2013-10-15 37 views
1

我有一個HTML表在PHP鏈接從SQL變量

<?php 
     $result = mysqli_query($con,"SELECT * FROM recepten ORDER BY datum DESC"); 



     echo "<table border='1' class='ms-list8-main'> 
     <tr> 
     <th class='ms-list8-top'>Link</th> 
     <th class='ms-list8-top'>Naam</th> 
     <th class='ms-list8-top'>Beschrijving</th> 
     <th class='ms-list8-top'>Datum</th> 

     </tr>"; 

     while($row = mysqli_fetch_array($result)) 
      { 
      echo "<tr>"; 
      echo "<td><a href='$url'>link</td>"; 
      echo "<td class='ms-list8-even'>" . $row['naam'] . "</td>"; 
      echo "<td class='ms-list8-even'>" . $row['beschrijving'] . "</td>"; 
      echo "<td class='ms-list8-even'>" . $row['datum'] . "</td>"; 
      echo "</tr>"; 
      } 
     echo "</table>"; 
     $url = $row['url'];   
     ?> 

鏈接echo "<td><a href='$url'>link</td>";不起作用

鏈接的名字是叫「網址」 SQL表

鏈接在每一行不同

我怎樣才能得到這個工作?

+0

不起作用?你遇到了什麼錯誤? – Pascamel

+0

我收到了一個空白頁 –

+0

我先回答了這個問題...如何投票? – Gavin

回答

2

那是因爲$url從來沒有定義。 PHP應該給你一個錯誤並抱怨。你有警告嗎?

你的意思是$row['url']

如:

while($row = mysqli_fetch_array($result)) 
{ 
    echo "<tr>"; 
    // I've manually concatenated for clarity. 
    echo "<td><a href='". $row['url'] ."'>link</td>"; 
    echo "<td class='ms-list8-even'>" . $row['naam'] . "</td>"; 
    echo "<td class='ms-list8-even'>" . $row['beschrijving'] . "</td>"; 
    echo "<td class='ms-list8-even'>" . $row['datum'] . "</td>"; 
    echo "</tr>"; 
} 
+0

謝謝,我之前試過這個腳本,但是我在點之前和之後有一個空格。感謝您的快速回答 –

0

它應該是...

$row['url'] 

$url