I have a MySQL database ver 5.1. I have two tables Table_A will always keep the records in it. Table_B will be used to only compare the differences. Is there a MySQL query function via PHP that will compare the rows and tell me what rows are missing on Table_B.
Table_A
user name
1 bob
1 joe
1 sally
1 matt
Table_B
user name
1 bob
1 joe
1 sally
1 john
When the PHP query was run how would I compare Table_A vs Table_B?
All I need is to tell me that matt no longer exists and it will add john to Table_A.
What would be the best way to do this?
Thanks.
SELECT name FROM Table_A WHERE name NOT IN (SELECT name FROM Table_B)
Related
I need a php routine for my mysql database. I have made an example to illustrate the problem:
Lets say I have a table that registrates customers and how much money they spend. A customer can have more registrations:
Table1:
Name - Amount
Jane - 3
Mark - 4
Sara - 5
Jane - 5
Jane - 6
Sara - 2
I want a routine that goes trough Table1, and finds how much each person has spend. I want the result in Table2, like this:
Table2:
Jane - 14
Mark - 4
Sara - 7
Do you have a solution to this?
insert into table2 select name, sum(amount) from table1 group by name;
SELECT Name, SUM(Amount) FROM Table1 GROUP BY Name
I have created a database and website that will be used by football managers to select their team etc. Once a match has been completed events will be stored in the match_players table. Such events are Goal, Yellow Card, Red Card etc. I have no problem getting this information into php from SQL db.
I need to add up how many times a Goal appears (a '1' is placed in the SQL table) and for what team so that a final score can be displayed. So, for example, if Team A has 1 goal and Team B has 2 then I need to display that. I am trying to count the amount of times that a Goal is registered in the table. Any help will be greatly appreciated.
You can use MYSQL SUM
select SUM(Goal) from match_players where Team="A"
Or you can get the same for all teams by
select Team,SUM(Goal) from match_players group by Team
Why don't you demand this sum to SQL directly?
SELECT SUM(goals)
FROM match_table
WHERE team = 'Barcellona'
This should be much faster also than elaborate all data at "php-level"
If you want this detail for all teams
SELECT team,SUM(goals)
FROM match_table
GROUP BY team
Well if you store a 1 each time a goal is scored, your table looks like this:
TeamID goal
1 1
2 1
1 1
3 1
2 1
2 1
1 1
So you just want a count of how many times a team appears in that table:
select TeamID, count(*) from table group by TeamID
Will give you
TeamID | count(*)
1 | 3
2 | 3
3 | 1
I have a table like this:
ID build1 build2 test status
1 John ram test1 pass
2 john shyam test2 fail
3 tom ram test1 fail
The problem that I am facing is - on one of my webpage, only the values from the column "uild1" are available to me. Now in table there are 2 entries corresponding to "John". so, even if the user selects different "John", its showing the values for other values from the row only. On my webpage, in the drop down list, user can see 2 "John" but since query has been made using "John" condition, on both occasions, its showing the results from the first row only.
Try this:
SELECT t1.*
FROM Table1 t1
WHERE t1.build1 NOT IN(SELECT t2.build1
FROM table1 t2
GROUP BY t2.build1
HAVING COUNT(t2.build1) > 1);
SQL Fiddle Demo
This will give you only:
| ID | BUILD1 | BUILD2 | TEST | STATUS |
-----------------------------------------
| 3 | tom | ram | test1 | fail |
Since, it is the only row that has no duplicate build1.
If I'm understanding your question correctly, given a web page with 2 johns available to click on, how can you get each result accordingly? Unfortunately, there is no way of doing this with just SQL.
In your PHP code, if you can pass a parameter to your SQL code with either the ID or a counter/row number, then you could query the database to return a corresponding unique record.
Good luck.
You build1 is not unique or primary key so it is picking all the row matching your condition. You should use primary key or unique key to find the result. In your select drop-down your option value should be uniq/primary key so when you select particular "John" it will get result of that john.
select * from table_name where id=params[:id] ;
If you post some more information. It will be helpful to write better code for you.
select * from yourtable where build1 == 'john' limit 1;
There are two tables table1 and table2
table1 has got two columns name and rank
table2 has got only one column name
names in table2 are almost listed in table1
I want compare two table and pull rank info from table1 and update/alter table2 with rank
table1
name | rank
-------------
john | 2
mathews| 5
keyn | 4
emly | 25
yancy | 8
stewart| 9
kim | 12
chris | 19
table2
name
-------
john
mathews
keyn
emly
yancy
stewart
I want update/insert rank details to table2 from table1
thats it and sorry for the confusion
Seem like you want to do something like this:
update table2,table1 set table2.rank=table1.rank where table2.name=table1.name
This will update the second table with ranks from the first table where the names are equal.
Then put auto increament field of table one in table2. and after this apply left join using these id and pull the info
refer this link
http://www.wellho.net/solutions/mysql-left-joins-to-link-three-or-more-tables.html
I'm new to database structure. I'm trying to create an app that allows users to like certain entries, but I want to be able to tie likes to users so that I can change the visuals before/after the like action.
I think from research that I should have an 'entries' and 'users' table and then have a 'likes' table that ties the two to each other.
The only thing I'm unsure of is, when getting and displaying the contents... how would I write the queries? If I query for all the entries I need, do I then go back and individually query each to see if it has a like tied to it for the current user? That seems like it might be a costly operation. Is there a more efficient way?
Hope that makes sense,
Thanks.
I think you have the right database design in mind. As far as queries are concerned, assume tables as such:
Users
ID | Name
1 | Bob
2 | Sally
Entries
ID | Name
1 | Red
2 | Blue
3 | Yellow
Likes
UserID | EntryID
1 | 1
1 | 2
2 | 2
2 | 3
So we can say Bob likes Red and Blue while Sally likes Blue and Yellow. So a query to retrieve all entries, plus an indicator of what Bob likes would be:
SELECT
e.ID,
e.Name,
l.UserID
FROM Entries e LEFT JOIN Likes l ON l.EntryID = e.ID
WHERE l.UserID = 1 -- Bob's User ID
ORDER BY e.Name
This would return
ID | Name | UserID
2 | Blue | 1
1 | Red | NULL
3 | Yellow | 1
The UserID column indicates if Bob likes the entry or not - a NULL is No and a value is Yes.
Assuming you have a table Entries with a column entity_id (and whatever else you store about the entity) and a second table UserLikes that contains the columns user_id and entity_id, you would do the following:
SELECT Entries.col1, Entries.col1 . . ., UserLikes.user_id
FROM Entries LEFT OUTER JOIN UserLikes ON
Entries.entity_id = UserLikes.entity_id
WHERE UserLikes.user_id = :user_id
AND Entity.col_whatever = :whatever
In this example, Entries.col1, Entries.col2 . . . is the list of columns you want to get back about the Entries. The :user_id is a parameter that contains the id of the user you're currently trying to display Entries for. And the last line is standing in for whatever limitations you want to put on the Entries are returned.
This query will give you a row for each Entry you searched for. You can check the value the returned column user_id. If it's NULL then it was not liked by the user, if it contains the user's id, it was liked by the user.
I think u can retrieve the entries and query the likes table at the same time to get if the current user likes the entry performing a stored procedure. So u can control the value of the set of data returned by the query for example returning one colum for the entry text and one boolean column to evaluates the current user likes... In this way you will at least one parameter for the stored procedure to indicate who is the current user
I hope this idea help u...