[RESOLVED] Dynamically return record for given table and id

thekidscareya

I have a php page that goes something like this:

$sql = "SELECT * FROM '".$_GET['tbl']."' WHERE `id` = '".$_GET['id']."' LIMIT 1";
$query = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($query);

My question is how do I display the record and its corresponding column heading in a table without having to create the table and have a format for each table in the database?

Thanks a lot. I really appreciate your help in advance.

coreyp_1

psuedocode:

$query = mysql_query('FIND ALL THE CANDY BARS AT WAL-MART');

echo '<table>';

$headerprinted = false;

while ($row = mysql_fetch_array($query) {
  if (!$headerprinted) {
    echo '<tr>';
    foreach ($row as $key => $val) {
      echo '<th>'.$key.'</th>';
    }
    echo '</tr>';
    $headerprinted = true;
  }
  echo '<tr>';
  foreach ($row as $val){
    echo '<td>'.$val.'</td>';
  }
  echo '</tr>';
}
echo '</table>';

(not guaranteed to be perfect.)

Roger_Ramjet

1 additional point: remove the quotes around the table name and the back-ticks around the column name

//use
$sql = "SELECT * FROM ".$_GET['tbl']." WHERE id = ".$_GET['id']." LIMIT 1"; 

// which will output as 
"SELECT * FROM table1 WHERE id = 20 LIMIT 1"; 

// instead of 
$sql = "SELECT * FROM '".$_GET['tbl']."' WHERE `id` = '".$_GET['id']."' LIMIT 1"; 

//which will output as
$sql = "SELECT * FROM 'table1' WHERE `id` = '20' LIMIT 1";

thekidscareya

I get this:

0 id 1 name
2 2 My Software My Software

My table just has 2 columns call id & name.

How do I get it so it just shows the name column?

coreyp_1

can you post the code you are using?

thekidscareya

in the first foreach statement i used an if statement
if(!is_int($key))
{
// code is here
}

the second foreach had to be modified a little bit
i made it foreach($row as $key=>$val)

and i put in the same if statement as above.