MySql Table with different colors by value - php

Question

I am using mysql queries to fetch data from db. All my data are showing fine in tables. Now I want to color the status by value less than 2 or more than 3.
The below code is not working.
Need help.
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result) > 0)
{
$output .= '<div class="table-responsive">
<table class="table table bordered">
<tr>
<th>name</th>
<th>genre</th>
<th>time</th>
<th>status</th>
<th>more...</th>
</tr>';
?>
<?php
function status_style($row) {
if ($row < 2) return 'background-color: #ff0000'; //red
if ($row > 3) return 'background-color: #33cc33'; //green
return '';
}
?>
<?php
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td>'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';
}
echo $output;
{
echo 'Data Not Found';
}
?>

Answer 1

You have already created function to color the rows but haven't called the function during showing output. That's why it is not working.
Change your $output variable with:
$output .= '
<tr>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td style="'.status_style($row["status"]).'">'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';

Answer 2

You can try something like this:
<?php
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result) > 0)
{
$output .= '<div class="table-responsive">
<table class="table table bordered">
<tr>
<th>name</th>
<th>genre</th>
<th>time</th>
<th>status</th>
<th>more...</th>
</tr>';
function status_style($row) {
if ($row < 2) return $color = '#ff0000'; //red
if ($row > 3) return $color = '#33cc33'; //green
return $color = '';
}
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr style='"'background-color: '"' . $color . '"'>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td>'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';
}
echo $output;
{
echo 'Data Not Found';
}
You didn't do anything with the background color.