cant display the database information , why?

[zgkhoo]

cant display the database information , why?

 

<?php
include 'config.php';
include 'opendb.php';

session_start();
echo "testing ".$_SESSION['username'];

$result = mysql_query("SELECT * FROM person");
while($row = mysql_fetch_array($result))
  {
  echo $row['fullname'] ;
  echo "<br />";
  }


include 'closedb.php';



?>

[Chevy]

put this after the mysql_query and see what it says:

 

or die(mysql_error());

[zgkhoo]

put this after the mysql_query and see what it says:

 

or die(mysql_error());

 

("SELECT username,password FROM person"

i changed to this code, problem solved, why? weird..

[MadTechie]

that change will NOT fix the problem, as your pulling 'fullname', add the

or die(mysql_error());

would of revealed the problem

 

also try

replacing

echo $row['fullname'] ;

with

print_r($row);

and check that fullname exists (it will not with your last change)

[Barand]

that change will NOT fix the problem, as your pulling 'fullname', add the

or die(mysql_error());

would of revealed the problem

 

 

AH! the clairvoyant version of mysql_error() that knows you are going to attempt to echo a non-existent column name subsequent to calling the query

[Chevy]

I never said it would fix it, I was just wondering if there was an error in the MySQL

[zgkhoo]

wat type of mysql error u meant? wat possible error might be..

thanks.

[MadTechie]

@Chevy: i was refering to zgkhoo "

i changed to this code, problem solved, why? weird..

 

@zgkhoo: any error, wrong field, invalid SQL statement.. have you eveb tried it yet ?

[zgkhoo]

nope

[Guest]

In your first code, you used this line

 

  echo $row['fullname'];

 

then when you changed your SQL to

 

  "SELECT username,password FROM person"

 

There was only the username and password fields being selected. We do not see any fullname field, so we assume you may be trying to get a field that doesn't exist when you did:

 

  echo $row['fullname'];