Question

我正在尝试检查用户表的访问列表中是否存在用户访问过的页面。

像这样我从URL

获取页面名称

$q1 =  basename($_SERVER['REQUEST_URI'], '?' . $_SERVER['QUERY_STRING']);
$var1 = "/".$q1;

从数据库中我获取登录用户的页面名称并使用in_array函数检查页面名称

$m1 = "select pages.page_id, pages.code, pages.page, pages.href, access_level.page_id, access_level.user_id FROM pages INNER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.user_id=".$_SESSION['user_id']."";
//echo $m1;

$m2 = mysqli_query($con, $m1) or die (mysqli_error($con));

while($m3 = mysqli_fetch_array($m2)) {
print_r($m3['href']);

if(in_array($var1, $m3['href'])) {

}
else
{
    header("location:no_access.php");
}
}

但我无法弄清楚in_array的问题是什么，因为所有页面都重定向到no_access.php页面，而且页面内容也没有显示但是显示页面不是正确重定向。

被修改

$q1 =  basename($_SERVER['REQUEST_URI'], '?' . $_SERVER['QUERY_STRING']);
$var1 = "/".$q1;

$m1 = "select pages.page_id, pages.code, pages.page, pages.href, access_level.page_id, access_level.user_id FROM pages INNER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.user_id=".$_SESSION['user_id']."";
//echo $m1;

$m2 = mysqli_query($con, $m1) or die (mysqli_error($con));

while($m3 = mysqli_fetch_array($m2)) {
    $href[] = ($m3['href']); 
//print_r($href);
//echo basename($_SERVER['REQUEST_URI'])."/".$var1;
if(in_array($var1, $href)) {
 header("location:".BASE_URL."".$var1);
}
else

{
    header("location:no_access.php");
}


}

Answer 1

尝试这样......

$m1 = "select pages.page_id, pages.code, pages.page, pages.href, access_level.page_id, access_level.user_id FROM pages INNER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.user_id=".$_SESSION['user_id']."";
    //echo $m1;

    $q1 =  basename($_SERVER['REQUEST_URI'], '?' . $_SERVER['QUERY_STRING']);

$ var1 =“/".$ q1;

$ m1 =“select pages.page_id，pages.code，pages.page，pages.href，access_level.page_id，access_level.user_id FROM pages INNER JOIN access_level ON pages.page_id = access_level.page_id WHERE access_level.user_id =” 。$ _ SESSION [ 'user_ID的'] “”。 // echo $ m1;

$m2 = mysqli_query($con, $m1) or die (mysqli_error($con));

while($m3 = mysqli_fetch_array($m2)) {
    $href[] = ($m3['href']); 
} //close while loop here
//print_r($href);
//echo basename($_SERVER['REQUEST_URI'])."/".$var1;
if(in_array($var1, $href)) {
 header("location:".BASE_URL."".$var1);
}
else

{
    header("location:no_access.php");
}

用字符串检查数组

1 个答案: