mysql的多在條款和計數

Question

我有以下SQL查詢：

SELECT field_number, COUNT(*) AS c FROM lead_detail WHERE field_number < 3 AND field_number > 2 AND lead_id IN (147,146,145,127,113) GROUP BY field_number 

它返回：

field_number c 
2.1   5 
2.2    5 
2.3    1 
2.4    1 
2.5    5 

我只希望根據用戶的選擇，所以我用來顯示一些字段：

SELECT field_number, COUNT(*) AS c FROM lead_detail WHERE field_number < 3 AND field_number > 2 AND lead_id IN (147,146,145,127,113,109,91,65,57,56,49) AND field_number IN (2.1,2.3) GROUP BY field_number 

但是現在有了這個查詢即時得到0個結果。我試圖把AND field_number = 2.1沒有成功。 任何線索爲什麼會發生這種情況，或者我可以如何避免這種情況？ lead_id id`s不都field_numbers，我想這也許就是這個原因。例如，在lead_id 147我們可以發現field_numbers 2.1和2.3，並在lead_id 146，我們發現2.4和2.5 感謝您的幫助很大

Answer 1

呦在查詢一個錯字錯誤是假設是COUNT(*) AS c 也有兩個不必要的聲明沒有使用field_number < 3 AND field_number > 2或field_number IN (2.1,2.3) 所以最終的查詢應該是這樣的：

SELECT field_number, COUNT(*) AS c 
FROM lead_detail 
WHERE lead_id IN (147,146,145,127,113,109,91,65,57,56,49) 
    AND field_number IN (2.1,2.3) 
GROUP BY field_number