Question

我从这个sql中绘制一个图表：

SELECT ljj.job_id,
       Sum(CASE
             WHEN ljj.job_type = "0" THEN 1
             ELSE 0
           END)                                        AS jobcount,
       Sum(CASE
             WHEN ljj.job_type = "1" THEN 1
             ELSE 0
           END)                                        AS interncount,
       Monthname(From_unixtime(ljj.job_timepublished)) AS month
FROM   {local_jobs_job} ljj
       INNER JOIN {local_jobs_location} ljl
               ON ljj.job_location = ljl.location_id
       INNER JOIN {local_companydetail} lc
               ON ljj.job_company_userid = lc.userid
WHERE  lc.link = "1"

目前，它仅返回ljj.job_timepublished中记录的月份。我希望显示从1月到12月的所有月份。如果它没有数据，它将为该月提供0值。

怎么做？

这是我查询数据的php文件：

<?php

require_once('../../config.php');

$data = optional_param('data', null, PARAM_RAW);
$key = optional_param('key', null, PARAM_RAW);
$user = optional_param('user', 0, PARAM_INT);
$id = optional_param('id', 0, PARAM_INT);

global $DB;

///query by location total post
$sql = 'SELECT ljj.job_id, 
SUM(CASE WHEN ljj.job_type = "0" THEN 1 ELSE 0 END) AS jobcount, 
SUM(CASE WHEN ljj.job_type = "1" THEN 1 ELSE 0 END) AS interncount, 
MONTHNAME(FROM_UNIXTIME(ljj.job_timepublished)) AS month FROM {local_jobs_job} ljj 
INNER JOIN {local_jobs_location} ljl ON ljj.job_location = ljl.location_id  
INNER JOIN {local_companydetail} lc ON ljj.job_company_userid = lc.userid 
WHERE lc.link = "1" 
GROUP BY MONTH(FROM_UNIXTIME(ljj.job_timepublished))';


//get the query into record
$data = $DB->get_records_sql($sql);

//put the query into array
$rows = array();

$rows = array_map(function($item) {
return (object) ['c' => [
    (object) ['v' => $item->month, 'f' => null],
    (object) ['v' => intval($item->jobcount), 'f' => null],
    (object) ['v' => intval($item->interncount), 'f' => null]
]];
}, array_values($data));


// prepare return data
$cols = [
(object) ['id' => '', 'label' => 'Month', 'pattern' => '', 'type' => 'string'],
(object) ['id' => '', 'label' => 'Job', 'pattern' => '', 'type' => 'number'],
(object) ['id' => '', 'label' => 'Internship', 'pattern' => '', 'type' => 'number'],

];

$returndata = new stdClass;
$returndata->cols = $cols;
$returndata->rows = $rows;

echo json_encode($returndata);

我使用ajax调用从php文件中调用数据表来绘制图表。

这是sql查询的输出。

{"cols":[{"id":"","label":"Month","pattern":"","type":"string"}, 
{"id":"","label":"Job","pattern":"","type":"number"},
{"id":"","label":"Internship","pattern":"","type":"number"}],
"rows":[{"c":[{"v":"July","f":null},{"v":6,"f":null},{"v":2,"f":null}]},
{"c":[{"v":"August","f":null},{"v":0,"f":null},{"v":3,"f":null}]}]}

Answer 1

您可以使用日历表将每个月引入结果，并将其作为子查询加入到您当前拥有的内容中。我还认为您应该按月汇总数据，这是您目前没有做的。记住所有这些，我们可以编写以下查询：

SELECT
    t1.monthname,
    COALESCE(t2.jobcount, 0)    AS jobcount,
    COALESCE(t2.interncount, 0) AS interncount
FROM
(
    SELECT 'January' AS monthname UNION ALL
    SELECT 'February'  UNION ALL
    SELECT 'March'     UNION ALL
    SELECT 'April'     UNION ALL
    SELECT 'May'       UNION ALL
    SELECT 'June'      UNION ALL
    SELECT 'July'      UNION ALL
    SELECT 'August'    UNION ALL
    SELECT 'September' UNION ALL
    SELECT 'October'   UNION ALL
    SELECT 'November'  UNION ALL
    SELECT 'December'
) t1
LEFT JOIN
(
    SELECT
        MONTHNAME(FROM_UNIXTIME(ljj.job_timepublished)) AS monthname,
        SUM(CASE WHEN ljj.job_type = "0" THEN 1 ELSE 0 END) AS jobcount,
        SUM(CASE WHEN ljj.job_type = "1" THEN 1 ELSE 0 END) AS interncount   
    FROM {local_jobs_job} ljj
    INNER JOIN {local_jobs_location} ljl
        ON ljj.job_location = ljl.location_id
    INNER JOIN {local_companydetail} lc
        ON ljj.job_company_userid = lc.userid
    WHERE lc.link = '1'
    GROUP BY
        MONTHNAME(FROM_UNIXTIME(ljj.job_timepublished))
) t2
    ON t1.monthname = t2.monthname;

请注意，正如@Gordon所指出的，更合理的聚合可能包括月份和年份。但这会使日历表变得更加复杂，并且可能需要一个脚本来生成它。

Answer 2

添加GROUP BY。我建议：

GROUP BY YEAR(From_unixtime(ljj.job_timepublished)), 
         MONTH(From_unixtime(ljj.job_timepublished))

但是，您的代码仅使用MONTHNAME()，因此组合来自不同年份的数据似乎是您的意图。如果那是你的意图：

GROUP BY Monthname(From_unixtime(ljj.job_timepublished))

否则将年份放在SELECT。

中

如何在sql中返回12个月？

2 个答案: