Question

我想在获取数据库时使用+=函数，但是当我使用group by条件时，最终结果不是我想要的，所以请帮我查看我的编码。

我有3个表，表1 = t1：

+-----------+-------------+-------------+-------------+
| ID        | areacode    | landstatus  | pictureid   |
+-----------+-------------+-------------+-------------+
| 1         | 1           | 0           | 0           |
| 2         | 1           | 0           | 0           |
| 3         | 1           | 4           | 1           |
| 4         | 1           | 4           | 2           |
| 5         | 1           | 4           | 1           |
| 6         | 1           | 2           | 1           |
| 7         | 1           | 4           | 4           |
| 8         | 1           | 1           | 0           |
| 9         | 2           | 0           | 0           |
| 10        | 2           | 4           | 1           |
+-----------+-------------+-------------+-------------+

表2 = t2：

+-------+-------------+------------+
| ID    | population  | other      |
+-------+-------------+------------+
| 1     | 10          | 0          |
| 2     | 20          | 0          |
| 3     | 30          | 0          |
| 4     | 40          | 0          |
+-------+-------------+------------+

通常我会像这样查询+-：

让我们说bid=1

$bid = intval($_GET['bid']);
$queryAP = DB::query("
SELECT t1.*
      , t2.population 
  FROM ".DB::table('t1')." t1 
  LEFT 
  JOIN ".DB::table('t2')." t2 
    ON t1.pictureid = t2.id
 WHERE t1.areacode = '$bid' 
   AND t1.landstatus = 4
");
while($rowAP = DB::fetch($queryAP)) { //search all landstatus == 4
    $totalareaP += $rowAP['population'];
}

当用户$totalareaP 80时，query输出将为bid=1。现在我的问题是，如果我想添加服务器任务（自动运行查询时间为＃39; up）将$totalareaP更新为t3 where t2.arecode = t3.id而不$_GET['bid']。

表3称为：t3。

+------+------------+-----------+
| ID   | population | timesup   |
+------+------------+-----------+
| 1    | 0          | timestamp |
| 2    | 0          | timestamp |
+------+------------+-----------+

我尝试编码：

$queryPPADD = DB::query("SELECT t1.*,t2.population FROM ".DB::table('t1')." t1 LEFT JOIN ".DB::table('t2')." t2 ON (t1.pictureid = t2.id) GROUP BY t1.areacode WHERE t1.landstatus = 4");
while($rowPPADD = DB::fetch($queryPPADD )) { //search all landstatus == 4
    $totalareaAAP += $rowPPADD ['population'];
}

当我打印$totalareaAAP没有显示任何值时，我想通过$totalareaAAP更新为t1.areacode更新t3.areacode WHERE t1.areacode = t3.id

谢谢。

Answer 1

“group by”需要一个组功能（在这种情况下为“sum”）

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（请注意vvv $queryPPADD = DB::query("SELECT t1.areacode,sum(t2.population) as population FROM " . DB::table('t1') . " t1 LEFT JOIN " . DB::table('t2') . " t2 ON (t1.pictureid = t2.id) GROUP BY t1.areacode WHERE t1.landstatus = 4");）

在PHP中你必须创建一个数组，

sum(t2.population) as population

PHP代码

array(areacode => population)

感谢 sum / group by ，每个“areacode”在结果中只出现一次。在PHP中，$result = array(); $queryPPADD = DB::query("SELECT t1.areacode,sum(t2.population) as population FROM " . DB::table('t1') . " t1 LEFT JOIN " . DB::table('t2') . " t2 ON (t1.pictureid = t2.id) GROUP BY t1.areacode WHERE t1.landstatus = 4"); while($rowPPADD = DB::fetch($queryPPADD)) { $areacode = $rowPPADD ['areacode']; $result[$areacode] = $rowPPADD ['population']; // just a = }数组有一个条目，其总人口由MySQL总结为“areacode”。

显示结果

$result

Answer 2

$queryPPADD = DB::query("SELECT t1.*,t2.population FROM ".DB::table('t1')." t1 LEFT JOIN ".DB::table('t2')." t2 ON (t1.pictureid = t2.id) GROUP BY t1.areacode WHERE t1.landstatus = 4");
$totalareaAAP = 0;
while($rowPPADD = DB::fetch($queryPPADD )) { //search all landstatus == 4
    echo $land = $rowPPADD ['landstatus'];// see what it is printing
    echo $pic = $rowPPADD ['pictureid'];// see what it is printing
    echo $pop = $rowPPADD ['population'];// see what it is printing
    echo $totalareaAAP += $rowPPADD ['population'];// see what it is printing
}

mysql如何按条件分组？

2 个答案: