can't passing value

[skyloon]

the code below show the result from table, when i click the view button, it should pass the value to the next form, but the next form does not show any result, don't know why. can anybody help me?

<? $Connect = mysql_connect("localhost","root","");
mysql_select_db("mw");
$result=mysql_query("select * from LeaveApplication");
$number_of_array = mysql_num_rows($result);

			print "<TABLE CELLSPACING=\"0\" CELLPADDING=\"0\" BORDER=\"1\" align=\"center\">\n";
			print "  <TR><TD width=\"25%\" align=\"center\">Staff ID</TD><TD width=\"25%\" align=\"center\">Leave Type</TD><TD width=\"25%\" align=\"center\">Job Title</TD><TD width=\"25%\" align=\"center\">Action</TD></TR>\n";

			while ($number_of_array = mysql_fetch_array($result)){
			$StaffID = "number_of_array[$StaffID]";
			echo "<tr>\n"; 
			echo "<TD width=\"25%\" align=\"center\">$number_of_array[StaffID]</TD>\n"; 
			echo "<TD width=\"25%\" align=\"center\">$number_of_array[LeaveType]</TD>\n"; 
			echo "<TD width=\"25%\" align=\"center\">$number_of_array[JobTitle]</TD>\n"; 
			echo "<TD width=\"25%\" align=\"center\"><form name=\"form1\" method=\"post\" action=\"view_staffleave.php\"><input type=\"submit\" name=\"Submit\" value=\" View \"></TD>\n";
			echo "<tr>\n"; 
			}
			mysql_close();
		?>

---

next form

<? $Connect = mysql_connect("localhost","root","");
mysql_select_db("mw");
$result=mysql_query("select * from LeaveApplication where StaffID = '$StaffID'") ;
$number_of_array = mysql_num_rows($result);
while ($number_of_array = mysql_fetch_array($result)){
echo "$number_of_array[StaffID]";
}
mysql_close();
?>

<? $Connect = mysql_connect("localhost","root","");
mysql_select_db("mw");
$result=mysql_query("select * from LeaveApplication where StaffID = '$StaffID'") ;
$number_of_array = mysql_num_rows($result);
while ($number_of_array = mysql_fetch_array($result)){
echo "$number_of_array[LeaveType]";
}
mysql_close();
?>

<? $Connect = mysql_connect("localhost","root","");
mysql_select_db("mw");
$result=mysql_query("select * from LeaveApplication where StaffID = '$StaffID'") ;
$number_of_array = mysql_num_rows($result);
while ($number_of_array = mysql_fetch_array($result)){
echo "$number_of_array[JobTitle]";
}
mysql_close();
?>

[tsinka]

Hi,

From what I see in your code you want to create n forms where n is the number of results.

But yor code creates just empty forms so I'd suggest to change it to something like this:

while ($number_of_array = mysql_fetch_array($result)){ 
    $StaffID   = number_of_array['StaffID']; 
    $LeaveType = number_of_array['LeaveType']; 
    $JobTitle  = number_of_array['JobTitle']; 

echo <<< EOHTML
<form name="form1" method="post" action="view_staffleave.php">
<input type="hidden" name="StaffID" value="$StaffID">
<tr>
    <td width="25%" align="center">$StaffID</td>
    <td width="25%" align="center">$LeaveType</td>
    <td width="25%" align="center">$JobTitle</td>
    <td width="25%" align="center">
        <input type="submit" name="Submit" value=" View ">
    </td>
</tr>
</form>
EOHTML;
} 

Besides that you don't really need the mysql_close() since php automatically closes mysql connections (non persistent connections) at the end of the script.

[tsinka]

One more ...

Depending on the version of php or the setting of register_globals you might need to use $_POST['StaffID'] instead of $StaffID on the second page.