PHP-循環按鈕和表單輸入

Question

我想創建公司列表（從sql）和要刪除的按鈕。 它看起來應該像：

 ------------------------------
 |name_firm1   'delete_button'|
 |name_firm2   'delete_button'|
 |name_firm3   'delete_button'|
 ------------------------------

我創建了表單（在這里我為您刪除了CSS元素）：

echo '<form method="post" name="remove_firm_to_targs" action="">';

         while($saf = mysql_fetch_assoc($show_added_firm)) 
         {
            $query = mysql_query ("SELECT id_firm, name_firm FROM firms WHERE id_firm ='".$saf['id_firm']."'");
            $q = mysql_fetch_assoc($query);

            <input type="hidden" name="id_firm[]" value="'.$q['id_firmy'].'" />
            <div> '.$q['name_firm'].' </div>
            <div><input type="submit" id="button" value="delete firm" name="remove_firm_to_targ"/></div></div>';
        }
echo '</form>';

我無法從隱藏的輸入中$ _post id_firm。 這是代碼：

if(isset($_POST['remove_firm_to_targ'])) {
    $id_targ = $_GET['id']; 

    $id_firmy = array();
    foreach ($_POST['id_firm'] as $idid)
    {
        $id_firm[] = array ('idid' => $idid);
    }

    mysql_query ("DELETE FROM firm_to_targ WHERE id_firm = '".$id_firm.' && id_targ = '".$id_targ."'");
}

我多次更改了此代碼，所以現在沒有意義了。

Answer 1

您的表單應如下所示：我將公司ID放入刪除按鈕值中。

while($saf = mysql_fetch_assoc($show_added_firm))
{
echo '<form method="post" name="remove_firm_to_targs" action="">';

    $query = mysql_query ("SELECT id_firm, name_firm FROM firms WHERE id_firm ='".$saf['id_firm']."'");
$q = mysql_fetch_assoc($query);

<div> '.$q['name_firm'].' </div>
<div><input type="submit" id="button" value="DELETE" name="remove_firm_to_targ"/></div></div>
    <input type="hidden" name="delete_this" value="'.$q['id_firmy'].'">
    ';

    echo '</form>';
}

因此，當您為公司提交按鈕時，它將刪除在刪除按鈕值上具有ID的公司：

if(isset($_POST['remove_firm_to_targ'])) {
$id_targ = $_GET['id'];

mysql_query ("DELETE FROM firm_to_targ WHERE id_firm = '".$_POST['delete_this'].' && id_targ = '".$id_targ."'");
}

Answer 2

while($saf = mysql_fetch_assoc($show_added_firm)) { 
    $query = mysql_query ("SELECT id_firm, 
             name_firm FROM firms WHERE id_firm ='".$saf['id_firm']."'"); 
        $q = mysql_fetch_assoc($query); 
<div> '.$q['name_firm'].' </div> 
<a href="delete.php?id=$q['name']</div>'; 
 } 
 echo '</form>'; 

在你的情況下