獲得最接近的日期爲當前日期SQL

Question

我知道，可問題是this重複，但是當我試圖解答他們提供我得到了不同的答案......

，所以我有一個表，今天是06/02/2017 tblpres

patID  pres  presDate 
    1  | asd  | 05/29/2017 
    2  | qwe  | 05/20/2017 
    3  | zxc  | 06/01/2017 
    4  | ety  | 05/27/2017 
    5  | ttt  | 05/18/2017 

，所以我做了一個查詢來獲取最接近的日期爲當前日期，而是我得到ID 3界河是最接近的日期爲當前日期，我收到了差異答案..

SELECT pres,presDate 
FROM tblpres 
WHERE patID = '20150518384' 
ORDER BY presDate DESC 
LIMIT 1 

SELECT pres, presDate 
FROM tblpres 
WHERE patID = '20150518384' 
ORDER BY ABS(DATEDIFF(presDate, `06/02/2017`)) 
LIMIT 1 

SELECT pres, presDate 
from tblpres 
WHERE patID = '20150518384' 
order by abs('06/02/2017' - presDate) desc 
limit 1 

SELECT pres,presDate 
FROM tblpres 
WHERE `patID` = '20150518384' AND presDate < '06/02/2017' 
ORDER BY presDate 
LIMIT 1 

這是我試過的查詢，但似乎沒有任何工作...我得到這個查詢的答案，但結果不是我所期待..

Answer 1

使用ISO/ANSI標準日期格式：

SELECT pres, presDate 
FROM tblpres 
WHERE patID = '20150518384' 
ORDER BY ABS(DATEDIFF(presDate, '2017-06-02')) 
LIMIT 1; 

或者，更好的是，使用CURDATE()：

SELECT pres, presDate 
FROM tblpres 
WHERE patID = '20150518384' 
ORDER BY ABS(DATEDIFF(presDate, CURDATE())) 
LIMIT 1; 

如果presdate實際上不是一個約會，你可能需要將其轉換爲一個：

SELECT pres, presDate 
FROM tblpres 
WHERE patID = '20150518384' 
ORDER BY ABS(DATEDIFF(str_to_date(presDate, '%m/%d/%Y'), CURDATE())) 
LIMIT 1;