Php 无法从数据库获取链接的URL

我正在做一个项目，并将永远感谢帮助我的URl的链接。我试着四处看看，但没有结果。我有一个数据库（4列）。最后一个链接（link1）应该链接到具有指定URL的视频。当表格出现时，URL不可单击（有没有办法简化此操作，比如“单击我”？）。这是我的密码。我还附上了一张桌子的图片。这真是伤了我的脑筋，谢谢

<?php
    $con = mysqli_connect("localhost","feedb933_charles","pass100","feedb933_test");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql = "SELECT * FROM videos";
    $result = mysqli_query($con, $sql); 

    echo "<table>"; 
    echo "<tr> 
    <th>topic1</th> 
    <th>subject1</th> 
    <th>link1</th> 
    </tr>"; 
    while( $row = mysqli_fetch_array( $result)) { 
        $topic1 = $row["topic1"]; 
        $subject1 = $row["subject1"]; 
        $link1 = $row["link1"]; 

        echo "<tr> 
        <td>$topic1</td>
        <td>$subject1</td>
        <td>$link1</td>
        </tr>";
    }
    echo "</table>";
    mysqli_close($con);
?>

试试这个：

<?php
    $sql = "SELECT * FROM `videos`";
    $result = mysqli_query($con, $sql);
?> 
<table>
<?php 
   while($row = mysqli_fetch_assoc( $result)) { 
 ?>       
    <tr> 
        <td><?php echo $row['topic1'];?></td>
        <td><?Php echo $row['subject1'];?></td>
        <td><a href="<?php echo $row['link1']; ?>" target="_blank">Click me</td>
  </tr>
  <?php } ?>
<table>

---

我添加了

target

属性以在新窗口中打开链接。

我查看了您的代码，发现了几个错误

更改

$con=mysqli_connect（“本地主机”、“feedb933_测试”、“pass100”、“feedb933_测试”）

至

$con=new mysqli（“本地主机”、“feedb933\u测试”、“pass100”、“feedb933\u测试”）

然后将
if（mysqli\u connect\u errno（））
更改为
if（mysqli\u connect\u error（））{

然后改变

$sql = "SELECT * FROM videos";

到

或者如果要选择一行

$differentvalue = ""; // value to run
$sql = "SELECT topic1, subject1, link1 FROM videos WHERE difvalue = ?";

difvalue
是与其他值不同的值，因此php代码知道要获取什么

使用stmt意味着没有sql注入

加：

然后在side
if（something）{}else{in HERE}

（如果您有
其中diffvalue
）添加：

然后

加：

然后在
if（something）{code}else{code}

加：

echo“主题1主题1链接1”；
foreach（$列为$行=>$new）{
$html='.$new['topic1'].'；
$html.=''.$new['subject1'].'；
$html.='.$new['link1'].'；
echo$html；
}
回声“；

如果您仍然有问题，请转到列出的链接




希望这有帮助

<?php
$con=mysqli_connect("localhost","feedb933_charles","pass100","feedb933_test");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql = "SELECT * FROM videos";
    $result = mysqli_query($con, $sql); 

    echo "<table>"; 
    echo "<tr> 
    <th>topic1</th> 
    <th>subject1</th> 
    <th>link1</th> 
    </tr>"; 
    while( $row = mysqli_fetch_array( $result)) { 
        $topic1 = $row["topic1"]; 
        $subject1 = $row["subject1"]; 
        $link1 = $row["link1"]; 

        echo "<tr> 
        <td>$topic1</td>
        <td>$subject1</td>
        <a href="'.$link.'"><td>$link1</td></a>
*//this href will give u the link as u asked. //be sure you store proper url. In case of exact video name saved in column thn can make the url for your web output. //ex:<a href="http://example.com/'.$link.'.extension">*

        </tr>";
    }
    echo "</table>";
    mysqli_close($con);
?>

Hi Mawia。谢谢，但它不起作用。link1列中的所有数据都消失了。看这里，我更新了我的答案。检查一下。你在问题中只显示了三列。谢谢你Mawia！！它起作用了。你是个明星。一开始它不起作用（但可能是缓存问题）.有趣的是，我也尝试了Do版本，但没有成功（没有出现任何问题）。令人惊讶的是，整个代码也小得多。谢谢阿加恩塔克斯提供的目标信息，很好的一个。谢谢尼古拉斯的快速回复。我会看看这个，但现在表中的链接正在工作。好的。我花了一段时间把这个答案放在一起，但这个代码对我有效。谢谢德拉维亚的快速回复
$differentvalue = ""; // value to run
$sql = "SELECT topic1, subject1, link1 FROM videos WHERE difvalue = ?";

$stmt = $con->stmt_init();
if (!$stmt->prepare($sql))
{
    print "Failed to prepare statement\n";
}
else
{

}

$stmt->bind_param("s", $differentvalue); // $stmt->bind_param("s", $differentvalue); if text, $stmt->bind_param("i", $differentvalue); if integer

$stmt->execute();
$list = $stmt->fetchAll();

echo "<table><tr><th>topic1</th><th>subject1</th><th>link1</th></tr><tr>";

foreach ($list as $row => $new) {
    $html = '<td>' . $new['topic1'] . '</td>';
    $html .= '<td>' . $new['subject1'] . '</td>';
    $html .= '<td>' . $new['link1'] . '</td>';
    echo $html;
}

echo "</tr></table>";

<?php
$con=mysqli_connect("localhost","feedb933_charles","pass100","feedb933_test");

    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql = "SELECT * FROM videos";
    $result = mysqli_query($con, $sql); 

    echo "<table>"; 
    echo "<tr> 
    <th>topic1</th> 
    <th>subject1</th> 
    <th>link1</th> 
    </tr>"; 
    while( $row = mysqli_fetch_array( $result)) { 
        $topic1 = $row["topic1"]; 
        $subject1 = $row["subject1"]; 
        $link1 = $row["link1"]; 

        echo "<tr> 
        <td>$topic1</td>
        <td>$subject1</td>
        <a href="'.$link.'"><td>$link1</td></a>
*//this href will give u the link as u asked. //be sure you store proper url. In case of exact video name saved in column thn can make the url for your web output. //ex:<a href="http://example.com/'.$link.'.extension">*

        </tr>";
    }
    echo "</table>";
    mysqli_close($con);
?>