Question

我想通过单个查询获得总ROW类型。让我解释一下。

通常我使用此查询来获取其中一个：

$database->query("SELECT COUNT(*) as c FROM workers WHERE job = 1");
$totalWorkers = $database->fetchObject()->c;

所以，它完美无缺，但还有7种工作类型（0-8）。这只是（1）。

我可以这样做：[太多要求执行，不好]

$database->query("SELECT COUNT(*) as c FROM workers WHERE job = 1");
$totalWorkers1 = $database->fetchObject()->c;

$database->query("SELECT COUNT(*) as c FROM workers WHERE job = 2");
$totalWorkers2 = $database->fetchObject()->c;

但是我不想执行这些多个查询，有没有简单的方法呢？我想得到那样的计数结果：

$totalWorkers2 = $database->fetchObject()->type_1;
$totalWorkers2 = $database->fetchObject()->type_2;

我到处搜索过但我找不到＆amp;想想任何逻辑...

Answer 1

item1[0];

或

SELECT sum(IF(job=1,1,0)) as job1count, sum(IF(job=2,1,0)) as job2count FROM workers;

取决于您是否希望它们在同一记录中

Answer 2

您可以按作业分组结果

$database->query("SELECT job, COUNT(*) as c FROM workers WHERE job in (1,2) group by job");

$tmp = $database->fetchAll();

$result = array();

foreach ($tmp as $o) $result[$o->job] = $o->c;

然后使用数组$result作为$result[1]作业= 1，$result[2]作业= 2

如果您需要所有作业的计数

，您可以按作业WHERE job in (1,2)删除过滤器

单个查询的MYSQL多个COUNT（）

2 个答案: