Php/sql Script Will Not Display First Result

[ryanfilard]

I made this script:

 <?PHP
    while($row_userfriends = mysql_fetch_array($userfriends))
 {
 echo '$row_userfriends['user']';
 }
 ?>

problem is it works but does not display the first result. Whats wrong? :-\

[KevinM1]

EDIT: Two things:

 

1. You're using a single-quoted string to output the value. That won't work as only double-quoted strings extrapolate values.

2. You can't output an array value in a string unless you either concatenate or surround it in braces.

 

In short you want either:

 

echo $row_userfriends['user'];

 

or:

 

echo "{$row_userfriends['user']}";

 

If that doesn't solve it, show your query and table structure.

Edited September 27, 2012 by KevinM1

[ryanfilard]

User is text, friend is text, and accepted is one

 

This is what I am using to display the data

mysql_select_db($database_main, $main);
$query_userfriends = "SELECT * FROM friends WHERE `friend` = '$uid' LIMIT 9";
$userfriends = mysql_query($query_userfriends, $main) or die(mysql_error());
$row_userfriends = mysql_fetch_assoc($userfriends);

[ryanfilard]

I made a mistake when retyping the script, it should be this:

echo $row_usersfriends['user'];

Edited September 27, 2012 by ryanfilard

[DavidAM]

I made this script:

 <?PHP
       while($row_userfriends = mysql_fetch_array($userfriends))
 {
 echo '$row_userfriends['user']';
 }
 ?>

problem is it works but does not display the first result. Whats wrong?

mysql_select_db($database_main, $main);
$query_userfriends = "SELECT * FROM friends WHERE `friend` = '$uid' LIMIT 9";
$userfriends = mysql_query($query_userfriends, $main) or die(mysql_error());
$row_userfriends = mysql_fetch_assoc($userfriends);

 

Instead of showing snippets, you should show all of the code that is pertinent. If your latest code comes before the code in your first post, then it is:

 

mysql_select_db($database_main, $main);
$query_userfriends = "SELECT * FROM friends WHERE `friend` = '$uid' LIMIT 9";
$userfriends = mysql_query($query_userfriends, $main) or die(mysql_error());
$row_userfriends = mysql_fetch_assoc($userfriends);
       while($row_userfriends = mysql_fetch_array($userfriends))
 {
 echo '$row_userfriends['user']';
 }

 

If you look at that, you are fetching the first row just before the while. Since you don't do anything with $row_userfriends before the while statement, you are effectively thowing away the first record. Remove the fetch statement (before the while) and you should be golden.

[Pikachu2000]

You have a call to mysql_fetch_assoc() before the while() loop begins. That will move the internal data pointer to the second record, and the first one will never be echoed.