我正在做一个项目,并将永远感谢帮助我的URl链接。我试过四处寻找无济于事。我有一个数据库(4columns)。最后一个(link1)应该链接到具有指定URL的视频。当表格出现时,URL不可点击(有没有办法简化这个说“点击我”?)。这是我的代码。我还附上了桌子的图像。这真的是我的大脑,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);
?>
答案 0 :(得分:0)
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 $row['link1']; ?>" target="_blank">Click me</td>
</tr>
<?php } ?>
<table>
或者你也可以使用do while
循环:
do{
echo '<tr>';
echo '<td>'.$row['topic1'].'</td>';
echo '<td>'.$row['subject1'].'</td>';
echo '<td><a href="'.$row['link1'].'" target="_blank">Click me</td>';
echo '</tr>';
} while($row = mysqli_fetch_assoc( $result);
我添加了target
属性以在新窗口中打开链接。
答案 1 :(得分:0)
我查看了你的代码,发现了一些错误。
将$con = mysqli_connect("localhost","feedb933_charles","pass100","feedb933_test");
更改为$con = new mysqli("localhost", "feedb933_charles", "pass100", "feedb933_test");
然后将if (mysqli_connect_errno())
更改为if (mysqli_connect_error()) {
然后改变
$sql = "SELECT * FROM videos";
到
$sql = "SELECT topic1, subject1, link1 FROM videos";
或者如果您想选择一行
$differentvalue = ""; // value to run
$sql = "SELECT topic1, subject1, link1 FROM videos WHERE difvalue = ?";
difvalue
是其他值不同的值,因此php代码知道要抓取什么。
使用stmt意味着没有sql注入
添加:
$stmt = $con->stmt_init();
if (!$stmt->prepare($sql))
{
print "Failed to prepare statement\n";
}
else
{
}
然后在if (something) { } else { IN HERE }
边
(如果您有WHERE diffvalue
)添加:
$stmt->bind_param("s", $differentvalue); // $stmt->bind_param("s", $differentvalue); if text, $stmt->bind_param("i", $differentvalue); if integer
然后
添加:
$stmt->execute();
$list = $stmt->fetchAll();
然后在if (something) { code } else { code }
添加:
echo "<table><tr><th>topic1</th><th>subject1</th><th>link1</th></tr><tr>";
foreach ($list as $row => $new) {
$html = '<td>' . $new['topic1'] . '</td>';
$html .= '<td>' . $new['subject1'] . '</td>';
$html .= '<td>' . $new['link1'] . '</td>';
echo $html;
}
echo "</tr></table>";
如果您仍然遇到问题,请转到列出的链接
http://php.net/manual/en/pdostatement.fetchall.php
http://php.net/manual/en/mysqli-stmt.get-result.php
http://php.net/manual/en/mysqli-result.fetch-all.php
希望这有帮助
答案 2 :(得分:0)
<?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);
?>