2013-10-15 38 views
1

我有一个HTML表在PHP链接从SQL变量

<?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'];   
     ?> 

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

链接的名字是叫“网址” SQL表

链接在每一行不同

我怎样才能得到这个工作?

+0

不起作用?你遇到了什么错误? – Pascamel

+0

我收到了一个空白页 –

+0

我先回答了这个问题...如何投票? – Gavin

回答

2

那是因为$url从来没有定义。 PHP应该给你一个错误并抱怨。你有警告吗?

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

谢谢,我之前试过这个脚本,但是我在点之前和之后有一个空格。感谢您的快速回答 –

0

它应该是...

$row['url'] 

$url