helgun
helgun

Reputation: 67

Display value instead of ID mysqli

When I add $elan_category, I get category_id, instead category_title.

Tried to apply "left join" but no success. I have following tables in database:

enter image description here

function getElanDetail()
{
    global $con;

    if (isset($_GET['elan_id'])) {

        $elan_id = $_GET['elan_id'];
        $get_elan = "select * from elan where elan_id='$elan_id'";
        $run_elan = mysqli_query($con, $get_elan);

        while ($row_elan = mysqli_fetch_array($run_elan)) {

            $elan_id = $row_elan['elan_id'];
            $elan_category = $row_elan['elan_category'];
            $elan_title = $row_elan['elan_title'];
            $elan_description = $row_elan['elan_description'];
            $elan_image = $row_elan['elan_image'];
            $elan_contact = $row_elan['elan_contact'];

            echo "
                $elan_category //Getting ID of category instead Title :(
                $elan_title
                $elan_description
                $elan_image
                $elan_contact
";

        }
    }
}

Upvotes: 0

Views: 57

Answers (2)

u_mulder
u_mulder

Reputation: 54831

With join you can do something like:

$elan_id = $_GET['elan_id'];
$get_elan = "SELECT * FROM `elan` 
    JOIN `categories` ON `categories`.category_id = `elan`.elan_category 
    WHERE `elan`.elan_id='$elan_id'";
$run_elan = mysqli_query($con, $get_elan);
while ($row_elan=mysqli_fetch_array($run_elan)){
    print_r($row_elan);
    // see the keys in $row_elan and use them accordingly
}

For subcategories try this query:

SELECT * FROM `elan` 
JOIN `categories` ON `categories`.category_id = `elan`.elan_category        
JOIN `subcategories` ON `subcategories`.subcategory_id = `elan`.elan_subcategory 
WHERE `elan`.elan_id='$elan_id'

Upvotes: 1

Arjan Shrestha
Arjan Shrestha

Reputation: 193

$elan_category = $row_elan['elan_category'];

add these two lines after above code

$cat = mysqli_fetch_row(mysqli_query($con,"SELECT category_title FROM categories WHERE category_id = $elan_category"));
$cat_name = $cat[0];

$cat_name is your category name enjoy

Upvotes: 0

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