顯示來自MySQL中PHP路徑的圖片

Question

我正處於開發招標系統的初期階段，用戶可以插入他們想要出售的物品。以下是從臨時路徑複製圖像並將圖像路徑存儲到另一個文件夾的腳本。

define ("MAX_SIZE","10000"); 

    //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. 
    function getExtension($str) { 
     $i = strrpos($str,"."); 
     if (!$i) { 
      return ""; 
     } 
     $l = strlen($str) - $i; 
     $ext = substr($str,$i+1,$l); 
     return $ext; 
    }; 

    //This variable is used as a flag. The value is initialized with 0 (meaning no error found) and it will be changed to 1 if an errro occures. If the error occures the file will not be uploaded. 
    $errors = 0; 

    //Checks if the form has been submitted 
    if (isset($_POST['Submit'])) 
    { 
     //Reads the name of the file the user submitted for uploading 
     $image=$_FILES['image']['name']; 

     //If it is not empty 
     if ($image) 
     { 
      //Get the original name of the file from the clients machine 
      $filename = stripslashes($_FILES['image']['name']); 

      //Get the extension of the file in a lower case format 
      $extension = getExtension($filename); 
      $extension = strtolower($extension); 

      //If it is not a known extension, we will suppose it is an error and will not upload the file, otherwize we will do more tests 
      if (($extension != "jpg") && 
       ($extension != "jpeg") && 
       ($extension != "png") && 
       ($extension != "gif")) 
      { 
       //Print error message 
       echo '<h1>Unknown extension!</h1>'; 
       $errors=1; 
      } 
     else 
     { 
      //Get the size of the image in bytes 
      //$_FILES['image']['tmp_name'] is the temporary filename of the file in which the uploaded file was stored on the server 
      $size=filesize($_FILES['image']['tmp_name']); 

      //Compare the size with the maxim size we defined and print error if bigger 
      if ($size > MAX_SIZE*111111111111024) 
      { 
       echo '<h1>You have exceeded the size limit!</h1>'; 
       $errors=1; 
      } 

      //We will give an unique name, for example the time in Unix time format 
      $image_name=time().'.'.$extension; 

      //The new name will be containing the full path where will be stored (images folder). 
      $imagepath='C:\\xampp\\htdocs\\biddingsystem\\Images\\' . $image_name; 

      //We verify if the image has been uploaded, and print an error instead 
      $copied = copy($_FILES['image']['tmp_name'], $imagepath); 
      if (!$copied) 
      { 
       echo '<h1>Picture upload failed!</h1>'; 
       $errors=1; 
      } 
     } 
    } 
} 

//If no errors registred, print the success message 
if(isset($_POST['Submit']) && !$errors && isset($_POST['image'])) 
{ 
    echo "<h1>Picture Uploaded Successfully! Try again!</h1>"; 
} 

然後，我插入與其他數據沿着圖像路徑MySQL數據庫與下面的腳本，並試圖顯示它們回到使用一個表中的用戶。其他數據工作得很好，但對於圖像顯示（圖像路徑），只顯示圖像的路徑，而不是圖像本身。

mysql_query("INSERT INTO items 
    (username, item, price, description, start_date, start_time, imagepath) 
    VALUES ('$username', '$_POST[item]', '$_POST[price]', '$_POST[description]','$_POST[start_date]', '$_POST[start_time]', '$imagepath') ") 
    or die ("Error - Couldn't add item"); 

    echo "Item added successfully"; 
    echo "<h1>You have added the following item:</h1>"; 
    $sql = "SELECT item, price, description, start_time, start_date, imagepath FROM items WHERE username = '$username' AND item='$_POST[item]'"; 
    $result = mysql_query($sql); 

$row = mysql_fetch_assoc($result); 
echo"<table border=2> 

     <tr><td>Item</td> 
      <td>Price</td><td>Description</td><td>Start time</td><td>Start date</td><td>Picture</td></tr> 

     <tr><td> $row[item]</td> 
       <td> $row[price]</td> 
       <td> $row[description]</td> 
      <td> $row[start_time]</td> 
      <td> $row[start_date]</td> 
      <td> $row[imagepath]</td> 
     </tr> 
    </table></br>"; 

我試過使用<img src="<?php $imagepath ?>">來顯示圖像，但無濟於事。我甚至嘗試使用BLOB類型將實際圖像本身存儲在數據庫中。然而，結果是一個充滿奇怪字符的頁面。我該如何解決這個問題？

Answer 1

你在混淆文件系統路徑和web URL。您只需要存儲/biddingsystem/Images/部分或甚至只顯示名稱並在顯示時間動態生成完整路徑。

另請注意，您沒有正確格式化數據，這會導致一些錯誤和安全漏洞。我在前面的答案中解釋了格式化規則，以Stack Overflow問題How to include a PHP variable inside a MySQL insert statement。

Answer 2

如果您想要使用第一種解決方案，請從路徑中進行顯示，請確保您指向的是可訪問路徑中的圖像。從您列出的代碼中，它看起來像$ imagepath是一個file：//路徑。在Web服務器上，您必須將其映射到相對的Web路徑。

如果您決定使用BLOB，最好的方法是創建一個單獨的頁面來提供圖像。您的輸出圖像標記應指向該標記，並且您需要更改圖像服務頁面標題中的內容類型。