duplicate-data

This shows me all the first names and last names that have exactly two entries that are identical SELECT `firstname`,`lastname`,COUNT(*) AS Count FROM `people` GROUP BY `firstname`,`lastname` HAVING Count = 2 How do I turn this into a DELETE FROM WHERE statement with a LIMIT to only remove one of each of the entries and leave the other one. okay this appears to be way to technical i'm just going to do it in a php while loop You can create a table with 1 record of each of the duplicates: Then delete all the dup records from the people table and then re-insert the dup records. -- Setup for

I want to merge data. Following are my MySQL tables. I want to use Python to traverse though a list of both Lists (one with dupe = 'x' and other with null dupes). This is sample data. Actual data is humongous. For instance : a b c d e f key dupe -------------------- 1 d c f k l 1 x 2 g h j 1 3 i h u u 2 4 u r t 2 x From the above sample table, the desired output is : a b c d e f key dupe -------------------- 2 g c h k j 1 3 i r h u u 2 What I have so far : import string, os, sys import MySQLdb from EncryptedFile import EncryptedFile enc = EncryptedFile( os.getenv("HOME") + '/.py-encrypted-file

I have a Database storing details of products which are taken from many sites, and gathered through the individual sites API's. When I call the feed, the details are stored in a database table. The problem I'm having is that because the exact same product is listed on many sites by the seller I end up having duplicate items in my database, and then when I display them on a web page there are many duplicates. The problem is that the item doesn't have any obvious unique identifier, it has specific details of the item (of which there could be many), and then a description of the item from the

Can I use a SQL query to find records where one field is identical in both? That is, can I use the following table and return 1,3 (the ids) by comparing the name columns (and ignoring the phone)? ID | Name | Phone 1 | Bob | 5555555555 2 | John | 1234567890 3 | Bob | 1515151515 4 | Tim | 5555555555 To get all names that exist more than once you can execute this statement: SELECT Name FROM People GROUP BY Name HAVING COUNT(*)>1; To get the IDs of the duplicates "1,3" concatenated that way use GROUP_CONCAT : SELECT GROUP_CONCAT( ID SEPARATOR ',' ) FROM Table GROUP BY Name HAVING COUNT(*) > 1

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

I want to find the indices all the rows of a matrix which have duplicates. For example A = [1 2 3 4 1 2 3 4 2 3 4 5 1 2 3 4 6 5 4 3] The vector to be returned would be [1,2,4] A lot of similar questions suggest using the unique function, which I've tried but the closest I can get to what I want is: [C, ia, ic] = unique(A, 'rows') ia = [1 3 5] m = 5; setdiff(1:m,ia) = [2,4] But using unique I can only extract the 2nd,3rd,4th...etc instance of a row, and I need to also obtain the first. Is there any way I can do this? NB: It must be a method which doesn't involve looping through the rows, as I'm