Display selected amount of records from database - mysql

Question

I need a page that displays records from a database, sorted by their jobs. So the database holds different kind of persons with different jobs. For example, on the page "teacher" I just want to display all the teachers, not the other persons. I want to have 3 persons on a page and a button "previous" and "next" beneath it. If an user clicks the next-button I want the next 3 records to be shown. When the user reaches the last records, I need the next-button to disappear. Same goes for "previous".
What I have so far:
This piece of code creates a value named startrow.
if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) {
//give the value of the starting row 0 because nothing was found in URL
$startrow = 0;
//otherwise take the value from the URL
} else {
$startrow = (int)$_GET['startrow'];
}
The query I have:
$sql = mysql_query("SELECT * FROM $tbl_name WHERE jobs='teacher' LIMIT $startrow, 3")or
die(mysql_error());
$sql2 = "SELECT COUNT(*) AS TotalJobs FROM $tbl_name WHERE jobs='teacher'";
$result_count = mysql_query($sql2);
$count = mysql_fetch_array($result_count);
I created the $count for the if / else function in the next part (the part that doesn't seem to work the way I want it to):
if ($startrow < $count )
echo 'Next';
$prev = $startrow - 3;
//only print a "Previous" link if a "Next" was clicked
if ($prev >= 0)
echo 'Previous';
So the previous-button seems to work the way it should. The next doesn't. I've created the
if ($startrow < $count)
so basicly, if the value of startrow is smaller than the total number of records, it puts a next-button. But if I test this, it displays the next-button anyhow, no matter the value of startrow.
What am I missing here?

Answer 1

mysql_fetch_array returns an array containing the count, not the value itself. Use $count['TotalJobs'] instead.
Also, you shouldn't use the mysql extension anymore, instead use PDO or MySQLi.
Edit
You can change:
$count = mysql_fetch_array($result_count);
to
$countArr = mysql_fetch_array($result_count);
$count = $countArr['TotalJobs'];