PHP的Mysql圖像顯示

Question

我正在研究一個php畫廊。我使用php顯示來自mysql數據庫的圖像，但我的圖像逐個顯示。這意味着第一行中的第一個圖像和第二行中的第二個圖像。但我想顯示我的圖像每行3或4。我可以做什麼編碼修改。我的php代碼如下所示。

<?php 
    include_once("config.php"); 
    $result=mysql_query("SELECT * FROM images"); 
    while($res=mysql_fetch_array($result)){ ?> 
     <table width='200'> 
      <tr> 
       <td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td> 
      </tr> 
      <tr> 
       <td> 
        <div id="news-image"> 
         <?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?> 
         <?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?> 
        </div> 
       </td> 
      </tr> 
     </table> 
<?php } ?> 

Answer 1

您正在輸出每張圖片的表格。至少，你的代碼應該是更多這樣的：

<table> 
<tr> 
<?php while($res etc...) { ?> 
<td> 
    <img src="<?php echo ......?>" /> 
</td> 
<?php } ?> 
</tr> 
</table> 

現在你會得到所有的圖像在一個表中的一行。讓它有多行是作爲OP的練習。

Answer 2

使用下面的代碼。

<?php 
include_once("config.php"); 
$result=mysql_query("SELECT * FROM images"); 
?> 
<table> 
<tr> 
<? 
$varcount=0; 
while($res=mysql_fetch_array($result)) 
{ 
$varcount++; 
if($varcount == 4) // Count of images per row. 3 or 4 
{ 
     $varcount=0; 
?> 
    </tr><tr> 
<? 
} 
?> 

<td> 
<table width='200'> 
<tr> 
<td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td> 
</tr> 
<tr> 
<td> 
<div id="news-image"> 
<?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?> 
<?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?> 
</div> 
</td> 
</tr> 
</table> 
</td> 
<?php 
} 
?> 
</tr> 
</table> 

Answer 3

array_chunk()是爲你拆分數組的X項目的集合，以循環的功能，通過無需保持計數器（那麼你可以使用array_pad（）的最後一個項目在列表中，如果你需要填充）

if($array = array_chunk(mysql_fetch_assoc($result),4)) 
{ 
    foreach($array as $row) 
    { 
    echo '<div class="row">'; 
    foreach($row as $col) 
    { 
     echo '<div class="item">' . $col['image'] . '</div>'; 
    } 
    echo '</div>'; 
    } 
} 

Answer 4

遵循這個例子

<table> 
<tr> 
<?php 
$i = 1; 
do{ 
    echo "<td>" . $i . "</td>"; 
    //Num of Columns 
    if($i%3 == 0){ 
     echo "</tr><tr>"; 
    } 

    $i++; 
}while($i<=10); 
?> 
</tr> 
<table> 

將返回喜歡你想要的結果東西..

Answer 5

<?php 
include_once("config.php"); 
$result=mysql_query("SELECT * FROM images"); 
?> 
<table> 
<tr> 
    <?php 
     $cnt = 0; 
     while($res=mysql_fetch_array($result)) 
     { 
     if($cnt == 3){ 
     echo "</tr><tr>"; 
     } 
      ?> 
     <td> 
    <table width='200'> 
     <tr> 
      <td><?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?><?php echo $res['imagename']?><?php echo"</a>"?></td> 
     </tr> 
     <tr> 
      <td> 
       <div id="news-image"> 
        <?php echo"<a href='indimage.php?imageid=$res[imageid]'>"?> 
        <?php echo'<img src='.$res['image'].' width="250" height="100">'?><?php echo"</a>"?> 
       </div> 
      </td> 
     </tr> 
    </table> 
</td> 
<?php 
$cnt++; 
    } 
?> 
</tr> 
</table> 

