json and php If, else statement - php

Question

This php code generates a table with some information like: temperature, water health, ph level and other things ...
if (count($auxGateway->getMaquina()) > 0)
{
foreach($auxGateway->getMaquina() as $maq)
{
echo '<div class="cont-maq-uni">';//Cria a div para cada maquina
echo ' <table class="maq">';//Cria uma tabela
echo ' <tr>';
echo ' <td colspan="2">';
echo ' <h2 class="maq-head">'.$maq->getNome().'</h2>';
echo ' </td>';
echo ' </tr>';
foreach($maq->getVariavel() as $var)
{
echo ' <tr class="maq">';
echo ' <td class="maq">';
echo ' <p class="n_font maq">'.$var->getNome().'</p>';
echo ' </td>';
echo ' <td class="maq">';
echo ' <p class="n_font maq">'.(floatval($var->getValor()) * floatval($var->getMultiplicador()) + floatval($var->getSomador())).' '.$var->getUnidadeValor().'</p>';
echo ' </td>';
echo ' </tr>';
}
echo ' </table>';
echo '</div>';
}
}
so I need to create a code to show some images if that information is bigger or less. For example: if 'temperature' <30 then it will show the image (notgood.jpeg). The problem is: how do I name these values since they are inside the MYSQL database? How will I name each row in the table?
Thank you guys!

Answer 1

You can use if..else.. or ternary operators
echo '<img src="' . ($var->getTemperature() < 30 ? 'notgood.jpeg' : 'good.jpeg') . '">' # using ternary operators;
This will give, if $var->getTemperature() < 30 <img src="notgood.jpeg">
else <img src="good.jpeg">