26

我收到以下错误

#1690 - BIGINT UNSIGNED 值超出 '( legends. spawns. quantity- tmp_field)'的范围

这是我的查询

SELECT drops.common, drops.uncommon, drops.rare, drops.legendary, spawns . *
     , ( quantity - COUNT( game_moblist.spawn_id ) ) AS quantity_to_spawn
     , mobs . * 
FROM spawns
     LEFT JOIN mobs
          USING ( mob_id ) 
     LEFT JOIN game_moblist
          USING ( spawn_id ) 
     LEFT JOIN drops ON ( 
               SELECT MAX( level ) 
                 FROM drops
                WHERE drops.type = mobs.drop_list
                  AND drops.level <= spawns.level ) 
GROUP BY spawn_id
HAVING quantity_to_spawn >=0
       AND next_spawn <=0

我一直盯着它有一段时间查询很长对不起。

spawns table - countgame_moblist.spawn_id适用0于所有可能的行，但 1 （我删除了一行来测试查询）

否则数据很长，我认为与我的问题无关

知道如何解决这个错误吗？

4

10 回答 10

39

请阅读“超出范围和溢出处理”。
它说：

As of MySQL 5.5.5, overflow during numeric expression evaluation results in an error. For example, the largest signed BIGINT value is 9223372036854775807, so the following expression produces an error.

mysql> SELECT 9223372036854775807 + 1;

ERROR 1690 (22003): BIGINT value is out of range in '(9223372036854775807 + 1)'

To enable the operation to succeed in this case, convert the value to unsigned;

mysql> SELECT CAST(9223372036854775807 AS UNSIGNED) + 1;
+-------------------------------------------+
| CAST(9223372036854775807 AS UNSIGNED) + 1 |
+-------------------------------------------+
|                       9223372036854775808 |
+-------------------------------------------+

A change to part of your query, as following, would solve the issue. 

( CAST( quantity AS SIGNED ) - COUNT( game_moblist.spawn_id ) ) AS quantity_to_spawn

Otherwise you may require to change the sql_mode on unsigned operations.

mysql> SET sql_mode = 'NO_UNSIGNED_SUBTRACTION';

and then run your query to get desired output.

See also a similar posting answered on a forum here.

于 2012-07-28T19:54:03.307 回答
8

To generalise the rule, MySQL will now refuse to substract an UNSIGNED operand from a SIGNED one.

Example : SELECT A - B; will fail if A is SIGNED whereas B is UNSIGNED.

Workarounds: Add 1.0 factor to the signed operand, so it implicitly casts it to FLOAT, or use CAST (B AS SIGNED), or even swap (B - A) and change the algorithm accordingly.

于 2017-01-17T03:13:47.257 回答
3

I had the same problem, it occurred on a JOIN and couldn't figure out what was going on, in the end it was typo in the ON clause where I placed a minus sign instead of an equal sign. Might be stupid but I just didn't see it for about 30 minutes and maybe this could help someone!!!

于 2015-03-08T10:17:12.450 回答
3

I don’t quite understand why everyone is saying unsigned. I have a special value in sql and also reported this error. I did it by converting this value to decimal.

cast(1000000000000000000 AS DECIMAL ( 35, 2 ))
于 2020-03-12T05:36:33.560 回答
2

I actualy found that question why I was searching for solution. If you have same problem as I do, try disabling "unsigned" parameter.

It is quite possible that your code fails here:

(
quantity - COUNT( game_moblist.spawn_id )
)

because if result of that matematic operation is less than zero it will fail with "unsigned" parameter.

于 2013-02-07T16:58:04.920 回答
1

Additional way is to use MySQL IF operator. This helped me when both columns were BIGINT and Unsigned.

于 2013-03-25T16:06:57.080 回答
1

I had similar problem, This error also come if our column have 0 value and we try to update it with -1 value.

In my case MySQL query were Failing if column1 is already have value 0, i.e column1 = 0

For example:

UPDATE `table1` SET `column1` = `column1` - 1 WHERE `column2` = XYZ

This give error

BIGINT UNSIGNED value is out of range in ....

To counter this problem

UPDATE `table1`
SET `column1` = (
    CASE WHEN `column1` < 1
    THEN 0
    ELSE (`column1` - 1)
end)
WHERE `column2` = 1
LIMIT 1
于 2019-02-09T18:54:41.607 回答
0

Another possible cause seems to be how the type of the result variable is allocated.

eg.

mysql> select (total_balance_06 * 0.045/(1-(1/1.045)^term_06) + unsec_instalments)/income from tbl_EUR_PDH;

fails with

ERROR 1690 (22003): BIGINT UNSIGNED value is out of range in '(1 - ((1 / 1.045) ^ markov.tbl_EUR_PDH.term_06))'

whereas

mysql> select (total_balance_06 * 0.045/(1.0-(1.0/1.045)^term_06) + unsec_instalments)/income from tbl_EUR_PDH;

does what one would expect (note that I simply replace "1" by "1.0")

Philippe

于 2013-12-11T12:56:50.860 回答
0

sql_mode worked in the MySQL client and Adminer, but not in CodeIgniter, where it counts. Casting didn't help either.

A simple arithmetic operation did the trick:

error

id - id_ex*1000000000 = id_sm

works

id_sm + id_ex*1000000000 = id
于 2018-04-17T20:15:10.190 回答
0

In my case, I was subtracting two columns that were UNSIGNED and the result of some of the subtractions was negative. Setting both columns to SIGNED INT resolved it for me but this is just a temporary fix

于 2021-07-08T22:43:18.147 回答