Question

我就是无法正常工作。我有一个名为“类别”的表：

table name:  categories
id
name

和这样的帖子表：

table name: posts
id
category_id
title
body

我正在编写一个控制器/模型/视图以列出category.name和属于at类别的帖子数。我正在崩溃并且正在燃烧该怎么做。这是部分尝试。

$this->db->select('
                posts.*,
                categories.*,
                COUNT(posts.category_id) as num_posts
            ');
$this->db->from('posts');
$this->db->join('categories', 'categories.id = posts.category_id');
$xx = $this->db->get();

    echo '<pre>';
    print_r($xx);
    echo '</pre>';
    exit();

非常感谢任何帮助。搜索了我的stackoverflow，但是看不到CI3解决方案。

新更新。此sql有效：

SELECT
    categories.id,
    categories.name,
    Count(posts.id) as cnt
FROM
    categories
LEFT JOIN
    posts
ON
    categories.id=posts.category_id
GROUP BY
    categories.name

当我运行它时，我得到的是以下期望值：

id     name                              cnt
---------------------------------------------
80  |  24354234234                    |  2
76  |  Associations and Organizations |  9

但是当我这样更改为活动记录时：

$this->db->select('categories.id, categories.name, COUNT(posts.id) as num_posts');
$this->db->from('categories');
$this->db->join('posts', 'categories.id = posts.category_id');
$this->db->group_by('categories.name');
$xx = $this->db->get();

我print_r（$ xx）却看不到我的任何数据。这就是我得到的：

CI_DB_mysqli_result Object
(
[conn_id] => mysqli Object
    (
        [affected_rows] => 2
        [client_info] => 5.5.59
        [client_version] => 50559
        [connect_errno] => 0
        [connect_error] => 
        [errno] => 0
        [error] => 
        [error_list] => Array
            (
            )

        [field_count] => 3
        [host_info] => Localhost via UNIX socket
        [info] => 
        [insert_id] => 0
        [server_info] => 5.5.58-0ubuntu0.14.04.1
        [server_version] => 50558
        [stat] => Uptime: 7546485  Threads: 2  Questions: 523149  Slow queries: 0  Opens: 257  Flush tables: 1  Open tables: 200  Queries per second avg: 0.069
        [sqlstate] => 00000
        [protocol_version] => 10
        [thread_id] => 42310
        [warning_count] => 0
    )

[result_id] => mysqli_result Object
    (
        [current_field] => 0
        [field_count] => 3
        [lengths] => 
        [num_rows] => 2
        [type] => 0
    )

[result_array] => Array
    (
    )

[result_object] => Array
    (
    )

[custom_result_object] => Array
    (
    )

[current_row] => 0
[num_rows] => 
[row_data] => 
)

仍然不知道如何获取数组。任何帮助表示赞赏。

Answer 1

我发现了这一点，这解释了为什么值是空白的：Codeigniter query returning blank

更多文档在这里。 https://www.codeigniter.com/user_guide/database/results.html

因此，这将回显值：

... extension of previous code
$xx = $this->db->get();
print_r($xx->result_array());

CI3-计算和加入活动记录

1 个答案: