Retrieve data from database using session variable

Question

<?php

{
session_start();
include "dbconnect.php";

$email = $_SESSION['email'];
echo $email;
$query = "SELECT uid FROM master WHERE emailid = '$email' ";
$result = mysql_query($query);
$row=mysql_fetch_array($result);
while($row)
 {
  $uid=$row[1];

  echo $uid;

   }

} whats wrong in this code.it is giving the email but not able to retrieve uid using session variable email. please help

What's stray { for?

Mr. Alien
– Mr. Alien

2014-01-14 07:53:06 +00:00
Commented Jan 14, 2014 at 7:53 — Mr. Alien
– Mr. Alien, Commented Jan 14, 2014 at 7:53

9edge · Accepted Answer · 2014-01-16 07:22:07Z

4

while($row) {
   $uid = $row[1];
   echo $uid;
}

should be

while($row) {
   $uid=$row['uid'];  
   echo $uid;    
}

Try if this code works for you

session_start();
include "dbconnect.php";

$email = $_SESSION['email'];
//echo $email;
$query = "SELECT uid FROM master WHERE emailid = '$email' ";
$result = mysql_query($query);

while($row=mysql_fetch_array($result)){
    $uid=$row['uid'];
    echo $uid;
}

edited Jan 16, 2014 at 7:22

answered Jan 14, 2014 at 7:54

9edge

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Comments

Alen · Accepted Answer · 2014-01-14 07:53:26Z

0

I don't know why you have { at beginning of your code, you should delete that. And second thing, delete echo $email; and you should have working code

answered Jan 14, 2014 at 7:53

Alen

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2 Comments

Mr. Alien Over a year ago

What? Why delete echo?

Kumar V Over a year ago

you can use comment section

Varun Davda · Accepted Answer · 2014-01-14 07:55:57Z

0

try $uid=$row[0] instead of $uid=$row[1]; Because $row is an array.Array always starts its counter form 0. here you have defined $uid=$row[1] which means it will display the result from second row of the result set..But there is no second row there is only one row which is "uid" so try $uid=$row[0] instead of $uid=$row[1];

answered Jan 14, 2014 at 7:55

Varun Davda

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Comments

Anil R · Accepted Answer · 2014-01-14 07:57:36Z

0

try this:

<?php
session_start();
include "dbconnect.php";

$email = $_SESSION['email'];

$query = "SELECT uid FROM master WHERE emailid = '".$email."'";
$result = mysql_query($query);
if(mysql_num_rows($result)>0)
{
    $row = mysql_fetch_array($result);
    $uid = $row["uid"];
    echo $uid;
}
else
{
    echo "No record found";
}
?>

answered Jan 14, 2014 at 7:57

Anil R

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2 Comments

Yogendra Karekar Over a year ago

thank u anil for the code but it is giving o/p as no record found, while the record is being added

Anil R Over a year ago

check database connection then or try to execute the query (echo query in code and execute it in phpmyadmin)

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