mysqli没有选择正确的数据[关闭]

Question

I have this fiddle file http://sqlfiddle.com/#!9/78b31/20 tho the issue i face is that i wont now just not Display the right data.

For eg. when i try date = 2018-01-24 for the 3rd data (C) is still showing up.

later on i would like to then have the date updated with this function date:

$date = ("Y-m-d");.

so that it would only Show the bowCode /BowID which have not been inserted into comeandtradaysparticipant

Kind regards

Markus

Answer 1

You are making a LEFT JOIN. What you are looking for is a simple JOIN. LEFT JOIN retrieve the data of the joined table if the ON condition is correct, otherwise it return null values.

If you try to display a.date you will see that with the LEFT JOIN it will be NULL values.

 SELECT 
       b.id, b.BowCode, b.Info, a.date
    FROM Equipment AS b
    JOIN comeandtrydaysparticipant AS a
             ON a.BowID = b.BowCode
    WHERE a.BowID IS NULL
    OR(a.BowID=b.BowCode AND a.date<>"2018-01-27");

    SELECT 
       b.id, b.BowCode, b.Info, a.date
    FROM Equipment AS b
    JOIN comeandtrydaysparticipant AS a
             ON a.BowID = b.BowCode
    WHERE a.BowID IS NULL
    OR(a.BowID=b.BowCode AND a.date <> "2018-01-24");