Php 按条件分组

我必须在php中显示一些数据，数据结构如下

id  channel_id  added_by

1     1          9999
2     1          195
3     8          9999
4     1          180
5     1          195
6     8          9999
7     1          9999
8     1          195
9     8          9999

channel_id   total  added_by

1            2           9999
1            4          -1

我只需要像这样显示通道id=1的总计数

id  channel_id  added_by

1     1          9999
2     1          195
3     8          9999
4     1          180
5     1          195
6     8          9999
7     1          9999
8     1          195
9     8          9999

channel_id   total  added_by

1            2           9999
1            4          -1

所有通道id的类似计数（通过9999相加）和所有其他相加的值均假定为（-1）

我能像你一样做这件事

SELECT channel_id, added_by, sum(id) as total 
from table 
where channel_id = 1 
group by channel_id  ,added_by;

但它给出的结果不是我所需要的

channel_id  added_by total 
1            9999      2
1             195      3
1             180      1

使用表达式，您可以确定“修改的”

added\u by

值，并改为对修改后的值进行分组

另外，

SUM（id）

也不会给出正确的行数。它宁愿将

id

值相加。您可以改为使用来计算行数：

SELECT channel_id, 
       CASE 
         WHEN added_by <> 9999 THEN -1 
         ELSE 9999 
       END AS modified_added_by, 
       COUNT(*) as total 
FROM your_table_name 
WHERE channel_id = 1 
GROUP BY channel_id, modified_added_by

---

使用表达式，您可以确定由添加的“修改的”值，并对修改后的值进行分组

另外，

SUM（id）

也不会给出正确的行数。它宁愿将

id

值相加。您可以改为使用来计算行数：

SELECT channel_id, 
       CASE 
         WHEN added_by <> 9999 THEN -1 
         ELSE 9999 
       END AS modified_added_by, 
       COUNT(*) as total 
FROM your_table_name 
WHERE channel_id = 1 
GROUP BY channel_id, modified_added_by

---

您可以尝试以下方法：

SELECT channel_id,
if(9999,9999,-1) as added_by,
sum(id) as total
from table
where channel_id = 1
group by channel_id,added_by;

您可以尝试以下方法：

SELECT channel_id,
if(9999,9999,-1) as added_by,
sum(id) as total
from table
where channel_id = 1
group by channel_id,added_by;

---

希望这对你有帮助

---

选择count（*）作为总计，channel_id，added_by from table group by channel_id，added_by
希望这对您有所帮助

选择count（*）作为总计，channel_id，added_by from table group by channel_id，added_by


它将如何在按列添加的\u中生成-1。它将如何在按列添加的\u中生成-1。