checking if there are any rows to display

Anon

I want PHP to keep the user on the same search page (or bring them back to the search page) if there are no rows to display from the user's search. My search page lets the user search by employee name, but I can't figure out how to check to see if there are rows to display first.

I tried using the snippit below (Kirk, if you're reading this, this is what you recommended earlier)

" if ($numrow < 1)
echo "No records found";
else
{
// go ahead and display them... "

But where do I put this? and how do I redirect the user back to the original search page and display a "no records found" message so that the user can try again? *

I've pasted most of the code below from the results page that I'm using.
If anyone can tell me exactly what to put and where to put it, it would be greatly appreciated.
-Thanks.

$dbh=mysql_connect("localhost","username","password");

mysql_select_db("databasename",$dbh);

if(empty($offset))

$offset=0;

$query="
select distinct
employees.employeeID as employeeID,
employees.name as name,
employees.scheduleID as scheduleID,
schedule.schedule_details as schedule_details
from
employees,
schedule
where (
employees.scheduleID=schedule.scheduleID and
employee.name like '%$name%'

)

order by name
limit $offset,20
";

$result = mysql_query($query,$dbh);
$numrow = mysql_numrows($result);

<BODY>
* graphics, header, table go here

$counter=0;
$linenumber=1;

do {

\for alternating row colors\
if (is_int($counter/2)):

$bgcolor="#eeeeee";
else:
$bgcolor="#ffffff";
endif;

$employeeID=mysql_result($result,$counter,"employeeID");

$name=mysql_result($result,$counter,"name");

$scheduleID=mysql_result($result,$counter,"scheduleID");

$schedule_details=mysql_result($result,$counter,"schedule_details");

start displaying results here

<? echo $name ?> <BR>
<? echo $schedule_details ?>

</BODY>

Anon

$result = mysql_query($query,$dbh);
$numrow = mysql_numrows($result);

You need to check if your sql query return any rows.

if ($numrow != 0)
{
//there is at least one record that match your query
code to show all records that match your query...
}
else
{
//no match for your query
code to add a new record or present the search box again
}

Kai

Anon

but what is the code to present the search box again?

I know the "if" part, I think I just need to know what the "else" part would be.

Anon

if ($numrows == 0)
{
echo "<script language=\"javascript>location.href='first_page.php?status=no_records_found';</script>";
}
else
{
//show results
}

Then, on the first_page, have a part like this

if ($status == "no_records_found")
{
echo "Your search did not return any records. Please search again";
}

This won't show the first time you show the search because $status won't be set.

Hope that helps.

---John Holmes...

Anon

I'm getting a parse error with the script below.
What do I have wrong?

if ($numrows == 0)
{
echo "<script language=\"javascript>location.href='search.php?status=no_records_found';</script>";

else
{
//show results
}

Anon

Several things: the if-true body doesn't have a terminating } and also
you'll probably get undesired results at the browser, too since the
language="javascript is missing a closing quote.

Actually though, you can achieve the same effect without Javascript
just by sending a location header:

  header( "Location: $PHP_SELF" );
  exit; // don't output anything else

The advantage is that this works even if the user has
turned Javasript off. On the other hand, it only works
if you haven't output anything yet, so it doesn't lend
itself very well to the in-and-out-of-PHP style you're using.

Anon

Thanks Kirk,

I'll give that a try.