SQL

find duplicates mysql column

SELECT name, COUNT(*) c FROM table GROUP BY name HAVING c > 1;

finding duplicate rows mysql

SELECT varchar_col
FROM table
GROUP BY varchar_col
HAVING COUNT(*) > 1;

find duplicate keys in mysql

SELECT col, COUNT(col) FROM table_name GROUP BY col HAVING COUNT(col) > 1;

myql find duplicates

SELECT 
    col, 
    COUNT(col)
FROM
    table_name
GROUP BY col
HAVING COUNT(col) > 1;
Code language: SQL (Structured Query Language) (sql)

select where duplicate mysql

SELECT 
    col1, COUNT(col1),
    col2, COUNT(col2)
FROM
    table_name
GROUP BY 
    col1, 
    col2
HAVING 
       (COUNT(col1) > 1) AND 
       (COUNT(col2) > 1);

mysql find duplicates in same table

## Find ALL duplicate recods by value (without grouping them by value) ##
# to find the duplicate, 
# replace all instances of tableName with your table name
# and all instances of duplicateField with the field name where you look for duplicates
SELECT t1.*
FROM tableName AS t1
INNER JOIN(
	SELECT duplicateField
	FROM tableName
	GROUP BY duplicateField
	HAVING COUNT(duplicateField) > 1
)temp ON t1.duplicateField = temp.duplicateField
order by duplicateField

MySQL repeated values

            
                
            
         SELECT 
    col, 
    COUNT(col)
FROM
    table_name
GROUP BY col
HAVING COUNT(col) > 1;
Code language: SQL (Structured Query Language) (sql)

finding duplicate rows mysql

SELECT name, COUNT(*) c FROM table GROUP BY name HAVING c > 1;

mysql find duplicates

SELECT column, COUNT(column) c FROM table GROUP BY column HAVING c > 1;

mysql query to find duplicate records

SELECT firstname, 
   lastname, 
   list.address 
FROM list
   INNER JOIN (SELECT address
               FROM   list
               GROUP  BY address
               HAVING COUNT(id) > 1) dup
           ON list.address = dup.address;

duplicate record mysql

SELECT firstname, 
   lastname, 
   list.address 
FROM list
   INNER JOIN (SELECT address
               FROM   list
               GROUP  BY address
               HAVING COUNT(id) > 1) dup
           ON list.address = dup.address;