CODEWITHSUNDEEP
phpjoinmysqli
Rick Skeels
Rick Skeels

Reputation: 513

Using JOIN to display data in a table

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();

?>

Upvotes: 0

Views: 3493

Answers (2)

DevlshOne
DevlshOne

Reputation: 8457

<?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();
?>

Upvotes: 1

Sean
Sean

Reputation: 12433

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>";

Upvotes: 2

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