搜索表单选择结果[重复]

I have two tables, one called users with user_id, user_name, school_id and subcat_id, and the other called schools with school_id, school_name, and school_decile. I've made a search form with select statement which gathers results from the first table based on which subcat_id and/or school_id the user has chosen on the search form. Now, I want to display the results with the user_name, school_name and school_decile. But as in my select statements, I only selected from the first table and not from the second one. How can I display my results of the users with the school_decile and school_name if they weren't in the select statement?

if ($subcat_number !== '0') {
    if ($school_number === '0') { //if a school isn't selected
        $sql = "SELECT * FROM users WHERE subcat_id=$subcat_number";
    } else { //if a school is selected
        $sql = "SELECT * FROM users WHERE subcat_id=$subcat_number AND school_id=$school_number ";
    }
}

$result = mysqli_query($con, $sql);
$found = mysqli_num_rows($result);

//results table
echo "<table>
<tr>
<th id='namecol'>Search Results:</th>
<th id='schoolcol'></th>
</tr><tr>";

if ($found > 0) {
    while ($row = mysqli_fetch_array($result)) {
        echo "<td id='namecol'>Name: " . $row['name'] . " </td><td>School: " . $row['school_name'] . " <br>Decile of School: " . $row['school_decile'] . "</td>";
//obviously, the $row['school_name'] and $row['school_decile'] doesn't work as they weren't selected in the select statements        
    }
} else {
    echo "<td>No Expert Found</td></tr>";
}
echo "</table>";

New Edit After following the suggestion below of doing a JOIN, this warning came up and the shown result is wrong.

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /var/www/projectv5/search.php on line 74

the line 74 is

$found = mysqli_num_rows($result);

</div>

As @Rasclatt suggested, you need to use JOIN

if ($subcat_number !== '0') {
    if ($school_number === '0') { //if a school isn't selected
        $sql = "SELECT name, school_name, school_decile  FROM users INNER JOIN schools on users.school_id = schools.school_id  WHERE subcat_id=$subcat_number";
    } else { //if a school is selected
        $sql = "SELECT name, school_name, school_decile FROM users INNER JOIN schools on users.school_id = schools.school_id WHERE subcat_id=$subcat_number AND school_id=$school_number ";
    }
}

$result = mysqli_query($con, $sql);
$found = mysqli_num_rows($result);

//results table
echo "<table>
<tr>
<th id='namecol'>Search Results:</th>
<th id='schoolcol'></th>
</tr><tr>";

if ($found > 0) {
    while ($row = mysqli_fetch_array($result)) {
        echo "<td id='namecol'>Name: " . $row['name'] . " </td><td>School: " . $row['school_name'] . " <br>Decile of School: " . $row['school_decile'] . "</td>";

    }
} else {
    echo "<td>No Expert Found</td></tr>";
}
echo "</table>";

You should be doing something like this:

<?php
    if($subcat_number !== '0') {
            $sql = "SELECT users.name, users.school_name, users.school_decile FROM users INNER JOIN schools on users.school_id = schools.school_id  WHERE users.subcat_id = $subcat_number";

            if($school_number === '0')
                $sql .= " AND schools.school_id = '$school_number'";

            $result =   mysqli_query($con, $sql);
            $found  =   mysqli_num_rows($result); ?>
    <table>
        <tr>
            <th id='namecol'>Search Results:</th>
            <th id='schoolcol'></th>
        </tr><?php

            if($found > 0) {
                    while ($row = mysqli_fetch_array($result)) { ?>
        <tr>
            <td id='namecol'>Name: <?php echo $row['name']; ?></td>
            <td>School: <?php echo $row['school_name']; ?><br>Decile of School: <?php echo $row['school_decile']; ?></td>
        </tr>
                <?php   }
                }
            else { ?>
        <tr>
            <td>No Expert Found</td>
        </tr><?php 
                } ?>
    </table><?php
        } ?>