CODEWITHSUNDEEP
phpmysqljoin
wobsoriano
wobsoriano

Reputation: 13434

2 columns on a left join

Hello so I have 2 tables. tbl_records and tbl_guards. On tbl_guards I have guard_id and on tbl_records I have guard_id and guard_id_in. And here is my current code:

try
    {
    $stat = "0";
    $query = "SELECT rec.*, tbl_guard.fname, tbl_guard.lname
FROM tbl_records as rec
LEFT JOIN tbl_guard ON tbl_guard.guard_id = rec.guard_id
LEFT JOIN tbl_records ON tbl_records.guard_id_in = tbl_guard.guard_id
    WHERE rec.stud_id=? AND rec.status=?";

    $stmt  = $dbc->prepare($query);
    $stmt->bindParam(1, $_GET['id']);
    $stmt->bindParam(2, $stat);
    $stmt->execute();
    echo "<table cellpadding='3' class='searchTbl'>";
    echo "<thead>";
    echo "<tr>";
    echo "<th>Actual Date</th>";
    echo "<th>Purpose</th>";
    echo "<th>Destination</th>";
    echo "<th>Exact TO</th>";
    echo "<th>Expected TI</th>";
    echo "<th>Guard</th>";
    echo "<th>Actual TI</th>";
    echo "<th>Guard IN</th>";
    echo "</tr>";
    echo "</thead>";
    while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
        extract($row);
        $guard = $fname . " " . $lname;
        echo "<tbody>";
        echo "<tr>";
        echo "<td>$act_date</td>";
        echo "<td>$purpose</td>";
        echo "<td>$destination</td>";
        echo "<td>$exact_timeout</td>";
        echo "<td>$exp_timein</td>";
        echo "<td>$guard</td>";
        echo "<td>$act_timein</td>";
        echo "<td>$guard</td>";
        echo "</tr>";
        echo "</tbody>";
        }
    }
catch (PDOException $e)
    {
    echo "Error: " . $e->getMessage();
    }
echo "</table>";

Here is tbl_records data. And here is tbl_guard data. Here is the current output.

My problem is it shows the same guard in guard_id and guard_id_in in my code.

Upvotes: 0

Views: 79

Answers (5)

Flunch
Flunch

Reputation: 826

You should select twice the guard table and use aliases for your fields :

SELECT tr1.*, tg1.fname AS FNAME, tg1.lname AS LNAME, 
       tg2.fname AS FNAME_IN, tg2.lname AS LNAME_IN 
FROM tbl_records AS tr1 
LEFT JOIN tbl_guard AS tg1 
     ON tg1.guard_id = tr1.guard_id, 
LEFT JOIN tbl_guard AS tg2 
     ON tg2.guard_id = trl.guard_id_in

then in PHP you'll have more vars : $FNAME, $LNAME, $FNAME_IN, $LNAME_IN

Upvotes: 1

PKeidel
PKeidel

Reputation: 2589

You can use LEFT JOIN multiple times, like:

SELECT tbl_records.*, tbl_guard.fname, tbl_guard.lname
FROM tbl_records as rec
LEFT JOIN tbl_guard ON tbl_guard.guard_id = rec.guard_id
LEFT JOIN tbl_records ON tbl_records.guard_id_in = tbl_guard.guard_id

Upvotes: 4

harry
harry

Reputation: 1007

You can use:

SELECT tbl_records.*, tbl_guard.fname, tbl_guard.lname 
FROM tbl_records 
LEFT JOIN tbl_guard ON (tbl_records.guard_id OR tbl_records.guard_id_in) = tbl_guard.guard_id

Upvotes: 2

Mohit maru
Mohit maru

Reputation: 827

SELECT tr1.*, tg1.fname, tg2.lname FROM tbl_records AS tr1 LEFT JOIN tbl_guard AS tg1 ON tg1.guard_id = tr1.guard_id, LEFT JOIN tbl_guard AS tg2 ON tg2.guard_id = trl.guard_id_in

Upvotes: 1

Flunch
Flunch

Reputation: 826

Just to be sure, your tbl_records table can have a link to two different tbl_guards via guard_id and guard_id_in ?

Maybe you can try : 

SELECT tbl_records.*, tbl_guard.fname, tbl_guard.lname FROM tbl_records LEFT JOIN tbl_guard ON tbl_guard.guard_id IN (tbl_records.guard_id, tbl_records.guard_id_in)

Upvotes: 1

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