2

我正在使用 php 和 Mysql 来查询数据库。我想要做的是创建以下内容：

Project Name 1
List todo
List todo
List todo 

Project Name 2
List todo
List todo
List todo

我有以下mysql表：

Project
project.projectId
project.projectName

Todo
todo.todoId
todo.todoEntry
todo.todoProjectid

我可以查询表格并浏览整个项目列表并离开加入待办事项。但是，我想创建一个标题（项目名称），然后在下面进行循环。

我可以根据父项为每个待办事项执行嵌套 sql 查询，但这对我来说听起来不够高效。有人建议改进吗？

4

2 回答 2

2

我让 Gordons 查询更简单一些。它现在完全符合您的要求。

SELECT
    `name`
FROM 
    (
        (SELECT 'project' as `type`, `name`, `id` as `projectid` FROM `Project`)
        UNION ALL
        (SELECT 'todo' as `type`, `name`, `projectid` FROM `Todo`)
    ) as `combined`
ORDER BY
    `project_id`, `type`

PHP 只获取所有行的列表：

$q = $db->query("SELECT `name` FROM ((SELECT 'project' as `type`, `name`, `id` as `project_id` FROM `projects`) union all (SELECT 'todo' as `type`, `name`, `project_id` FROM `todos`)) as `combined` ORDER BY `project_id`, `type`");

while($row = $q->fetch_object()) {
    echo $row->name . '<br />';
}

PHP 使用“复杂”查询获取嵌套列表：

$q = $db->query("SELECT `name`, `type` FROM ((SELECT 'project' as `type`, `name`, `id` as `project_id` FROM `projects`) union all (SELECT 'todo' as `type`, `name`, `project_id` FROM `todos`)) as `combined` ORDER BY `project_id`, `type`");

echo '<ul>';

$needToBeClosed = false;

while($row = $q->fetch_object()) {
    if($row->type == 'project' AND $needToBeClosed) {
        echo '</ul></li>';
        $needToBeClosed = false;
    }

    echo '<li>' . $row->name;

    if($row->type == 'project') {
        echo '<ul>';
        $needToBeClosed = true;
    } else {
        echo '</li>';
    }
}

if($needToBeClosed) {
    echo '</ul></li>';
}

echo '</ul>';

但正如您所看到的，您在查询中尝试做的越多。更多的 PHP 需要以简单的方式使用它。所以你需要在 SQL 和 PHP 之间找到一个平衡点来获得最好的代码。我通常不会采用上述方法，而只是执行多个查询，如下所示：

PHP 在没有“复杂”查询的情况下获取嵌套列表：

$projects = $db->query('SELECT * FROM `projects`');

echo '<ul>';

while($project = $projects->fetch_object()) {
    echo '<li>' . $project->name . '<ul>';
    $todos = $db->query('SELECT * FROM `todos` WHERE `project_id` = ' . $project->id);

    while($todo = $todos->fetch_object()) {
        echo '<li>' . $todo->name . '</li>';
    }

    echo '</ul></li>';
}

echo '</ul>';

您仍然需要根据自己的需要（表名等）修改查询。

于 2013-07-30T12:55:20.857 回答
0

在这种情况下，您有相同的列从查询中出来。有些来自Project，有些来自Todo。挑战在于以正确的方式订购它们：

select which, name
from ((select 'Project' as which, name, 1 as ordering, projectid, NULL as id
       from  Project p
      ) union all
      (select 'List' as which, entry, 2 as ordering, projectid, id
       from Todo t
      )
     ) t
order by projectid,
         ordering,
         id;
于 2013-07-30T12:37:18.937 回答