2016-05-09 298 views
2

您好我創建一個動態圖像滑塊,但圖像不顯示。我可以在文件目錄中上傳圖片。我不知道如果數據庫是問題還是什麼是什麼問題。我是新來的HTML和CSS也PHP。有人可以給我想法是什麼造成的不顯示圖像?圖像不顯示在圖像滑塊

這裏是圖像沒有顯示,但我可以上傳文件到我的圖庫文件目錄。 enter image description here

這裏是我的數據庫sql。 enter image description here

這裏是我的php代碼。

<?php 
//for connecting db 
include('connect.php'); 
if (!isset($_FILES['image']['tmp_name'])) { 
echo ""; 
} 
else 
{ 
$file=$_FILES['image']['tmp_name']; 
$image= addslashes(file_get_contents($_FILES['image']['tmp_name'])); 
$image_name= addslashes($_FILES['image']['name']); 
move_uploaded_file($_FILES["image"]["tmp_name"],"gallery/" . $_FILES["image"]["name"]); 
$photo="gallery/" . $_FILES["image"]["name"]; 


$query = mysqli_query($mysqli, "INSERT INTO images(photo)VALUES('$photo')"); 
$result = $query; 

echo '<script type="text/javascript">alert("image successfully uploaded ");window.location=\'index.php\';</script>'; 
} 
?> 
<!DOCTYPE html> 
<html> 
    <head> 
     <link href="css/style.css" rel="stylesheet" /> 
     <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script> 
     <script src="js/slider.js"></script> 
     <script> 
     $(document).ready(function() { 
     $('.flexslider').flexslider({ 
     animation: 'fade', 
     controlsContainer: '.flexslider' 
     }); 
     }); 
     </script> 
    </head> 
    <body> 
    <div class="container"> 
    <form class="form" action="" method="POST" enctype="multipart/form-data"> 
     <div class="image"> 
      <p>Upload images and try your self </p> 
     <div class="col-sm-4"> 
       <input class="form-control" id="image" name="image" type="file" onchange='AlertFilesize();'/> 
       <input type="submit" value="image"/> 
      </div> 
     </div> 
     </form> 
     <div class="flexslider"> 
      <ul class="slides"> 
       <?php 
        // Creating query to fetch images from database. 
        $query = mysqli_query($mysqli, "SELECT * from images order by id desc limit 5"); 
        $result = $query; 
        while($r = mysqli_fetch_array($result)){ 
       ?> 
        <li> 
        <img src="<?php echo $r['photo'];?>" width="400px" height="300px"/> 
        </li> 
       <?php 
       } 
       ?> 
      </ul> 
     </div> 
    </div> 
    </body> 
</html> 

這裏是我的connect.php

<?php 
// hostname or ip of server 
$servername='localhost'; 
// username and password to log onto db server 
$dbusername='root'; 
$dbpassword=''; 
// name of database 
$dbname='pegasus'; 

////////////// Do not edit below///////// 
$mysqli = new mysqli($servername,$dbusername,$dbpassword,$dbname); 
if($mysqli->connect_errno){ 
    printf("Connect failed: %s\n", $mysql->connect_error); 
    exit(); 
} 

?> 
+0

圖像顯示你的列名是'image',你是取由列名'photo'記錄?這是什麼? – Apb

+0

嘿人。我如何將你的評論標記爲答案?有效。 – nethken

+0

現在接受它作爲回答 – Apb

回答

1

圖片顯示您的列名的圖像和你取由列名的照片記錄。

只要改變

<img src="<?php echo $r['photo'];?>" width="400px" height="300px"/> 

<img src="<?php echo $r['image'];?>" width="400px" height="300px"/>