2013-05-10 38 views
-1
<?php 

$id = (int)$_GET['id']; 

$uyeler = array(
    1 => 'Ahmet', 
    2 => 'Mehmet', 
    3 => 'Ali', 
    4 => 'Veli'); 



echo '<select name="id">'; 
foreach($uyeler as $k=>$v){ 
echo '<option value="'.$k.'" '.($k==$id?' selected="selected"':'').'>'.$v.'</option>'; 

} 
echo '</select>'; 
?> 

我的代碼這個,我的問題如何動態地將數據id和名字寫入數組代碼? 1,2,3,4靜態數據id veli,ali,ahmet satatic數據名稱。我想要轉動我的數據庫表名etkinlikler動態系統。我的專欄ID和名稱。特別感謝你寫的答案...如何從mysql動態數據中排列代碼結果

+0

什麼是你的問題和/或問題? – SOfanatic 2013-05-10 15:52:43

+0

我在數組代碼中動態數據查詢的問題。我不能查詢動態ID和名稱mysql數據庫 – 2013-05-10 19:42:21

回答

0

如果我明白你的問題,你想從數據庫中獲得選擇的數據和表名是etkinlikler有標識和名稱的列...

<?php 

$id = (int)$_GET['id']; 


//Connect to database.. change mysql_user with your username and mysql_password with your password 

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password'); 
if (!$link) { 
    die('Could not connect: ' . mysql_error()); 
} 

//select database.. change YOUR_DB with your database name 
mysql_select_db('YOUR_DB', $link); 

// run SQL Query and get result handle 
$result = mysql_query('SELECT id, name FROM etkinlikler'); 
if (!$result) { 
    die('Invalid query: ' . mysql_error()); 
} 


echo '<select name="id">'; 

// fetch data from result and show it 
while ($row = mysql_fetch_assoc($result)) { 
    echo '<option value="'.$row['id'].'" '.($row['id']==$id?' selected="selected"':'').'>'.$row['name'].'</option>'; 
} 
echo '</select>'; 

// Close Connection 
mysql_close($link); 

?> 
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謝謝你好這個代碼運行,非常特別感謝;) – 2013-05-13 07:11:42

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@ user2370613 ywc;)點擊答案被接受:D – 2013-05-13 11:06:46