drj14664 2017-03-30 18:28
浏览 49

从数据库显示图像的破碎图像

Hello i am trying to retrieve image stored in database as BLOB however it is coming back as broken so the image doesnt display what am i doing wrong code below.

<?php  
$connect = mysqli_connect("localhost", "root", "", "testing");  
$query = "SELECT * FROM item ORDER BY item_id DESC";  
$result = mysqli_query($connect, $query);  

while($row = mysqli_fetch_array($result)){   

    echo '<tr>';
    echo '<td>';
    echo '<img src="data:image/jpeg;base64,'.base64_encode($row['photo'] ).'" height="100" width="100" class="img-thumnail" />';
    echo '</td>'; 
    echo '</tr>';  
}  
?> 
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1条回答 默认 最新

  • doudou130216 2017-03-30 19:20
    关注

    Use addslashes function while save BLOB image into database. Because apostrophe ' may broken your image, addslashes function replace ' with \' which solve the problem.

    For Inserting BLOB image into DataBase

    $connect = mysqli_connect("localhost", "root", "", "testing");
    $image = addslashes(file_get_contents($_FILES['images']['tmp_name']));
    $query = "INSERT INTO item (item_id,photo) VALUES('','$image')";  
    $result = mysqli_query($connect, $query);
    

    For Accessing BLOB image From DB

    $connect = mysqli_connect("localhost", "root", "", "testing");  
    $query = "SELECT * FROM item ORDER BY item_id DESC";  
    $result = mysqli_query($connect, $query);
    while($row = mysqli_fetch_array($result)){   
    
        echo '<tr>';
        echo '<td>';
        echo '<img src="data:image/jpeg;base64,'.base64_encode($row['photo'] ).'" height="100" width="100" class="img-thumnail" />';
        echo '</td>'; 
        echo '</tr>';  
    }
    
    评论

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