to put it simple, her is what i have:

PHP Code:

 mysql_query("UPDATE layout SET `set` = '$set', 


that works, but what i want to do is this:

PHP Code:

 mysql_query("UPDATE $layout SET `set` = '$set', 


this dose not work, what can i do?

Try surrounding your vars with{}

I.E.
mysql_query("UPDATE layout SET `set` = '{$set}',

ibbo

no that dident work for the tabel, i need the tabel name to be a var like:

mysql_query("UPDATE $layout SET `set` = '$set',

You should be able to build this string just like any other string. Make sure the variables are assigned before you build the string.

What do you get if you :

PHP Code:

echo "UPDATE $layout SET `set` = '$set'"; 


it echos the resalts from the var $layout

but it still dosent work in the normal script.

What it sounds like you are saying is you can "hard-code" the sql statement, and it works, but when you substitute a variable for one word in the sql statement, it does not work.

if (hard_coded_statement === code_generated_string) {
you won't get an error
}

Are you getting an error message? Is the variable $layout returning a SQL keyword? Is code before or after this line affecting something? Sometimes it seems that the error must be in one line, when in fact, it's in another.

no, there is no error, it just dosent work. whin I use the var it simply dosent update the tabel. the var works evrywhere else just not for the tabel name. here I will post my whole update code, maybe that would help.

PHP Code:

 function update_db($data_array, $col_check)
{

  foreach($data_array AS $set => $items)
  {
     $i = 0;
     foreach($items AS $item)
     {
       $item = mysql_escape_string($item);
       $set  = mysql_escape_string($set);
       
       mysql_query("UPDATE $username SET `set` = '$set', `order` = '$i'  WHERE `item` = '$item' $col_check");
       $i ++;
     }
  }
} 


I think you are going about it correctly. And this is a very good way to do it. But you just need to make sure that the string you are sending to mysql_query() is proper SQL syntax, and has the correct (existing) tablename, etc.

I understand the need to protect code, but if you are really stuck, then the best way to get help is to show a little code, the error message (if any), and in this example, show the SQL string that gets echoed (after PHP replaced the variable $layout with the string "user", or whatever the case may be).

Also, if $layout is the prefix to a SQL table, then you definitely want to surround it with {}, like ibbo said, so it isn't looking for a different variable.

PHP Code:

//set var
$layout = "user"

//this will NOT return "user_layout_table";
echo "$layout_password_table";

//this WILL return "user_layout_table";
echo "{$layout}_password_table"; 


this is not a big part of my scripting job I am working on, just a little side job so here is all the affected sql update/post script.

PHP Code:

$username = layout;

mysql_connect('localhost', 'username', 'password');
mysql_select_db('database');
function parse_data($data)
{
  $containers = explode(":", $data);
  foreach($containers AS $container)
  {
      $container = str_replace(")", "", $container);
      $i = 0;
      $lastly = explode("(", $container);
      $values = explode(",", $lastly[1]);
      foreach($values AS $value)
      {
        if($value == '')
        {
            continue;
        }
        $final[$lastly[0]][] = $value;
        $i ++;
      }
  }
    return $final;
}

function update_db($data_array, $col_check)
{

  foreach($data_array AS $set => $items)
  {
     $i = 0;
     foreach($items AS $item)
     {
       $item = mysql_escape_string($item);
       $set  = mysql_escape_string($set);
       
       mysql_query("UPDATE '{$username}' SET `set` = '$set', `order` = '$i'  WHERE `item` = '$item' $col_check");
       $i ++;
     }
  }
}

// Lets setup Sajax
require_once('Sajax.php');
sajax_init();
// $sajax_debug_mode = 1;

function sajax_update($data)
{
  $data = parse_data($data);
  update_db($data, "AND (`set` = 'sajax1' OR `set` = 'sajax2'  OR `set` = 'sajax3')");
  return 'y';
}

sajax_export("sajax_update");
sajax_handle_client_request(); 


why I dont understand, is because it dose work here:

PHP Code:

 <ul id="sajax3" class="sortable boxy">
  <?php
$r = mysql_query("SELECT * FROM $username WHERE `set` = 'sajax3' ORDER BY `order` ASC");
while($rw = mysql_fetch_array($r))
{
  echo '<li id="'.$rw['item'].'">'.$rw['item'].'</li>';
}
?>

</ul>

How about changing to:

mysql_query("UPDATE `$username` SET `set` = '$set', `order` = '$i' WHERE `item` = '$item' $col_check");

nope, still no go. :-(

Got It!!!!!

here is the gode that did it:

PHP Code:

function update_db($data_array, $col_check)
{

  foreach($data_array AS $set => $titles)
  {
     $i = 0;
     foreach($titles AS $title)
     {
       $title = mysql_escape_string($title);
       $set  = mysql_escape_string($set);
       $username = layout;
       
       mysql_query("UPDATE `$username` SET `set` = '$set', `order` = '$i'  WHERE `title` = '$title' $col_check");
       $i ++;
     }
  }
}