[SOLVED] MySql Query help

[blackwolf23]

Hello,

 

I'm having some difficulty formulating  query that will allow me to retrieve the details of a specific user in the database. The user's information is stored in one table called Users and another called UserProfile. Both tables are linked by the UserID field.

 

My current query returns a Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource and my knowledge of Mysql is not great(Still learnng)

 

Here is my code:

function view_profile($Username)
{

  $conn = connect();
  $query ="SELECT Users.Firstname,Users.Surname,Users.Email,Profile.Age FROM
  Users,UserProfile WHERE  Username  = '".$Username."' AND  Users.UserID =
  UserProfile.UserID";

  $result = mysql_query($query);
  $num = mysql_numrows($result);

  mysql_close();

echo "<b><center>Your profile</center></b><br>";

$i=0;
while ($i < $num) {

$Firstname =mysql_result($result,$i,"Firstname");
$Surname=mysql_result($result,$i,"Surname");
$Email=mysql_result($result,$i,"Email");
$Age=mysql_result($result,$i,"Age");


//echo "<<br>Firstname: $Firstname<br>Surname: $Surname<br>Email: $Email<br>Password: $Password<br> <br>Age: $Age<br>";

$i++;
}
}

[chocopi]

I think you are missing as underscore

 

try replacing

 

mysql_numrows

 

with

 

mysql_num_rows

 

Hope that helps 

 

~ Chocopi

[blackwolf23]

I did some error checking and fixed a mistake where Profile.Age should be UserProfile.Age which removed the error message from appearing but my View_Profile function displays nothing when I test it by going to the View Profile page.

 

This query works and allows me to display the user's profile

  $query = "SELECT * FROM Users WHERE Username  = '".$Username."'"; 

  but I need to display the User's profile from both the users able and the UserProfile table.

 

$query ="SELECT Users.Firstname,Users.Surname,Users.Email,UserProfile.Age FROM
  Users,UserProfile WHERE  Username  = '".$Username."' AND  Users.UserID =
  UserProfile.UserID";

  $result = mysql_query($query);
  $num = mysql_numrows($result);

 

I'm guessing my query is wrong so I will try and work it out.

 

[blackwolf23]

Silly me,I commented out

 echo "<<br>Firstname: $Firstname<br>Surname: $Surname<br>Email: $Email<br>Password: $Password<br> <br>Age: $Age<br>"; 

[chocopi]

Well done for fixing it