上传并显示图像php-mysqli


upload and display image php mysqli

我需要帮助我的显示图像,显示是正确的,但我不知道为什么图像没有出现。

CREATE TABLE `images` (
  `id` int(11) NOT NULL,
  `name` varchar(100) DEFAULT NULL,
  `size` int(11) DEFAULT NULL,
  `type` varchar(20) DEFAULT NULL,
  `content` mediumblob
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

这是我的"upload.php"

<?php
// Check for post data.
if ($_POST && !empty($_FILES)) {
    $formOk = true;
    //Assign Variables
    $path = $_FILES['image']['tmp_name'];
    $name = $_FILES['image']['name'];
    $size = $_FILES['image']['size'];
    $type = $_FILES['image']['type'];
    if ($_FILES['image']['error'] || !is_uploaded_file($path)) {
        $formOk = false;
        echo "Error: Error in uploading file. Please try again.";
    }
    //check file extension
    if ($formOk && !in_array($type, array('image/png', 'image/x-png', 'image/jpeg', 'image/pjpeg', 'image/gif'))) {
        $formOk = false;
        echo "Error: Unsupported file extension. Supported extensions are JPG / PNG.";
    }
    // check for file size.
    if ($formOk && filesize($path) > 500000) {
        $formOk = false;
        echo "Error: File size must be less than 500 KB.";
    }
    if ($formOk) {
        // read file contents
        $content = file_get_contents($path);
        //connect to mysql database
        if ($conn = mysqli_connect('localhost', 'root', '', 'upload')) {
            $content = mysqli_real_escape_string($conn, $content);
            $sql = "insert into images (name, size, type, content) values ('{$name}', '{$size}', '{$type}', '{$content}')";
            if (mysqli_query($conn, $sql)) {
                $uploadOk = true;
                $imageId = mysqli_insert_id($conn);
            } else {
                echo "Error: Could not save the data to mysql database. Please try again.";
            }
            mysqli_close($conn);
        } else {
            echo "Error: Could not connect to mysql database. Please try again.";
        }
    }
}
?>
<html>
    <head>
        <title>Upload image to mysql database.</title>
        <style type="text/css">
            img{
                margin: .2em;
                border: 1px solid #555;
                padding: .2em;
                vertical-align: top;
            }
        </style>
    </head>
    <body>
        <?php if (!empty($uploadOk)): ?>
            <div>
                <h3>Image Uploaded:</h3>
            </div>
            <div>
                <img src="image.php?id=<?=$imageId ?>" width="150px">
                <strong>Embed</strong>: <input size="25" value='<img src="image.php?id=<?=$imageId ?>">'>
            </div>
            <hr>
        <?php endif; ?>
        <form action="<?=$_SERVER['PHP_SELF']?>" method="post" enctype="multipart/form-data" >
          <div>
            <h3>Image Upload:</h3>
          </div>
          <div>
            <label>Image</label>
            <input type="hidden" name="MAX_FILE_SIZE" value="500000">
            <input type="file" name="image" />
            <input name="submit" type="submit" value="Upload">
          </div>
        </form>
    </body>
</html>

那么这就是我的"image.php"

<?php
    // verify request id.
    if (empty($_GET['id']) || !is_numeric($_GET['id'])) {
        echo 'A valid image file id is required to display the image file.';
        exit;
    }
    $imageId = $_GET['id'];
    //connect to mysql database
    if ($conn = mysqli_connect('localhost', 'root', '', 'upload')) {
        $content = mysqli_real_escape_string($conn, $content);
        $sql = "SELECT type, content FROM images where id = {$imageId}";
        if ($rs = mysqli_query($conn, $sql)) {
            $imageData = mysqli_fetch_array($rs, MYSQLI_ASSOC);
            mysqli_free_result($rs);
        } else {
            echo "Error: Could not get data from mysql database. Please try again.";
        }
        //close mysqli connection
        mysqli_close($conn);
    } else {
        echo "Error: Could not connect to mysql database. Please try again.";
    }   
    if (!empty($imageData)) {
        // show the image.
        header("Content-type: {$imageData['type']}");
        echo $imageData['content'];
    }
?>

每次我尝试上传图片的过程都是成功的,但我上传的图片没有出现。但这似乎只是图像错误。有谁能帮我吗?

或者你们中有人可以给我一个如何使用数据库上传和显示图像的教程吗?

    <?php
     error_reporting( ~E_NOTICE );
     require_once 'dbconfig.php';
 
     if(isset($_GET['edit_id']) && !empty($_GET['edit_id']))
      {
      $id = $_GET['edit_id'];
      $stmt_edit = $DB_con->prepare('SELECT userName, userProfession, userPic FROM 
        tbl_users 
         WHERE userID =:uid');
        $stmt_edit->execute(array(':uid'=>$id));
        $edit_row = $stmt_edit->fetch(PDO::FETCH_ASSOC);
        extract($edit_row);
      }
      else
       {
      header("Location: index.php");
      }
 
      if(isset($_POST['btn_save_updates']))
     {
      $username = $_POST['user_name'];// user name
      $userjob = $_POST['user_job'];// user email
   
      $imgFile = $_FILES['user_image']['name'];
      $tmp_dir = $_FILES['user_image']['tmp_name'];
      $imgSize = $_FILES['user_image']['size'];
     
      if($imgFile)
       {
         $upload_dir = 'user_images/'; // upload directory 
         $imgExt = strtolower(pathinfo($imgFile,PATHINFO_EXTENSION)); // get image extension
         $valid_extensions = array('jpeg', 'jpg', 'png', 'gif'); // valid extensions
         $userpic = rand(1000,1000000).".".$imgExt;
         if(in_array($imgExt, $valid_extensions))
         {   
          if($imgSize < 5000000)
         {
           unlink($upload_dir.$edit_row['userPic']);
          move_uploaded_file($tmp_dir,$upload_dir.$userpic);
       }
       else
       {
       $errMSG = "Sorry, your file is too large it should be less then 5MB";
      }
      }
      else
      {
      $errMSG = "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";  
      } 
     }
      else
      {
     // if no image selected the old image remain as it is.
     $userpic = $edit_row['userPic']; // old image from database
    } 
      
  
     // if no error occured, continue ....
     if(!isset($errMSG))
      {
        $stmt = $DB_con->prepare('UPDATE tbl_users 
              SET userName=:uname, 
               userProfession=:ujob, 
               userPic=:upic 
               WHERE userID=:uid');
              $stmt->bindParam(':uname',$username);
              $stmt->bindParam(':ujob',$userjob);
              $stmt->bindParam(':upic',$userpic);
              $stmt->bindParam(':uid',$id);
    
           if($stmt->execute()){
        ?>
                <script>
    alert('Successfully Updated ...');
    window.location.href='index.php';
    </script>
                <?php
   }
   else{
    $errMSG = "Sorry Data Could Not Updated !";
   }
  }    
 }
?>

您必须提取['content']一段时间。像这样:

    $rs = mysqli_query($conn, $sql);
    while ( $imageData = mysqli_fetch_assoc($rs) ) {
        $content = $imageData['content'];
    }
    header("Content-type: $content ");
    echo $content;

我会建议你检查并更改:

$sql = "SELECT type, content FROM images where id = {$imageId}"; 

对于

$sql = "SELECT type, content FROM images where id = '$imageId' ";