如何从MySql中的表中获取最小id

Question

I have written a MySql query to get the columns related with minimum id . Looks something like this

SELECT min(id) as ID,feed , idpropiedad FROM `registrofeed` WHERE feed=21

The table has 4 rows looks like this

So according to the function that I have written

function setLC()
 {

     $sql = "
SELECT min(id) as ID
  , feed 
  , idpropiedad 
  FROM `registrofeed` 
 WHERE feed=21
";
     $result = $this->localDb->execute($sql);
     $row=mysql_fetch_array($result);
     echo $sql;
     echo $row['idpropiedad'];
     $this->lastCode = $row['idpropiedad'];
 }

It returns empty string for idpropiedad

Can any one help me out where I am going wrong

Thanks in advance

Answer 1

I'd think the query you're actually looking for is this:

SELECT id, feed, idpropiedad 
  FROM registrofeed 
 WHERE feed = 21 
 ORDER BY id ASC
 LIMIT 1

MIN() is giving you the generally lowest value in the column, it does not affect the rest of the columns. If you want the whole row with the lowest id it doesn't help.

To illustrate, if you really wanted to use MIN here, you'd have to do:

SELECT id, feed, idpropiedad 
  FROM registrofeed 
 WHERE id = (SELECT MIN(id) FROM registrofeed WHERE feed = 21)

Answer 2

You can do a better query like this:

 $sql = "
SELECT id as ID
  , feed 
  , idpropiedad 
  FROM `registrofeed` 
 WHERE feed=21
 HAVING MIN(id)
";

This will return only one row with the minimum id number. It's more readable than using ORDERING AND LIMIT 1.

Answer 3

try your select query as

SELECT * FROM registrofeed WHERE feed ='21' ORDER BY id ASC LIMIT 1

this fetches the row having minimum id.

Hope it helps

Answer 4

试试这个

$sql = "SELECT min(id) as ID,feed , idpropiedad FROM `registrofeed` WHERE feed='21' order by id  asc";