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Editorial Team
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Asked: 2026-05-22T21:10:35+00:00

This is my person table: eid sex email 17 Female example@example.com This is my

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This is my “person” table:

eid sex email   
17  Female  example@example.com

This is my php page:

<?php
$loginquery = mysql_query('SELECT eid, email, sex FROM person WHERE email = "'example@example.com'" ;') or die("Query fail: " . mysql_error() );
$userlogin = mysql_fetch_array($loginquery);

if($userlogin){
$_SESSION['eid']= $userlogin['eid'];
$_SESSION['email']= $userlogin['email'];
$_SESSION['sex']= $userlogin['sex'];
}


echo "user ".$_SESSION['eid']." ".$_SESSION['email']." ".$_SESSION['sex']; 
?>

This is the output:

user example@example.com Female

Why the $_SESSION['eid'] is not printed ?

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1 Answer

  1. Editorial Team

    I see at least 2 bugs

    session_start(); is not called and it should.

    The query is weird:

    $loginquery = mysql_query('SELECT eid, email, sex FROM person WHERE email = "'example@example.com'" ;') or die("Query fail: " . mysql_error() );

    I don’t know why this does not produce an error, but your string is not build well. Try something like

    $loginquery = mysql_query("SELECT eid, email, sex FROM person WHERE email = 'example@example.com';") or die("Query fail: " . mysql_error() );
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