SQL查詢條件與日期

Question

因此，我在這里有此查詢：

$strSQL = "SELECT formas.*, SMS_SERVISI.IDTICKET, SMS_SERVISI.MBYLLUR,SMS_SERVISI.time_added,servis_furnitor.id_servis,servis_furnitor.furnitori,servis_furnitor.kohezgjatja
FROM formas  
LEFT JOIN servis_furnitor ON formas.furnitori = servis_furnitor.id_servis 
LEFT JOIN SMS_SERVISI ON formas.ID = SMS_SERVISI.IDTICKET 
ORDER BY formas.id DESC 
WHERE $today-formas.data_fillim > servis_furnitor.kohezgjatja";

我知道最后一行是錯誤的，我的意思是我可以。.我有這個訂單，他們的起始日期是formas.data_fillim ，我有今天的日期：

$today = date("Ymd"); 

因此， $today-formas.data_fillim之間的差異不應大於servis_furnitor.kohezgjatja ，后者本身是一個整數，它顯示了天數

formas.data_fillim

是一個日期類型。

我需要提取所有與今天的日期和開始日期的差異不超過“ kohezgjatja”中預定義的天數的數據

任何幫助請..謝謝

更新

$today = date("Y-m-d H:i:s", time());  


echo $strSQL = "SELECT  formas.*,
        SMS_SERVISI.IDTICKET,
        SMS_SERVISI.MBYLLUR,
        SMS_SERVISI.time_added,
        servis_furnitor.id_servis,
        servis_furnitor.furnitori,
        servis_furnitor.kohezgjatja
FROM formas
LEFT JOIN servis_furnitor
        ON formas.furnitori = servis_furnitor.id_servis
LEFT JOIN SMS_SERVISI
        ON formas.ID = SMS_SERVISI.IDTICKET
WHERE DATEDIFF ( day , '$today' , formas.data_fillim ) > servis_furnitor.kohezgjatja
ORDER BY formas.id DESC"

Answer 1

SELECT語句的正確格式為

SELECT ....
FROM...
WHERE ...
GROUP ....
ORDER BY...

所以就你而言

SELECT  formas.*,
        SMS_SERVISI.IDTICKET,
        SMS_SERVISI.MBYLLUR,
        SMS_SERVISI.time_added,
        servis_furnitor.id_servis,
        servis_furnitor.furnitori,
        servis_furnitor.kohezgjatja
FROM formas
LEFT JOIN servis_furnitor
        ON formas.furnitori = servis_furnitor.id_servis
LEFT JOIN SMS_SERVISI
        ON formas.ID = SMS_SERVISI.IDTICKET
WHERE DATEDIFF ( day , $TODAY , formas.data_fillim )  > servis_furnitor.kohezgjatja
ORDER BY formas.id DESC

Answer 2

這樣的事情應該起作用：

where formas.data_fillim <= DateAdd(day, servis_furnitor.kohezgjatja, getdate())

您可能必須在servis_furnitor.kohezgjatja前面加上減號，具體取決於您在那存儲的內容。