Question

我已经很长时间没有进行JOIN查询了，所以我很生锈。我所拥有的：一个称为触发器和用户的表。最初，我只是有一个查询，该查询从触发器表中选择了所有信息。这包括一个user_id。好吧，我认为不仅仅是将user_id显示为用户名，而是将其显示为用户名。

记录来自各种用户名，因此显示的用户名必须与触发器表中的user_id匹配到用户表的id。这是我迷路的地方。将t.user_id与u.id匹配后，我不知道该怎么办。如何获取这些记录的u.username？

我仍然希望显示所有记录，就像您执行SELECT * FROM触发器一样。我只想能够使用正确的用户名进行记录。

这是我尝试过的。我从这里开始出错：

SELECT ( u.username ):

任何帮助将不胜感激。数据库表如下所示。

$sql_triggers = "
    SELECT
        t.*,
        u.id,
        u.username
    FROM
        triggers AS t
    LEFT JOIN
        users AS u ON t.user_id = u.id
        SELECT (
            u.username
        )
";
$triggers_stmt = $con->prepare($sql_triggers);
$triggers_stmt->execute();
$triggers_rows = $triggers_stmt->fetchAll(PDO::FETCH_ASSOC);
$triggers_arr = array();
foreach ($triggers_rows as $triggers_row) {
    $trigger_id = $triggers_row['id'];
    $trigger_title = $triggers_row['trigger_name'];
    $trigger_description = $triggers_row['description'];
    $trigger_setting = $triggers_row['setting'];
    $trigger_user = $triggers_row['user_id'];
    $trigger_placement = $triggers_row['placement'];
    $trigger_date = $triggers_row['date_changed'];

触发器= t

CREATE TABLE `triggers` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `trigger_name` varchar(70) COLLATE utf8_unicode_ci NOT NULL,
 `setting` varchar(20) COLLATE utf8_unicode_ci NOT NULL,
 `description` text COLLATE utf8_unicode_ci NOT NULL,
 `user_id` int(11) NOT NULL,
 `placement` varchar(10) COLLATE utf8_unicode_ci NOT NULL,
 `date_changed` datetime NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci

用户= u

CREATE TABLE `users` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `firstname` varchar(55) COLLATE utf8_unicode_ci NOT NULL,
 `lastname` varchar(55) COLLATE utf8_unicode_ci NOT NULL,
 `email` varchar(55) COLLATE utf8_unicode_ci NOT NULL,
 `username` varchar(70) COLLATE utf8_unicode_ci NOT NULL,
 `password` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
 `salt` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
 `joined` datetime NOT NULL,
 `last_login` datetime NOT NULL,
 `group` int(11) NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=24 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_c

Answer 1

对于sql代码，您只需要

SELECT
    t.*,
    u.id,
    u.username
FROM  triggers AS t
LEFT JOIN  users AS u ON t.user_id = u.id

如何联接两个SQL表以基于匹配的ID获得特定的其他数据

1 个答案: