PHP elseif else - php

Question

With the following code, I would like to plot the image sun.png when sky is clear, cloud when the sky is cloudy and variable in other cases...but something fails...I always get the image variable.png
<?php
if($sky == "clear" ) {
echo '<img src="images/sun.png" width="40">';
}
elseif ($sky == "cloudy" ){
echo '<img src="images/cloud.png" width="40">';
}
else {
echo '<img src="images/variable.png" width="40">';
}
?>
I consult the database using this code #Jack Goodman
$data_query = mysqli_query($conexionbd,'select * from `weather` where `data` = "2017-03-22" and (`num` = "1" or `num` = "2" or `num` = "3")');
while($data = mysqli_fetch_assoc($data_query)){ ?>
At the end I have solved it, my code had a mistake, the right code is
<?php
if($data['sky'] == "clear" ) {
echo '<img src="images/sun.png" width="40">';
}
elseif ($data['sky'] == "cloudy" ){
echo '<img src="images/cloud.png" width="40">';
}
else {
echo '<img src="images/variable.png" width="40">';
}
?>

Answer 1

Your problem is somewhere in the variable sky. I downloaded pictures and tested it with this code by changing the variable's value and it all works fine.
$sky = "cloudy";
if($sky == "clear" )
{
echo '<img src="sun.jpg" width="40">';
}
elseif ($sky == "cloudy" )
{
echo '<img src="cloud.jpg" width="40">';
}
else
{
echo '<img src="variable.png" width="40">';
}
Show me the code where you get the sky's value please.

Answer 2

Might be clearer using a switch statement, then you could have a catch-all for unmet conditions and use it to debug your issue:
switch($sky){
case 'clear':
echo '<img src="images/sun.png" width="40">';
break;
case 'cloudy':
echo '<img src="images/cloud.png" width="40">';
break;
default:
echo '$sky is something unknown';
var_dump($sky);
break;
}