Question

I have the table name in the form as a hidden field, i have to do it this way for some reason but it’s not working when i do the following:

 // This is being processed from the previous page in a form
 $table = mysql_real_escape_string($_POST['just_field']);

//Here i am using the $table value to use in place of a table name
$sql = "select `Field` from `$table` where `another_field` = '$id' 
and `another_another_field` = '$something'";
$query = mysql_query($sql) or die ("Error: ".mysql_error());

while ($row = mysql_fetch_array($query)){

    $somefield = $row['field'];
  }
 mysql_free_result($query);

How can i make it work so the $table is being read by the command…

Answer 1

try to include the connection string

$sql = "select `Field` from `$table` where `another_field` = '$id' 
and `another_another_field` = '$something'";
$query = mysql_query($sql,$conn) or die ("Error: ".mysql_error());

 while ($row = mysql_fetch_array($query)){

    $somefield = $row['field'];
  }
 mysql_free_result($query);

if you are not using any framework