How can I mark the values from one table, that are not in another table (mySQL)?

Question

In mySQL this is my table animals :

+-----------+--------+--------+
| animal_ID |  name  | animal |
+-----------+--------+--------+
|         1 | alan   | dog    |
|         2 | sam    | frog   |
|         3 | marion | cat    |
|         4 | george | rabbit |
|         5 | bob    | bird   |
+-----------+--------+--------+

and this is my table orders

+----------+-----------+
|   date   | animal_ID |
+----------+-----------+
| 02.03.16 |         4 |
| 12.04.16 |         3 |
| 18.07.16 |         1 |
+----------+-----------+

I want to list all animals but mark the animals that are NOT in the orders table red. This is my expected result

 <table > <tr> <td>alan</td> </tr> <tr> <td style="color:red">sam</td> </tr> <tr> <td>marion</td> </tr> <tr> <td>george</td> </tr> <tr> <td style="color:red">bob</td> </tr> </table> 

I have now the problem, that I do not know how to list all animals from the animal table, I just get the animals in my list that are not in the orders table:

$sql = 'SELECT  *  FROM animals a WHERE EXISTS (SELECT * FROM orders o WHERE  o.animal_ID = a.animal_ID)';

Here my result is this:

  <table > <tr> <td>alan</td> </tr> <tr> <td>marion</td> </tr> <tr> <td>george</td> </tr> </table> 

Answer 1

You need to make this script using two query.

 1. Query for all the animals.
 2. Query with the animals ID in the orders.

The first one is:

$sql = 'SELECT  *  FROM animals";

Loop this and check in the order table inside the first loop-

$sql2 = "SELECT * FROM orders WHERE  animal_ID = fetched_animal_ID";
if(mysqli_num_rows($sql2) > 0){
    // print the normalrows here
}else{
    // print the red rows here
}

==============================================================

You can do this using the LEFT JOIN as well-

$sql = "SELECT *, o.animal_ID as o_animal FROM animals a LEFT JOIN orders o 
        ON a.animal_ID = o.animal_ID";

Online Example

Now you need to check the o_animal if it is null or empty then use the red rows and else the normal.

If you have any question please ask me.

Answer 2

You can try another method without using join query.

<?php
include 'conn.php';
$sq=mysqli_query($conn,'select * from orders');
$i=0;
$ar=array();
while($re=mysqli_fetch_array($sq))
{
$anid=$re['animal_ID'];
$ar[$i]=$anid;
$i++;
}
$sql=mysqli_query($conn,"select * from animals");
?>
<table >
    <?php

while($all=mysqli_fetch_array($sql))
{
    if(in_array($all['animal_ID'],$ar))
    {
       ?>

    <tr>

        <td><?php echo $all['name'];?></td>
    </tr
    <?php
    }

else
{
  ?>  <tr>

        <td style="color:red"><?php echo $all['name'];?></td>
    </tr>
<?php
}
}
?>
</table>

Answer 3

Simple negation?

$sql = 'SELECT  *  FROM animals a WHERE NOT EXISTS (SELECT * FROM orders o WHERE  o.animal_ID = a.animal_ID)';

Or maybe grab the count with a sub query, highlight the zeroes...

$sql = 'SELECT  a.*,
        (SELECT COUNT(*) FROM orders WHERE animal_ID = a.animal_ID)
        FROM animals a';