＃34;

<?php

if(isset($_POST['submit']))
{ 
    $host = "...";
    $db_name = "....";
    $username = "....";
    $password = ".....";
    $link=mysqli_connect($host, $username, $password, $db_name);
    $filename=$_FILES['file']['name'];
    $filetmp=$_FILES['file']['tmp_name'];
    $target="uploaded/".$filename;
    $image_url="http://holo-com.stackstaging.com/uploaded".$filename;
    $sql="INSERT INTO images (name,image_url) VALUES  ('$filename','$image_url')";
    mysqli_query($link,$sql);
    if(move_uploaded_file($filetmp,$target))
    {   echo "Image uploaded";  }
}
else
{   echo "Failed to upload";    }

?>

从MySQL数据库中检索图像

3 个答案: