[SOLVED] MySQL - is the column in table 1, or table 2? -> Revised and Revisited

[MasterACE14]

evening everyone,

 

I had a script a couple days ago to work out if a MySQL column was in 1 table or another. And it never ended up working, I have done some more research and tried some more things, and this is what I got:

 

<?php
// Check if the column in the database is in 1 table or another
function what_table($column) {

$fields = mysql_list_fields("database", "cf_users");
       
$columns = mysql_num_fields($fields);
       
for ($i = 0; $i < $columns; $i++) {
    $field_array[] = mysql_field_name($fields, $i);
}
       
if (!in_array($column, $field_array)) {
    
$table = "cf_users2";

}

return $table;

};

//and my function for selecting a column from the database for the logged in user

// Select fields from the user table
function player_table($select){
    $where = $_SESSION['playerid'];
  /*  $rs = mysql_connect("localhost", "ace_ACE", "shadow69");
    mysql_select_db("ace_cf", $rs); */
    $sql = "SELECT `" . what_table($select) . "`.`" . $select . "` FROM `cf_users` INNER JOIN `cf_users2` ON (`cf_users2`.`id` = `cf_users`.`id`) LIMIT 1";
    $rs = mysql_query($sql) or die('Query:<br />' . $sql . '<br /><br />Error:<br />' . mysql_error());
    while($row = mysql_fetch_assoc($rs)){
        if(isset($row[$select])){return $row[$select];}
    }
};

?>

 

But for some reason I am getting an error, and the SELECT field is empty  ???

Query:

SELECT ``.`id` FROM `cf_users` INNER JOIN `cf_users2` ON (`cf_users2`.`id` = `cf_users`.`id`) LIMIT 1

 

Error:

Column 'id' in field list is ambiguous

 

my what_table() function does seem to be putting the correct table in after the FROM.

 

any ideas on why the SELECT field is empty, would be greatly appreciated. 

 

Regards ACE

[btherl]

You set table to cf_users2 if the column is not found .. but what if it is found?

[MasterACE14]

haha, you got a point their, as soon as I added a ELSEIF, it is now working perfectly 

 

<?php
// Check if the column in the database is in 1 table or another
function what_table($column) {

$fields = mysql_list_fields("database", "cf_users");
       
$columns = mysql_num_fields($fields);
       
for ($i = 0; $i < $columns; $i++) {
    $field_array[] = mysql_field_name($fields, $i);
}
       
if(!in_array($column, $field_array)) {
$table = "cf_users2";
}
elseif(in_array($column, $field_array)) {
$table = "cf_users";
}

return $table;

};

 

Cheers 

 

Regards ACE