Getting Info From MYSQL Database.

[bradblog2]

Hey there everyone.

 

I have a little coding project that I am working on and cannot get the PHP part working.

 

Basically all I want to do is get the information (supplied by the user at the time of registration) to display when the server calls for it in their profile page.

 

Here is my code. Thanks

 

<?php
$link=mysql_connect("localhost","root","lolfunny1");
if(!$link) {
	die('Failed to connect to server: ' . mysql_error());
}
//Select database
$db=mysql_select_db("members");
if(!$db) {
	die("Unable to select database");
}
$result = mysql_query("SELECT * from members WHERE member_id='$_SESSION['SESS_MEMBER_ID']'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

?>

 

Any suggestions? I'm very N00B at this and could use any tips.

 

Brad

[whizard]

Try

 

$result = mysql_query("SELECT * from members WHERE member_id='" . $_SESSION['SESS_MEMBER_ID'] . "'");

 

HTH

Dan

[Wolphie]

You shouldn't really post your database connect details on forums perhaps a moderator will remove it.

Anyway, back to the point:

 

<?php
if($con = mysql_connect('servername', 'username', 'password')) 
 mysql_select_db('db_name') 
   or die('MySQL Error: ' . mysql_error());


if(isset($_SESSION['id'])) { // make sure the session is set
 if(is_int($_SESSION['id']))
   $uid = mysql_real_escape_string(htmlentities(htmlspecialchars($_SESSION['id']))); // secure the data for inserting into the database!
 
 $query = sprintf("SELECT * FROM `db_name` . `table_name` WHERE `id` = '%d' LIMIT 1", $uid); // Make sure it only calls one user
 $result = mysql_query($query) or die('MySQL Error: ' . mysql_error());
 while($row = mysql_fetch_array($result)) { // while loop
   if(mysql_num_rows($result) > 0) {
     print $row['username']; // etc...
     print $row['email']; // etc...
   }
   else
     print 'User does not exist!';
 }
}
else
 die 'You are not logged in!';

mysql_close($con);
?>

 

That's how I would do it personally, but your query should be:

 

$result = mysql_query("SELECT * from members WHERE member_id=' . $_SESSION['SESS_MEMBER_ID'] . '");

 

 

EDIT: whizard got there before me =[

[bradblog2]

Alright, I tried replacing the code with the example above and I got the following error:

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in xxxxxxxxxx on line 11

 

I tried the second set of code (replacing the first one) and simply nothing displayed.

 

Is there something I'm doing wrong?

[Stephen]

Try using:

<?php
$con = mysql_connect('servername', 'username', 'password')

if($con) {
  mysql_select_db('db_name') or die('MySQL Error: ' . mysql_error());
}

if(isset($_SESSION['id'])) { // make sure the session is set
  if(is_int($_SESSION['id'])) {
    $uid = mysql_real_escape_string(htmlentities(htmlspecialchars($_SESSION['id']))); // secure the data for inserting into the database!
  
  $query = "SELECT * FROM table_name WHERE id = '".$uid."' LIMIT 1"; // Make sure it only calls one user
  $result = mysql_query($query) or die('MySQL Error: ' . mysql_error());
  while($row = mysql_fetch_array($result)) { // while loop
    if(mysql_num_rows($result) > 0) {
      print $row['username']; // etc...
      print $row['email']; // etc...
    }
    else {
      print 'User does not exist!';
}
  }
  }
}
else
  die 'You are not logged in!';

mysql_close($con);
?>

 

I edited Wolphies code a little

[abdfahim]

I don't get how possibly could you use session variables in a page where you didn't start session. On top of the page, you should write

if (!isset($_SESSION)) {
session_start();
}

[bradblog2]

@ Stehpen - Parse error: syntax error, unexpected T_IF in xxxx on line 4

 

@ abdbuet - I have a login script that starts a session when they log in.

[Stephen]

Sorry, forgot to add a linebreak.

<?php
$con = mysql_connect('servername', 'username', 'password');

if($con) {
  mysql_select_db('db_name') or die('MySQL Error: ' . mysql_error());
}

if(isset($_SESSION['id'])) { // make sure the session is set
  if(is_int($_SESSION['id'])) {
    $uid = mysql_real_escape_string(htmlentities(htmlspecialchars($_SESSION['id']))); // secure the data for inserting into the database!
  
  $query = "SELECT * FROM table_name WHERE id = '".$uid."' LIMIT 1"; // Make sure it only calls one user
  $result = mysql_query($query) or die('MySQL Error: ' . mysql_error());
  while($row = mysql_fetch_array($result)) { // while loop
    if(mysql_num_rows($result) > 0) {
      print $row['username']; // etc...
      print $row['email']; // etc...
    }
    else {
      print 'User does not exist!';
}
  }
  }
}
else
  die 'You are not logged in!';

mysql_close($con);
?>

[abdfahim]

@ abdbuet - I have a login script that starts a session when they log in.

As of my knowledge, you have to start your session in every page to keep the session variables. It is not enough that you start it during one logs in and then don't start it in other pages.

[bradblog2]

@abduet - Alright I'll add that in.

[bradblog2]

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in xxxxx on line 26

 

 

[Stephen]

Hmm, try this:

<?php
$con = mysql_connect('servername', 'username', 'password');

if($con) {
  mysql_select_db('db_name') or die('MySQL Error: ' . mysql_error());
}

if(isset($_SESSION['id'])) { // make sure the session is set
  if(is_int($_SESSION['id'])) {
    $uid = mysql_real_escape_string(htmlentities(htmlspecialchars($_SESSION['id']))); // secure the data for inserting into the database!
  
  $query = "SELECT * FROM table_name WHERE id = '".$uid."' LIMIT 1"; // Make sure it only calls one user
  $result = mysql_query($query) or die('MySQL Error: ' . mysql_error());
  while($row = mysql_fetch_array($result)) { // while loop
    if(mysql_num_rows($result) > 0) {
      print $row['username']; // etc...
      print $row['email']; // etc...
    }
    else {
      print 'User does not exist!';
}
  }
  }
}
else {
  die("You are not logged in!");
}

mysql_close($con);
?>

 

Otherwise it looks like it'd work to me.

[bradblog2]

Thanks for your help. Stephen and everyone. I really appreciate it.

 

It now says I'm not logged in, which is a problem with the other code.

 

At least it is that far