2013-07-25 75 views
-2

我有一個窗體,顯示圖像誰的路徑存儲在MySQL中。該表單爲每個圖像提供「隱藏」和「顯示」提交按鈕。按下隱藏按鈕應將「狀態」字段的值設置爲2(隱藏)。但它什麼都不做,我的查詢似乎沒有執行。Mysql查詢似乎沒有工作

這裏是形式

<form class="removeform"action='headerimageadd.php' method='post' enctype='multipart/form-data'  
name='image_remove_form' > 
    <?php 
    include '../inc/connect.php'; 
    $q = "SELECT * FROM headerrotatorimage WHERE rotator = 1"; 
    $result = $link->query($q); 
    while($row=mysqli_fetch_array($result)){ 
     echo "<input type='submit' name='hide[{$row['id']}]' value='Hide'>", 
      "<input type='submit' name='show[{$row['id']}]' value='Show'>", 
      "<br />", 
      "<img src='{$row['filename']}' alt='{$row['name']}' />", 
      "<br />";       
     } 
     ?> 
     </form> 

這裏是當隱藏按鈕被按下

<?php 
    include '../inc/connect.php'; 
    if(isset($_POST['hide'])){ 
     $chk = (array) $_POST['hide']; 
     $p = implode(',',array_keys($chk)); 
     echo $p; 
     $t = "SELECT * FROM headerrotatorimage WHERE id IN ($p)"; 
     echo $t; 
     $s = "UPDATE headerrotatorimage SET status = 2 WHERE id IN ($p)"; 
     echo $s; 
    } 
    ?> 

誰能幫助時執行的PHP? 謝謝。

+5

我沒有看到你執行你的選擇更新查詢..,我看你呼應出來,但不使用它們。 –

+2

我總是看到這種類型的問題,您應該閱讀[用於PHP和MySQL的常見數據庫調試](http://jason.pureconcepts.net/2013/04/common-debugging-php-mysql/)。 –

回答

0

你沒有實際發行代碼中的查詢:

<?php 
include '../inc/connect.php'; 
if(isset($_POST['hide'])){ 
    $chk = (array) $_POST['hide']; 
    $p = implode(',',array_keys($chk)); 
    echo $p; 
    $t = "SELECT * FROM headerrotatorimage WHERE id IN ($p)"; 
    echo $t; 

    // Execute query and process result set for $t 

    $s = "UPDATE headerrotatorimage SET status = 2 WHERE id IN ($p)"; 
    echo $s; 

    // Execute query and process result set for $s 
} 
?>