Comparison of two tables in MYSQL

Question

I have two tables. One table (table1) has 28500 rows and the another (table2) has 17450 rows. I would like to compare these tables and find rows that do not exist in table1.

SELECT * FROM table1 WHERE ID NOT IN (SELECT DISTINCT(ID) FROM table2)

Any suggestions?

Answer 1

Try this:

SELECT table1.*
FROM table1
LEFT OUTER JOIN table2
ON table1.id = table2.id
WHERE table2.id IS NULL

LEFT OUTER JOIN link two table starting by table1, if table2 has no linked row all fields of table2 will be null. So, if you put in your WHERE condition table2.id is null, you get only rows in table1 not existing in table2

Answer 2

You could solve this by doing a left outer join and checking for all rows that don't exist. Try the following depending on if you want to find values not existent from table1 in table2 or table2 in table1.

SELECT *
FROM table1
LEFT OUTER JOIN table2 ON (table1.id = table2.id)
WHERE table2.id IS NULL;


SELECT *
FROM table2
LEFT OUTER JOIN table1 ON (table1.id = table2.id)
WHERE table2.id IS NULL;

SQL Fiddle: http://sqlfiddle.com/#!2/a9390/8

Answer 3

Make use of this query:

SELECT 
    * 
FROM 
    table2 
LEFT JOIN 
    table1
ON 
    table2.primary_key = table1 .primary_key
WHERE 
    table1 .primary_key IS NULL
;

Answer 4

well, if you want the answer in PHP, then here is it:

$sql=mysql_query("SELECT * FROM table1");
while($row=mysql_fetch_array($sql))
{
    $id=$row['id'];
    $sql2=mysql_query("SELECT * FROM table2 WHERE id='$id'");
    $check=mysql_num_rows($sql2);
    if($check==0)
    {
        echo $id." is not in table1<br>";
    }
}

I hope this help you