Question

我试图加入两张桌子。一个是类别，另一个是子类别。类别表中的id成为子类别表中的cat id

类别

id  name  catimage  
2   cat1  image1

子类

id  name     subcatimage  catid
1   subcat1   image2       2

代码是

<?php
ob_start();
require_once('config.php');

    $selectsubcategory = mysql_query("SELECT category.name, subcategory.name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");    
    $posts = array();
    if(mysql_num_rows($selectsubcategory)) 
        {
            while($post = mysql_fetch_assoc($selectsubcategory)) 
                {
                    $posts[] = $post;
                }
            header('Content-type: application/json');
            echo stripslashes(json_encode(array('subcategorylist'=>$posts)));
        } 
    else 
        {
            header('Content-type: application/json');
            echo stripslashes(json_encode(array('subcategorylist'=>'No subcategory')));
        }
?>

我收到所有细节，但问题是我没有在结果中获得category.name。任何人都可以帮忙

P.S我已经使用了sql，现在稍后会更改它我的关注点是功能部分

Answer 1

subcategory.name会覆盖您的类别名称。您必须使用别名作为子类别名称：

 $selectsubcategory = mysql_query("SELECT category.name, subcategory.name AS sub_name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");

Answer 2

这可能会对你有帮助......

此处 MainCatName 将显示类别名称字段数据...和 SubCatName 将显示为子类别名称字段数据...

SELECT
    `category`.`name`  AS MainCatName
    , `subcategory`.`name` AS SubCatName
FROM
    `category`
    INNER JOIN `subcategory` 
        ON (`category`.`id` = `subcategory`.`catid`);

Answer 3

SELECT
      Category.name, category.catimage,
      subcategory.name, subcategory.id,
      subcategory.catid
FROM
     category INNER JOIN subcategory
WHERE
      category.id = subcategory.catid

连接两个表来显示数据

3 个答案: