$_SESSION_Job101=mysql_fetch_array(mysql_query("SELECT progress FROM job_101 WHERE username='$_SESSION_User'"));
$_SESSION_Job102=mysql_fetch_array(mysql_query("SELECT progress FROM job_102 WHERE username='$_SESSION_User'"));
$_SESSION_Job103=mysql_fetch_array(mysql_query("SELECT progress FROM job_103 WHERE username='$_SESSION_User'"));
$_SESSION_Job104=mysql_fetch_array(mysql_query("SELECT progress FROM job_104 WHERE username='$_SESSION_User'"));
$_SESSION_Job105=mysql_fetch_array(mysql_query("SELECT progress FROM job_105 WHERE username='$_SESSION_User'"));
$_SESSION_Job106=mysql_fetch_array(mysql_query("SELECT progress FROM job_106 WHERE username='$_SESSION_User'"));
$_SESSION_Job107=mysql_fetch_array(mysql_query("SELECT progress FROM job_107 WHERE username='$_SESSION_User'"));
$_SESSION_Job108=mysql_fetch_array(mysql_query("SELECT progress FROM job_108 WHERE username='$_SESSION_User'"));
$_SESSION_Job109=mysql_fetch_array(mysql_query("SELECT progress FROM job_109 WHERE username='$_SESSION_User'"));
$_SESSION_Job1010=mysql_fetch_array(mysql_query("SELECT progress FROM job_110 WHERE username='$_SESSION_User'"));
for ($x=1;$x<=10;$x++)
{
$e = '$_SESSION_Job10'.$x;
//if ($e >= 100)
//{ $_SESSION_Job10.$x.['progress'] = 100; }
echo $e;
}
我可以让$e变量返回我想要的变量，但是我希望$e本质上充当if语句中的变量，而不仅仅是返回它的名称。当我从if语句中删除注释时，页面变为空白，而不是回显$e。我该如何做到这一点?
使用
$e = ${'_SESSION_Job10'.$x}; // $e = $_SESSION_Job101; actual variable here
<我>现场演示
不是$e = '$_SESSION_Job10'.$x; // $e = '$_SESSION_Job101'; // string here
编辑
你必须使用像这样的数组
${'_SESSION_Job10'.$x}['progress'];