Dynamic population error

[Anon]

I have a dropdown box that is dynamically populated from the db, it works perfectly until I place it into a variable i.e. $block
once placed inside the variable it fails to draw anything from the db.

Code:
$block = "
<form name=\"form1\" method=\"post\" action=\"InsertEmp2.php\">
<table width=\"100%\" border=\"0\" cellspacing=\"0\" cellpadding=\"0\">
<tr>
<td>Select the leader</td>
<td>
<select name=\"leader\">
<option value=\"\">Choose a Team Leader...</option>
$SQL = \"SELECT * FROM staff_table INNER JOIN leader ON leader.staff_id=staff_table.staff_id\";

$MySQLQuery = mysql_query($SQL) 
or die(mysql_error());

// loop thru the results
// print each result to a select option
while ($leaderdd = mysql_fetch_row($MySQLQuery)) {
print \"<option value=\"$leaderdd[9]\">$leaderdd[1]</option>\n\";
	}

      </select></td>
  </tr>
</table>

</form>
";

Can anyone explain this to me please. Why does it fail to work once inside the variable.

This code is used to display $block
<?
if ($position < "4") {
echo $block;
}

Any help would be great thanks.
Matt Thornhill

[biggus]

Matt,

you can't run a query inside a variable definition. For the above code, I'd close the double quotes just before $SQL=... and, instead of using print for the new options, append them to $block with $block.="<option>..." Then use $block.="</select></td>..." to tack on the end.

HTH

[Anon]

Hi

Thanks very much...your a life saver...I've been stuck on this for ages. 😃

Thanks again
Matt