我有20多个链接，它们看起来像：
<div>
<a href='writer.php?id=1'>1st Writer</a>
<a href='writer.php?id=2'>2nd Writer</a>
<a href='writer.php?id=3'>3d Writer</a>
<a href='writer.php?id=4'>4th Writer</a>
<a href='writer.php?id=5'>5th Writer</a>
</div>如何从每个href='writer.php获取"id"？id="并将其传递给$id writer.php？
writer.php:
<?$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test.com";
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * FROM articles WHERE id='$id'";
$result3 = mysqli_query($conn, $sql);
if (mysqli_num_rows($result3) > 0) {
while ($row = mysqli_fetch_assoc($result3)) {
echo $row[text];
}
}
mysqli_close($conn);?>我只想点击链接并在新页面（writter.php）上显示我来自数据库的数据，其中id=href中的数字。但是我不知道如何…
writer.php？id=1
这可以通过php代码中的$_GET['id']访问
$id = mysqli_real_escape_string($_GET['id']);
然后根据这个id 编写您的查询
$sql = "SELECT * FROM articles WHERE id='$id'";
$result3 = mysqli_query($conn, $sql);
我建议用面向对象的方式来尝试mysqli，而不是用程序化的方式
$con = new mysqli("host","user","password","database");
$id = $con->real_escape_string($_GET['id']);
$result3 = $con->query("SELECT * FROM articles WHERE id='$id'");
$id = htmlspecialchars((int)$_GET['id']);
$sql = "SELECT * FROM articles WHERE id='$id'";
http://phpform.net/dynamic.php