Answer 6

你可以代替輸出的圖像作爲div的列表，或者只是div的，然後用CSS來顯示兩列的圖像。你的佈局不應該那麼硬。

<style> 
    div.gallery { 
     width: 650px; 
    } 
    div.gallery ul li { 
     list-style: none; 
     float: left; 
    } 
    div.image { 
     height: 500px; 
     width: 300px; 
    } 
</style> 
<div class="gallery"> 
    <ul> 
     <li> 
      <div class="image"> 
       <span class="image_title">Some title</span><br/> 
       <img src="foo.png"/> 
      </div> 
     </li> 
     <li> 
      <div class="image"> 
       <span class="image_title">Another title</span><br/> 
       <img src="bar.png"/> 
      </div> 
     </li> 
     <li> 
      <div class="image"> 
       <span class="image_title">Another title</span><br/> 
       <img src="foo.png"/> 
      </div> 
     </li> 
     <li> 
      <div class="image"> 
       <span class="image_title">Another title</span><br/> 
       <img src="bar.png"/> 
      </div> 
     </li> 
     <li> 
      <div class="image"> 
       <span class="image_title">Another title</span><br/> 
       <img src="foo.png"/> 
      </div> 
     </li> 
    </ul> 
</div> 

結果：

你的代碼應該是這個樣子：

<div class="gallery"> 
    <ul> 

<?php 
include_once("config.php"); 
$result = mysql_query("SELECT * FROM images"); 
while($res = mysql_fetch_array($result)) { 
?> 
     <li> 
      <div class="image"> 
       <a class="image_title" href="indimage.php?imageid=<?php echo $res['imageid']?>"><?php echo $res['imagename']?></a><br/> 
       <a href="indimage.php?imageid=<?php echo $res['imageid']?>"><img src="<?php echo $res['image']?>" /></a> 
      </div> 
     </li> 
<?php 
} 
?> 
    </ul> 
</div> 

Answer 7

這裏是解決方案，我想你想的是： 有三頁。 1. index.php（具有上傳圖片的形式） 2. upload.php（將圖片保存在目錄及其在數據庫中的路徑） 3. showimage.php（最後會顯示圖片）

這裏是代碼 （的index.php）

<form method="post" action="upload.php" enctype="multipart/form-data"> 

<label>Choose File to Upload:</label><br /> 

<input type="hidden" name="id" /> 

<input type="file" name="uploadimage" /><br /> 
<input type="submit" value="upload" /> 

</form> 

（upload.php的）

<?php 
$target_Folder = "upload/"; // directory where images will be saved 



$target_Path = $target_Folder.basename($_FILES['uploadimage']['name']); 

$savepath = $target_Path.basename($_FILES['uploadimage']['name']); 

    $file_name = $_FILES['uploadimage']['name']; 

    if(file_exists('upload/'.$file_name)) 
{ 
    echo "That File Already Exisit"; 
    } 
    else 
    { 

     // Database 

     $con=mysqli_connect("localhost","user_name","pasword","database"); // Change it if required 
     //Check Connection 
     if(mysqli_connect_errno()) 
     { 
      echo "Failed to connect to database" .  mysqli_connect_errno(); 
     } 

     $sql = "INSERT INTO image (id,image, image_name) 
        VALUES ('$uid','$target_Folder$file_name','$file_name') "; 

     if (!mysqli_query($con,$sql)) 
     { 
      die('Error: ' . mysqli_error($con)); 
     } 
     echo "1 record added successfully in the database"; 
     echo '<br />'; 
     mysqli_close($con); 

     // Move the file into UPLOAD folder 

     move_uploaded_file($_FILES['uploadimage']['tmp_name'],  $target_Path); 

     echo "File Uploaded <br />"; 
     echo 'File Successfully Uploaded to:&nbsp;' . $target_Path; 
     echo '<br />'; 
     echo 'File Name:&nbsp;' . $_FILES['uploadimage']['name']; 
     echo'<br />'; 
     echo 'File Type:&nbsp;' . $_FILES['uploadimage']['type']; 
     echo'<br />'; 
     echo 'File Size:&nbsp;' . $_FILES['uploadimage']['size']; 

    } 
?> 

<a href="showimage.php">Show Image</a> 

（showimage。PHP）

<?php 


$con=mysqli_connect("localhost","user_name","password","database_name"); // Change it if required 

// Check connection 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$result = mysqli_query($con,"SELECT * FROM image "); 


while($row = mysqli_fetch_array($result)) 
{ 
echo '<img src="' . $row['image'] . '" width="200" />'; 
echo'<br /><br />'; 
} 


mysqli_close($con); 

?> 

特點

它會檢查文件名，如果該名稱的文件已經找不到，它將不會上傳下載文件，並提醒用戶。

數據庫結構

ID INT（4）自動遞增 - 圖像VARCHAR（100） - IMAGE_NAME VARCHAR（50）