PHP匹配错误

我正在使用php创建一个配置文件站点，我需要检查配置文件id是否与成员id不相等。因此，我有以下内容：

<?php
$get_personal_sql = "SELECT * FROM members WHERE username = '$thisuser'";
$get_personal_res = mysqli_query($con, $get_personal_sql);
while ($personal = mysqli_fetch_assoc($get_personal_res)){
    $yourid = $personal['id'];
    $yourtype = $personal['type'];
    $yoursubtype = $personal['subtype'];
    $yourusername = $personal['username'];
    $youremail = $personal['email'];
    $yourname = $personal['name'];
    $youractive = $personal['active'];
    $yourpro = $personal['pro'];
    $yourdel = $personal['del'];
    $yourdate = $personal['date'];
}
?>
<?php
$get_member_sql = "SELECT * FROM members WHERE username = '$thismember'";
$get_member_res = mysqli_query($con, $get_member_sql);
$count_member = mysqli_num_rows($get_member_res);
if($count_member == 0){
header("Location: notfound.php");
}
while ($member = mysqli_fetch_assoc($get_member_res)){
    $memberid = $member['id'];
    $membertype = $member['type'];
    $membersubtype = $member['subtype'];
    $memberusername = $member['username'];
    $memberemail = $member['email'];
    $membername = $member['name'];
    $memberactive = $member['active'];
    $memberpro = $member['pro'];
    $memberdel = $member['del'];
    $memberdate = $member['date'];
}
?>
<?php
if($memberid === $yourid){
    $yourprofile = "true";
}
else{
    $yourprofile = "false";
}
?>

---

我在页面中回显了$memberid和$yourid，它们完全不同，但回显$yourprofile时显示“true”。有人知道为什么会这样吗？

显示您的真实代码，此代码没有错误。您可以这样设置$yourprofile:

$yourprofile=“true”。$memberid。“$yourid并显示输出。在本例中，它们不可能“完全不同”。@steven是的，它们完全不同，但我会在后面的页面中添加另一个if语句，说if（$yourprofile=“true”）{…}，这显然改变了$yourprofile的值。现在修好了，不过还是谢谢你