duplicate-removal

Is there a way of removing record on duplicate key in MySQL? Say we have a record in the database with the specific primary key and we try to add another one with the same key - ON DUPLICATE KEY UPDATE would simply update the record, but is there an option to remove record if already exists? It is for simple in/out functionality on click of a button. It's a work-around, but it works: Create a new column and call it do_delete , or whatever, making it a tiny-int. Then do On Duplicate Key Update do_delete = 1; Depending on your MySQL version/connection, you can execute multiple queries in the

How to do this using standard methods (without manual iteration through source string)? PS: At final I want to get sorted characters of source string. I tried to use NSCharacterSet , but can't find a method to convert character set to string (without iterating the set). There is no built-in method for this, but it's pretty easy to iterate over the characters of the string and build a new string without duplicates: NSString *input = @"addbcddaa"; NSMutableSet *seenCharacters = [NSMutableSet set]; NSMutableString *result = [NSMutableString string]; [input enumerateSubstringsInRange:NSMakeRange(0

I am dynamically generating a query like below that creates different combinations of rules by left joining (any number of times) on itself and avoiding rules with some of the same attributes as part of the joins conditions e.g. SELECT count(*) FROM rules AS t1 LEFT JOIN rules AS t2 ON t1.id != t2.id AND ... LEFT JOIN rules AS t3 ON t1.id != t2.id AND t1.id != t3.id AND t2.id != t3.id AND ... I am currently removing duplicates by creating an array of ids from the joined rows then sorting and grouping by them: SELECT sort(array[t1.id, t2.id, t3.id]) AS ids ... GROUP BY ids I would like to know

I have a system that will generate messages sporadically, and I would like to only submit either zero or one message every 5 minutes. If no message is generated, nothing would be processed by the queue consumer. If a hundred identical messages are generated within 5 minutes I only want one of those to be consumed from the queue. I am using AMQP(RabbitMQ), is there a way to accomplish this within rabbitmq or the AMQP protocol? Can I inspect a queue's contents to ensure that I don't insert a duplicate? It seems that queue inspection is a bad idea and not typically what should be done for a

I have a column in openoffice calc with a set of codes. For example: B1 B1 Br Bh Ht C3 C3 So what I would like to do is delete all the duplicates so I am left with just: Br Bh Ht Any help much appreciated. Cheers Select the entire range containing data to filter, then click on the menu Data > Filter > Standard Filter and: Use a condition that is always TRUE, like field1 = Not empty Click on the button more, select Remove Duplicate, select Copy to and put the address of an empty cell The whole range (without duplicate) will be analyzed and copied at that new address. source Or For both Calc and

Consider the following table: TAB6 A B C ---------- ---------- - 1 2 A 2 1 A 2 3 C 3 4 D I consider, the records {1,2, A} and {2, 1, A} as duplicate. I need to select and produce the below record set: A B C A B C ---------- ---------- - ---------- ---------- - 1 2 A or 2 1 A 2 3 C 2 3 C 3 4 D 3 4 D I tried the below queries. But to no avail. select t1.* from t6 t1 , t6 t2 where t1.a <> t2.b and t1.b <> t2.a and t1.rowid <> t2.rowid / A B C ---------- ---------- - 1 2 A 2 1 A 2 1 A 2 3 C 3 4 D 3 4 D 6 rows selected. Or even this: select * from t6 t1 where exists (select * from t6 t2 where t1.a

Just want to remove duplicates. Pool is vector<pair<string, int>> but I seem to miss some elements at the start of the vector somehow. Can anyone verify the logic of the removal? Thanks :) Pool Master::eliminateDuplicates(Pool generation) { for(int i = 0; i < generation.size(); i++) { string current = generation.at(i).first; for(int j = i; j < generation.size(); j++) { if(j == i) { continue; } else { string temp = generation.at(j).first; if(current.compare(temp) == 0) { Pool::iterator iter = generation.begin() + j; generation.erase(iter); } } } } return generation; } This is a very common