Using JOIN to display data in a table - php

Question

Thanks for reading my question
i am trying to make *clients_id* from the table repair_jobs appear as the name from the table contacts
but i am having no luck
i have got 2 sql query's is this wrong?
the 1st
$query = "select * from repair_jobs";
this helps me display the information i need regarding the fields from repair_jobs and works
this is the 2nd
$query = "SELECT repair_jobs.client_id, contacts.name
FROM repair_jobs
INNER JOIN contacts
ON repair_jobs.client_id=contacts.name";
under that i have this to try to display the name of the client
echo "<td>{$client_id}</td>";
but it is only displaying the number and not the data (clients name) that i need
am i missing something?
Additional information
The client_id (repair_jobs) is a number and is the same as id (contacts) but wanting to display the name (contacts)
CLIENTS
Id – name – surname – phone – address
REPAIRS
Id – clients_id (same as id in clients) – unit – date – price
current code
<?php
//include database connection
include 'db_connect.php';
//query all records from the database
$query = "select * from repair_jobs";
//execute the query
$result = $mysqli->query( $query );
//get number of rows returned
$num_results = $result->num_rows;
//this will link us to our add.php to create new record
if( $num_results > 0){ //it means there's already a database record
//start table
//creating our table heading
echo " <table class='table_basic'>";
echo "<thead><tr>";
echo "<th>Job #</th>";
echo "<th>Name Of Unit</th>";
echo "<th>Client</th>";
echo "<th>Estimated Value</th>";
echo "</thead></tr><tbody><tr>";
//loop to show each records
while( $row = $result->fetch_assoc() ){
//extract row
//this will make $row['firstname'] to
//just $firstname only
extract($row);
//creating new table row per record
echo "<tr>";
echo "<td width='40px'><a href='rdetails.php?id={$id}'># {$id}</a></td>";
echo "<td>{$rmake} {$rmodel}</td>";
$query = "SELECT rj.client_id, c.name AS client_name FROM repair_jobs rj INNER JOIN contacts c ON rj.client_id=c.id";
echo "<td>{$client_name}</td>";
echo '<td align="center"><span class="badge badge-success">£';
$lhours = $labour;
$repaircosts = $ourcosts;
$labourpay = $labourcharge;
$sum_total = $repaircosts +($lhours * $labourpay);
print ($sum_total);
echo '</span></td>';
echo "</td>";
echo "";
}
echo "</tr></table>";//end table
}else{
//if database table is empty
echo "No records found.";
}
//disconnect from database
$result->free();
$mysqli->close();
?>

Answer 1

Change your 1st query to you join query, as there is no reason to do a 2nd query in the middle of your code. (also you never executed that query anyway).
//query all records from the database
$query = "SELECT repair_jobs.*, contacts.name as client_name
FROM repair_jobs
INNER JOIN contacts
ON repair_jobs.client_id=contacts.id";
Then in your table keep the $client_name
echo "<td>{$client_name}</td>";

Answer 2

<?php
include 'db_connect.php';
$query = "SELECT rj.Id AS job_number, rj.unit, rj.make, rj.model, c.name AS client_name, rj.price FROM repair_jobs rj INNER JOIN contacts c ON rj.clients_id = c.id ORDER BY c.date";
$result = $mysqli->query( $query );
$num_results = $result->num_rows;
if( $num_results > 0){ //it means there's already a database record
echo " <table class='table_basic'>";
echo "<thead><tr>";
echo "<th>Job #</th>";
echo "<th>Name Of Unit</th>";
echo "<th>Client</th>";
echo "<th>Estimated Value</th>";
echo "</tr></thead><tbody>";
while( $row = $result->fetch_assoc() ){
extract($row);
echo "<tr>";
echo "<td width='40px'><a href='rdetails.php?id={$job_number}'>#{$job_number}</a></td>";
echo "<td>{$make} {$model}</td>";
echo "<td>{$client_name}</td>";
echo "<td align='center'><span class='badge badge-success'>£";
$lhours = $labour;
$repaircosts = $ourcosts;
$labourpay = $labourcharge;
$sum_total = $repaircosts +($lhours * $labourpay);
echo $sum_total;
echo '</span></td>';
echo "</td>";
echo "</tr>";
}
echo "</tbody></table>";
} else {
echo "No records found.";
}
$result->free();
$mysqli->close();
?>