php - 连接两个表以显示数据

Question

我尝试连接两个表。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

子类别

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')));
        }
?>

我正在获取所有详细信息，但问题是我没有在结果中获取 category.name。谁能帮忙

P.S 我已经使用过 sql，现在稍后会更改它我关心的是功能部分

Answer 1

subcategory.name 会覆盖您的类别名称。您必须像这样为您的子类别名称使用别名:

 $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 ");