[英]How do I use variable names using a for loop to reference a number in the variable name?
$_SESSION_Job101=mysql_fetch_array(mysql_query("SELECT progress FROM job_101 WHERE username='$_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 $e;
}
我可以使$ e变量返回我想要的变量,但是我希望$ e本质上充当if语句中的该变量,而不仅仅是返回其名称。 当我从if语句中删除注释时,我的页面将变为空白而不是回显$ e。 我该如何完成?
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