Question

我有两个表（为简洁起见，删除了多余的列）。

动作：

$(document).on('click', '#export-csv', function(){
    $.ajax({
        url: export_url,
        type: 'post',
        data:{ validations: validations },
        dataType: 'json',
        cache: false,
        timeout: 1000000,
        success: function (data) {
            var resp = data.response;
            console.log(resp);
        },
        error: function (error) {
            console.log(error);
        }
    });
});

user_logs：

public function exportCsvAction()
{
    $request = $this->getRequest();
    if ($request->isPost()) {

        $input = $request->getPost();
        $response = $this->getResponse();

        $file = fopen('php://output', 'w');
        foreach ($input['validations'] as $line) {
            fputcsv($file, $line, chr(9), '"');
        }
        fclose($file);

        header('Content-type: text/csv; charset=utf-8');
        header('Content-Disposition: attachment; filename="validations-' . time() . '.csv"');

        return $response->setContent(\Zend\Json\Json::encode(array('response' => true)));
    }
    return $this->getResponse()->setStatusCode(400);
}

我希望将这些表加入用户ID以及user_logs时间与操作时间之间的最小差异（例如，用户首先会在记录其操作之前登录users_logs表。

在同一用户的多个操作中准确无误的最佳方式是什么？

在此示例中，理想输出为

id            user_id            action           created_at 
 1               1                open            2015-01-01 03:03:00
 2               1                closed          2015-02-03 02:13:00
 3               2                open            2015-02-04 03:23:00

Answer 1

这可以通过内部联接实现，如下所示：

SELECT 
   user_logs.user_id,
   actions.action,
   user_logs.source,
FROM user_logs

INNER JOIN actions ON user_logs.user_id = actions.user_id

INNER JOIN
(
    SELECT
        MIN(TIMESTAMPDIFF(SECOND, user_logs.created_at, actions.created_at)) as minTime
    FROM user_logs
    INNER JOIN actions ON user_logs.user_id = actions.user_id
    WHERE actions.created_at > user_logs.created_at
    GROUP BY actions.id
) as minSelQ ON TIMESTAMPDIFF(SECOND, user_logs.created_at, actions.created_at) = minSelQ.minTime

GROUP BY actions.id

匹配时间戳之间的最小差异

1 个答案: