Retrieving employee salaries with mysql and php using nymber_format - php

Question

I am returning employee information from a musql db using php. I am getting the salary to returned with a formatted comma using the number_format() method but when the data is returned, all employees have the same salary. How do I get php to return individual salaries from the employees table?
PHP/MySQL
<?php
require_once("db.php");
$sql = "SELECT `*` FROM `employees`";
$results = mysqli_query($connect, $sql) or die(mysql_error());
$row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error());
$salary = $row['salary'];
$rows_sal = number_format($salary);
echo("<table>");
while($row = mysqli_fetch_array($results, MYSQL_BOTH))
{
echo("<tr>");
echo "<td>" . $row['empid'] . '</td>' .
"<td>" . $row['lastname'] . '</td>' .
"<td>" . $row['firstname'] . '</td>' .
"<td>" . $row['department'] . '</td>' .
"<td>" . $row['position'] . '</td>' .
"<td>" . $rows_sal . '</td>';
echo '</br>';
echo('</tr>');
}
echo("</table>");
?>

Answer 1

number_format doesn't work on an array. Do it like this:
<?php
require_once("db.php");
$sql = "SELECT `*` FROM `employees`";
$results = mysqli_query($connect, $sql) or die(mysql_error());
$row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error());
//$salary = $row['salary'];
//$rows_sal = number_format($salary);
echo("<table>");
while($row = mysqli_fetch_array($results, MYSQL_BOTH))
{
echo("<tr>");
echo "<td>" . $row['empid'] . '</td>' .
"<td>" . $row['lastname'] . '</td>' .
"<td>" . $row['firstname'] . '</td>' .
"<td>" . $row['department'] . '</td>' .
"<td>" . $row['position'] . '</td>' .
"<td>" . number_format($row['salary']) . '</td>';
echo '</br>';
echo('</tr>');
}
echo("</table>");
?>

Answer 2

That's because you're working with the salary outside the foreach loop. (And, by the way, you're leaving the first row out of the table).
Before I suggest you how to do it, let me explain what it's happening with your actual code: (follow the comments)
<?php
require_once("db.php");
$sql = "SELECT `*` FROM `employees`";
$results = mysqli_query($connect, $sql) or die(mysql_error());
//here you're asking THE FIRST row:
$row = mysqli_fetch_array($results, MYSQL_BOTH) or die(mysql_error());
$salary = $row['salary'];
//so $rows_sal is the first salary. All this have nothing to do with the iteration below.
$rows_sal = number_format($salary);
echo("<table>");
//here you're asking for the next results (leaving 1st row outside the table!)
while($row = mysqli_fetch_array($results, MYSQL_BOTH))
{
echo("<tr>");
echo "<td>" . $row['empid'] . '</td>' .
"<td>" . $row['lastname'] . '</td>' .
"<td>" . $row['firstname'] . '</td>' .
"<td>" . $row['department'] . '</td>' .
"<td>" . $row['position'] . '</td>' .
//and here you're referencing the value set before the while.
"<td>" . $rows_sal . '</td>';
echo '</br>';
echo('</tr>');
}
echo("</table>");
?>
Now this is what you should do: (follow the comments)
<?php
require_once("db.php");
$sql = "SELECT `*` FROM `employees`";
$results = mysqli_query($connect, $sql) or die(mysql_error());
//don't ask for results yet!
echo("<table>");
while($row = mysqli_fetch_array($results, MYSQL_BOTH)) //including 1st row
{
//now $row is EACH row, NOW you can format your salary:
$salary = $row['salary'];
$rows_sal = number_format($salary);
echo("<tr>");
echo "<td>" . $row['empid'] . '</td>' .
"<td>" . $row['lastname'] . '</td>' .
"<td>" . $row['firstname'] . '</td>' .
"<td>" . $row['department'] . '</td>' .
"<td>" . $row['position'] . '</td>' .
"<td>" . $rows_sal . '</td>';
echo '</br>';
echo('</tr>');
}
echo("</table>");
?>
Hope it helps.