Question

我已经从一个月的表数据中编码了json数组，如下所示，但是如何在组合所有数据后将多个数据表与其他月组合在一个json数组和GROUP BY部门中？

<?php

include("dbconfig.php");

$sql = "SELECT dept, SUM(ttlot) AS sum_ot, SUM(ttlnorm) AS sum_norm
        FROM month_jan
        GROUP BY dept
        ORDER BY sum_ot DESC";

$result = mysqli_query($dbconfig, $sql) or die("Error in Selecting " . mysqli_error($dbconfig));

$category = array();
$category['name'] = 'Dept';

$series1 = array();
$series1['name'] = 'Normal';

$series2 = array();
$series2['name'] = 'OT';

$emparray = array();
while ($row = mysqli_fetch_assoc($result)) {

$category['data'][] = $row['dept'];
$series1['data'][] = $row['sum_norm'];
$series2['data'][] = $row['sum_ot'];

}

$result = array();
array_push($result,$category);
array_push($result,$series1);
array_push($result,$series2);

$json = json_encode($result,JSON_NUMERIC_CHECK);

echo $json;

mysqli_close($dbconfig);
?>

1月份的产出：

[{"name":"Dept","data":["CNC","MACH","ANOD","BUFF","CAST","POLISH","SL","EPT","TUMB","TOOL","SHOT","QC","LOG","MAIN","LC","WWT","OG","NPD","E-COAT","SFT"]},{"name":"Normal","data":[47429.1,39975.7,34553.8,49075.9,28316.3,21237.1,13492.5,5848.2,7691.1,6963.9,5636.1,7555.8,5821.9,2161.2,1812,1191.7,1479.1,1299.6,11542.6,602]},{"name":"OT","data":[20041,17874,14431,13535.5,8800.5,5613.5,3569.5,3101,2327,2278,2237,2142,1810,942,690,456,297,110.5,66,50.5]}]

结合四个月后我想要的结果是什么：

[{"name":"Month","data":["Jan","Feb","Mac","Apr"]},{"name":"Normal","data":[504291,409757,295538,430759]},{"name":"OT","data":[89041,96874,81431,80535]}]

有没有人可以帮我解决问题？

Answer 1

我通过打印它们来手动完成，如果您在对两个表的JSON进行编码时遇到问题，我认为这是一个替代解决方案。有时，具有相同名称的行会在 json_encode（）函数上产生错误。

a subclass of C

？>

如果您的行具有相同的名称，则应像在上面的示例中那样重命名另一个（questionaire_name，questionaire_description）。

它将产生以下输出：

<?php
session_start();

require("../config.php");
$output = array();

$id = $_GET['id'];

if(isset($_SESSION['user'])){
    $f_data ='';
    $sql = "SELECT * FROM quiz WHERE subject_id=$id ORDER BY id DESC";
    $query=$conn->query($sql);
    while($row=$query->fetch_array()){
        $sql2 = "SELECT * FROM q_details WHERE id=$row[2] ORDER BY id DESC LIMIT 1";
        $query2=$conn->query($sql2);
        while($row2=$query2->fetch_array()){
            $f_data .= '{
                    "id":"'.$row[0].'",
                    "subject_id":"'.$row[1].'",
                    "questionaire_id":"'.$row[2].'",
                    "name":"'.$row[3].'",
                    "description":"'.$row[4].'",
                    "start":"'.$row[5].'",
                    "end":"'.$row[6].'",
                    "date":"'.$row[7].'",
                    "questionaire_name":"'.$row2[2].'",
                    "questionaire_description":"'.$row2[3].'"
                    },';
        }
    }
     $f_data = substr($f_data, 0, -1);
     echo '['.$f_data.']';
}
else{
    echo"<script>window.open('http://localhost/lnhs_app/#/','_self')</script>";
}

Answer 2

使用MySQL的JSON功能和GROUP_CONCAT的组合，你可以通过组合表来做一些非常棒的事情：

您想要的结果：

[{"name":"Month","data":["Jan","Feb","Mac","Apr"]},{"name":"Normal","data":[504291,409757,295538,430759]},{"name":"OT","data":[89041,96874,81431,80535]}]

利用GROUP_CONCAT，您不需要连接表，而只需要按字段分组，以便汇总字段：

SELECT
  CONCAT('{\"NAME\" : \"NORMAL\",',
  '\"data\": [',  GROUP_CONCAT( '"', SUM(ttlnorm), '"'), ']},',
  '{\"NAME\" : \"OT\",',
  '\"data\": [',  GROUP_CONCAT( '"', SUM(ttlot), '"'), ']}',
  )
  from month_jan
  GROUP BY dept;

如果您使用按月存储的表结构（“month_jan”），则需要在几个月内执行联合或更改表结构，以便所需的所有月份都包含在同一个表中。您可以根据GROUP BY执行多个级别的串联。例如，如果您有两个按字段分组，则可以嵌套JSON：例如：GROUP BY DEPT，MONTH

SELECT
  CONCAT('{',
  '\"DEPT\" :\"', dept,'\",',
  '{\"NAME\" : \"Month\",',
  '\"data\": [',  GROUP_CONCAT( '"', month, '"'), ']},',
  '{\"NAME\" : \"NORMAL\",',
  '\"data\": [',  GROUP_CONCAT( '"', SUM(ttlnorm), '"'), ']},',
  '{\"NAME\" : \"OT\",',
  '\"data\": [',  GROUP_CONCAT( '"', SUM(ttlot), '"'), ']}',
  )
  from ttl_ot_norm
  GROUP BY dept, month;

因此，您的GROUP_CONCAT中的数据将根据您表达的格式计算出来。

另一种解决方案是创建一个基于cron作业或表触发器进行更新的表，并将此数据放入JSON字段类型中，以便以低廉的CPU成本继续为检索做好准备。这样，您就不会在每个查询上加入连接和表连接的额外开销。

您可以通过组合GROUP_CONCAT并对表进行分组来跳过大量PHP聚合。多年来我用这个以及MySQL的JSON功能组合（下面的链接）节省了数百小时的编程

版本8引入了这些功能，您无需使用JSON_ARRAYAGG和其他出色的JSON功能创建自己的JSON字符串。虽然，上述内容适用于版本5及更高版本。

https://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

https://dev.mysql.com/doc/refman/5.7/en/json-creation-functions.html#function_json-array

版本8： https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html#function_group-concat

如何组合多个表数据并将json编码为一个数组？

2 个答案: