Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/248.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181

Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/database/9.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 从会话中获取价值_Php_Database_Session_Mysqli - Fatal编程技术网

Php 从会话中获取价值

Php 从会话中获取价值,php,database,session,mysqli,Php,Database,Session,Mysqli,我已经创建了一个登录名,其中ID存储在$\u会话['Login\u user']中 身份验证后,用户将重定向到其主页。我想从会话中获取值,并在查询中使用它来检索数据库中与此userID对应的所有属性。这是我尝试过的代码 session_start(); .......... .......... .......... $uid= $_SESSION['login_user']; $result = mysqli_query($db,"SELECT * FROM users UID ='".$ui

我已经创建了一个登录名,其中ID存储在$\u会话['Login\u user']中

身份验证后,用户将重定向到其主页。我想从会话中获取值,并在查询中使用它来检索数据库中与此userID对应的所有属性。这是我尝试过的代码

session_start();
..........
..........
..........
$uid= $_SESSION['login_user'];
$result = mysqli_query($db,"SELECT * FROM users UID ='".$uid."'");
它应该获取存储在“users”表数据库中的所有值,但问题是,有时它工作,有时它不工作。大多数时候,它不起作用。 如果能帮上点忙,我将不胜感激

在执行此步骤之前,我还检查了会话是否为空

这是设置会话时登录的一部分:

$result=mysqli_query($db,"SELECT uid FROM users WHERE username='$username' and
password='$password'");
$count=mysqli_num_rows($result);

$row=mysqli_fetch_array($result,MYSQLI_ASSOC);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1)
{
$_SESSION['login_user']=$row['uid'];

您是否尝试过使用正确的SQL语法,如SELECT*FROM{table},其中something=someval?比如:

$result = mysqli_query($db,"SELECT * FROM users WHERE UID ='".$uid."'");

session_start()?然后您应该将其添加到问题代码中。在查询之前使用print\r($\u SESSION)查看$\u SESSION中的值。确保在同一页面中未修改
$\u SESSION['login\u user']
。感谢您的提示。这就是它给出的结果:数组([login\u user]=>)我相信登录没有正常工作,对吗?