如何从url中获取url并将其与php中的给定表进行比较

您正在PHP中打开无效的PHP标记，只需执行以下操作：

<button type="button" class="btn btn-primary">
    <?php echo $_GET['url']; ?>
</button>

<?php 
    $goturl=$_GET['url'];

    $con=mysqli_connect("localhost","stud_user","123456","stud_db");

    if (mysqli_connect_errno()){
        echo "Oops! We ran into some problem!" . mysqli_connect_error();
    }

    $result = mysqli_query($con,"SELECT * FROM users WHERE url='<?=$goturl;?>'");

    while($row = mysqli_fetch_array($result)){
        echo $row['verification_value'];
    }

    if($verification_value == 'UNVERIFIED'){ 
        $val = 'UNVERIFIED';
    } else { 
        $val = 'VERIFIED';
    }
?>

<button type="button" class="btn btn-info"> <?php echo $val; ?> </button>
<br />  
<?php mysqli_close($con); ?>
<hr />

另外，在尝试查询之前，最好先检查GET变量是否使用传递。

试试这个

result = mysqli_query($con,"SELECT * FROM users WHERE url='$goturl'");

---

$result=mysqli_query（$con，“从url=$goturl的用户中选择*）；你需要ajax。你不能在页面加载后调用php代码。我可以模糊地猜测为什么会有否决票，对那个人来说：你无法理解什么更重要->实际的解决方案或额外的安全性。正确的方法是给我留言，让我加上“it”。这两种解决方案都很有效！非常感谢paula和satish:）

$goturl = $_GET['url'];
$goturl = mysqli_real_escape_string($con, $goturl); 

$result = mysqli_query($con,"SELECT * FROM users WHERE url='$goturl' LIMIT 1");
if($row = mysqli_fetch_array($result))
{
    echo $row['verification_value'];
}
else
{
    echo "No in Table";
}