I'm trying to make a mySQL query that updates a field called "connected" to 1 ONLY IF user1 = derp AND user2=derpette or the opposite. I'm having ad difficult time with the syntax. I read some of the other threads on overflow, which led me to creating the following query. If anyone could provide me any suggestions towards the correct syntax or perhaps provides an alternate solution to this problem, it would be greatly appreciated. Thanks.
UPDATE xGames
SET connected = 1
WHERE ((user1 = 'derp' AND user2 = 'derpette')
OR (user2='derpette' AND user1='derp'))
I'm not sure if you're getting a syntax error, or if it just isn't behaving as you expect, but you didn't reverse the conditions - you only reversed the ORDER of the same conditions...
UPDATE xGames
SET connected = 1
WHERE ((user1 = 'derp' AND user2 = 'derpette')
OR (user1='derpette' AND user2='derp'));
Related
I am trying to update a record using Laravel.
I have gone through lot of StackOverflow Questions to check whether this question is already raised.
mysql query : UPDATE students SET total_marks = total_marks + 80 WHERE id = 1
I have to translate this mysql query into Laravel query builder, but couldn't get a solution yet.
Instead of getting the early value from DB before update, Can we update the table with one update query using Laravel Query Builder.
2 Queries way:
$student_marks = Students::select('total_marks')->where('id','=',1);
$current_mark = $student_marks['total_marks']+80;
$update_student_marks = Students::where('id','=',1)->update('total_marks','=',$current_mark);
How to update a record like this with single query builder in Laravel.
I think you need to make a few adjustments to your query.
First, you need to select the student correctly and than use Eloquent to call save method on it after setting the property to the correct value. I assume you are on Laravel 6.
$student_marks = Students::find($id);
$student_marks->total_marks += 80;
$student_marks->save();
Please, take a look at Laravel docs:
https://laravel.com/docs/6.x/eloquent
The reading takes some time but its definitely worth it. You will learn how to deal with eloquent and make your code better by using the most appropriate techniques.
You can use the save function for this.
$student_marks = Students::select('total_marks')->where('id','=',1);
$student_marks->total_marks += 80; //or $student_marks->total_marks = $student_marks->total_marks + 80;
$student_marks->save();
Pass update data as array
Try this way
$update = array('total_marks'=>$current_mark);
$update_student_marks = Students::where('id','=',1)->update($update);
I am looking for the way to execute MySQL statement checking if given parameter exists. As I remember I can do the following in Oracle to achieve that:
select s.* from Site s
where s.study = :study
and (:enabled is null or s.enabled = :enabled)
is anything like that possible in MySQL too? The same code executes without error but never return any records.
My goal here is to avoid multiple lfs and elses in my java code. It should work the way that the query looks like that when enabled parameter is null:
select s.* from Site s
where s.study = :study
and like that if the parameter is not null:
select s.* from Site s
where s.study = :study
and s.enabled = :enabled
and I want to do that with a single query
I believe this is what you are asking:
SELECT s.* from Site s
WHERE s.study = "some_study"
AND (s.enabled IS NULL OR s.enabled = '' OR s.enabled = "enabled");
Unfortunately it is highly dependent on database driver. My initial query works when run in database tools but doesn't have to when it comes to run it by JPA. So I'm to close this question as it doesn't require further answers. I'm sorry lads for wasting your time.
I am trying to execute a query into MySQL but it keeps telling me i am using the wrong syntax, I tried searching the MySQL community but I am not getting anything usefull.. most of the answers i find on google are for other databases yet they label them for "MySQL", yet it keeps failing.
This is the statement i am trying to execute:
$statement = "IF (SELECT ttb_id FROM timetable WHERE ttb_week = $i AND ttb_time = $j) THEN
BEGIN
UPDATE types SET typ_name = '$subj'
WHERE typ_name = 'student';
END;
ELSE
BEGIN
INSERT INTO types VALUES (null,`Yo`);
END;
";
error:
ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'IF (SELECT ttb_id FROM timetable WHERE ttb_week = 0 AND ttb_time = 0) THEN ' at line 1
I am using:
PHP Script Language Version 5.2.6
MySQL Database Version 5.0.51b
I have been looking around but to no avail, and the if condition stated on the MySQL dev website are not helping at all..
I am new to this and it is driving me mad! all the different queries i tried failed.. it is never the right syntax.
I found many answers for the problem on stackExchange and other websites but it is always wrong.. also I remember this structure from a VB.net lesson a while ago so maybe this is for MSSQL? then what about MySQL? everyone so far listed similar structure and said it works for MySQL, I took my answer from an answer on this community* labeled for MySQL and he claimed it worked. this is one of many i tried.
I would really appreciate your help
*: Usage of MySQL's "IF EXISTS"
My first thought would be you're not comparing your SELECT return to anything to actually utilize your conditional logic. Are you just looking to see if your query returns values? If it doesn't return a value then you insert a new record in otherwise you update.
Maybe use IS NOT NULL or a check to see count on the select to see how many rows and compare to see if that's greater than 0.
IF (SELECT ttb_id FROM timetable WHERE ttb_week = $i AND ttb_time = $j IS NOT NULL) THEN
IF (SELECT COUNT(*) FROM timetable WHERE ttb_week = $i AND ttb_time = $j) > 0 THEN
> *1. I need to write this in Active records *
i need to join these 3 tables , but the condition for join is very very selective
$this->db->select(name);
$this->db->from('table0');
$this->db->join('table1','(table1.id=0 AND table0.feild1 = table1.feild1) OR (table1.id=1 AND table0.feild2 = table1.feild2)') // <--- how to write this is my question
i could do a simple join but the main problem is achieving the condition in the join that i have mentioned above.Also this is a small part of a very , very ! big query so i really cant change it back to the native sql query like :
$this->db->query('//entire sql query'); //cant do this, need to write ACTIVE RECORDS
when i write the active records , firebug throws me an error saying :
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near ') OR
any suggestions ?
To be honest, I don't know that it is possible. CI doesn't use true Active Records, so each method has very rigid parameters, and I don't see any that can be tricked into performing the task you need.
What I would try is re-writing your query a bit:
$this->db->select(name);
$this->db->from('table0');
$this->db->join('table1','(table1.id=0 AND table0.feild1 = table1.feild1)', LEFT);
$this->db->join('table1','(table1.id=1 AND table0.feild2 = table1.feild2)', LEFT);
$this->db->where('(table1.id=0 AND table0.feild1 = table1.feild1)');
$this->db->or_where('(table1.id=1 AND table0.feild2 = table1.feild2)');
I believe this query would return the correct values, but you'll certainly want to thoroughly test it. Hope this at least points you in the right direction.
I have a query that tries to tell is a member had participated in any of 3 types of actions. Each action is represented by a table that I am joining. And then I am doing an or to see if there is a match in any of the tables. Here is the query:
select problems.problem_id, problem_title from
problems left join attempted_solutions
on roblems.creator_member_id = attempted_solutions.suggester_id
left join problem_comments
on problems.creator_member_id = problem_comments.user_id
left join suggested_solutions
on problems.creator_member_id = suggested_solutions.suggester_id
where attempted_solutions.suggester_id = 1
or problem_comments.user_id = 1 or suggested_solutions.suggester_id = 1;
But for some reason this gives a "MySQL client ran out of memory error." if I try it command-line, and a "database connection interrupted" error if I do it from a script.
Any idea where I am going wrong here?
This thread has some solutions maybe you should try them see if any of them is your problem
http://forums.mysql.com/read.php?37,94800,232288#msg-232288
also, try to simplify your query then add your joins and where statements back one by one so that you can detect the exact cause of your problem