Php mySQL在一个查询中进行两次计数，涉及3个表

我有三张桌子，分别是：城镇；村庄；农民…我需要计算每个村庄有多少农民，每个城镇有多少村庄。我不熟悉php和mysql，所以我的代码还是垃圾。。到目前为止，我已经完成了查询的代码：

$query = "SELECT *, COUNT(DISTINCT village.villageID) AS cnt_village, 
COUNT( farmer.farmerID ) AS cnt_farmers
FROM town LEFT JOIN village ON village.townID = town.townID
LEFT JOIN farmer ON farmer.villageID = village.villageID
GROUP BY town.townID
ORDER BY town.townName";
$result = mysql_query($query);

它已经输出了如下所示的正确计数：

----TOWN--------VILLAGES----FARMERS-----
| Caibiran   |      2     |    23       |
| Culaba     |      7     |    39       |
|      TOTAL |      9     |    62       |

在我尝试搜索某个特定城镇之前，它一直运行良好。发生的情况是，“农民”列的值显示了指定城镇的正确值。但整个表格仍然显示，其他城镇的农民栏显示为0

它表明：

 ----TOWN--------VILLAGES----FARMERS-----
| Caibiran   |      2     |    23       |
| Culaba     |      7     |    0        |
|      TOTAL |      9     |    23       |

但我真正想要的是：

----TOWN--------VILLAGES----FARMERS-----
| Caibiran   |      2     |    23       |
|      TOTAL |      2     |    23       |

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但我不知道怎么做。我已经没有主意了。请帮帮我。

这能解决问题吗

SELECT *, COUNT(DISTINCT village.villageID) AS cnt_village, 
COUNT( farmer.farmerID ) AS cnt_farmers
FROM town
JOIN village ON village.townID = town.townID
JOIN farmer ON farmer.villageID = village.villageID
GROUP BY town.townID
ORDER BY town.townName

---

由于没有任何farmer可以加入，因此应该从结果集中删除行。

我不能不测试就说，但请尝试删除

左连接，然后尝试普通的equijoin？