doubi4531 2016-01-17 18:37
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显示值而不是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
";

        }
    }
}
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2条回答 默认 最新

  • duanjiagu0655 2016-01-17 18:46
    关注

    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'
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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