Question

我有以下PHP，我的问题是;如何使href根据值生成自己 $行[＆＃39;类别＆＃39;] 如果有任何不清楚的地方，请告诉我。

<?php
    echo "<div class=News>";
$q = "SELECT * FROM post ORDER BY newsno DESC";
$r = mysqli_query($mysqli, "$q");
if($r)
{

while($row=mysqli_fetch_array($r)){
    echo "<form action=index.php method=post>";
    echo "<div class=Container>";
    echo "<div class=Newsheader>";
    echo "<h7>";
    echo $row['newstopic'];
    echo "</h7>";
    echo "</div>";
    echo "<a href=image.php?newsno=".$row['newsno']." data-lightbox=roadtrip><img src=image.php?newsno=".$row['newsno']." width=180 height=180/></a>";
    echo "<div class=tb>";
    echo nl2br(substr ($row['newsinfo'], 0, 50));
    echo "<a href='This/Is/My/problem"."'> ...Read more</a>";
    echo "</div>";
    echo "</div>";
    echo "<div class=newsfooter>";
    echo 'Category: ' . $row['category']; 
    echo "</div>";
    echo "</br>";
    echo "</br>";
    echo "</form>";
}
echo "</div>";

}

else
{
echo mysqli_error();
}

?>

Answer 1

如果您尝试根据类别制作动态网址，而不是放置

echo '<a href="https://www.vg.no/'.$row['category'].'"> ...Read more</a>';

根据MySQL表行中的值生成HREF链接

1 个答案: