Question

計數傳真由夜班晚上10點完成的 - 早上7點，

日期和時間，在不同的領域

$stmt = $db -> prepare("SELECT count(*) FROM fax WHERE 
       date BETWEEN CURDATE() and CURDATE() + INTERVAL 1 DAY 
        and time >= '22:00:00' and time <= '7:00:00' 
         and shift='GY' 
          and complete=1");     
    $stmt -> execute(); 
    echo $GY_COMP = $stmt -> fetchColumn(); 

這個查詢總是產生0，但有數據。

有人可以幫我查詢嗎？ 感謝提前:)

Answer 1

我認爲這個問題是你的時間比較：您有效地過濾掉一切與你的time >= '22:00:00' and time <= '7:00:00'

希望這有助於：http://sqlfiddle.com/#!2/45108/7/0

SELECT * FROM fax 
WHERE date BETWEEN CURDATE() and CURDATE() + INTERVAL 1 DAY 
and 
((time >= '22:00' and time <= '23:59') 
or 
(time >= '0:00' and time <= '7:00')) 
and shift='GY' 
and complete=1 

Answer 2

感謝爲最大時間比較， 我已經對語法做了更改

SELECT count(*) FROM fax 
    WHERE ((date = CURDATE() and (time >='22:00' and time <= '23:59')) 
      or (date=CURDATE() + INTERVAL 1 DAY and (time >= '00:00' and time <= '7:00'))) 
        and shift = 'GY' 
          and complete = 1 

現在我的問題是移位的範圍是今天晚上10點到湯姆7am。

但是，當更改日期

HELP在使用我的語法上面仍然得到錯誤的if語句MYSQL

IF YES（選擇此）ELSE IF（選擇此） 如何正確IF ELSE在使用MYSQL語句

SELECT count(*) FROM fax 
    IF 
    (date = CURDATE() and (time >='22:00' and time <= '23:59') 
    THEN 
     SELECT count(*) FROM fax 
     WHERE ((date = CURDATE() and (time >='22:00' and time <= '23:59')) 
     or (date=CURDATE() + INTERVAL 1 DAY and (time >= '00:00' and time <= '7:00'))) 
     and shift = 'GY' and complete = 1 
    ELSE IF 
    (date = CURDATE() and (time >= '00:00' and time <= '7:00') 
      THEN 
      SELECT count(*) FROM fax 
      WHERE ((date = CURDATE() and (time >= '00:00' and time <= '7:00')) 
      or (date=CURDATE() - INTERVAL 1 DAY and (time >='22:00' and time <= '23:59'))) 
      and shift = 'GY' and complete = 1