PHP duplicate staffID - php

Question

staff table
code
<?php
//db configuration
$q = "select * from staff";
$r = mysqli_query($dbc, $q);
$num_rows = mysqli_num_rows($r);
$staffID = array();
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
$staffID[] = $row['staffID'];
}
for($i = 0; $i < $num_rows; $i++)
{
if($staffID[$i] == $staffID[$i+1])
{
$remark = "Not OK. Multiple staff ID selected.";
$i++;
}
else
{
$remark = "OK.";
}
$data[$i] = $staffID[$i].','.$remark.'<br />';
}
$list = array($data);
print_r($list);
?>
From above code, I want to check whether the staffID got duplicate.
Expected output
0001, OK.
0002, Not OK. Multiple staffID.
0003, OK.
0004, OK.
However, I get another error as below:
Undefined offset: 5 in play.php on line X.
How do I solve it?

Answer 1

By performing the correct query in the first place.
SELECT staffID
FROM staff
GROUP BY staffID
HAVING COUNT(staffID) > 1
Or if you need to know about all staff IDs:
SELECT staffID, COUNT(staffID) as qty
FROM staff
GROUP BY staffID

Answer 2

Just updating your code and i think you have to check the existence first :
if (isset($staffID[$i]) && isset($staffID[$i+!])) {
if($staffID[$i] == $staffID[$i+1])
{
$remark = "Not OK. Multiple staff ID selected.";
$i++;
}
else
{
$remark = "OK.";
}
$data[$i] = $staffID[$i].','.$remark.'<br />';
}

Answer 3

You get the error at this line:
if($staffID[$i] == $staffID[$i+1])
Say for example your $num_rows variable is 3. Then the $staffID array would have indexes 0, 1 and 2. But what happens at this line I noted is that once $i reaches 2 (which is a valid since it's less than 3) you go and do $i+1 which equals 3. And, as you can see, the index 3 is out of the bounds of the array.
If you wish to do this like you wrote your for loop should be like this:
for($i = 0; $i < $num_rows-1; $i++)
But, there is a better way without even doing the for loop. You can do it all in the while loop by making use of the in_array function. Or as Ignacio pointed out, you could do it all on the SQL level.

Answer 4

Not a good way to solve this problem. You should use GROUP BY query for this.
SELECT staffID, COUNT(staffID) as count
FROM staff
GROUP BY staffID
HAVING count > 1
And also about your error. It's because of
if($staffID[$i] == $staffID[$i+1])
on last iteration $i is equal 4 and $i+1 is 5. But $staffID array has no element with such key