mysql返回结果，即使ID是错误的
在ID值仍然查询返回结果之后，我将面临问题。
理想情况下，它应该返回一个空的结果。
mysql> select * from pricelists where id = '1abcd';
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
| id | name | markup | routing_type | quality_base | initially_increment | inc | shadow_billing_id | status | reseller_id | pricelist_id_admin | routing_prefix | call_count | creation_date | last_modified_date | decimal_value |
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
| 1 | default | 0 | 0 | 1 | 0 | 60 | 0 | 0 | 0 | 0 | | 0 | 2016-07-25 00:00:00 | 2022-07-06 10:36:31 | 4 |
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
1 row in set, 1 warning (0.00 sec)
I am facing an issue when adding a random string after the id value still query return result.
Ideally, it should return an empty result.
mysql> select * from pricelists where id = '1abcd';
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
| id | name | markup | routing_type | quality_base | initially_increment | inc | shadow_billing_id | status | reseller_id | pricelist_id_admin | routing_prefix | call_count | creation_date | last_modified_date | decimal_value |
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
| 1 | default | 0 | 0 | 1 | 0 | 60 | 0 | 0 | 0 | 0 | | 0 | 2016-07-25 00:00:00 | 2022-07-06 10:36:31 | 4 |
+----+---------+--------+--------------+--------------+---------------------+-----+-------------------+--------+-------------+--------------------+----------------+------------+---------------------+---------------------+---------------+
1 row in set, 1 warning (0.00 sec)
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实际上毫无疑问，但原因是此ID列是一个int列，并且您的条件中的值会自动解析为int。因此，它的“ 1”最终。
Its actually no question, but the reason is that this id column is an int column and the value in your condition is automatically parsed to int. So its "1" in the end.
这意味着
ID列是数字的，并且比较具有数字上下文。字符串文字被隐式转换为数字值，ID值与数字1值进行了比较。您必须为比较设置字符串上下文。例如，您可以使用显式铸造（
cast（id as char）='1abcd'）或任何隐式转换（例如，whese concat（id，''''）='1abcd '）。This means that
idcolumn is numeric one, and the comparing have numeric context. The string literal is converted to the numeric value implicitly, andidvalue is compared with numeric1value.You must set string context for the compare. For example, you may use explicit CAST (
where CAST(id AS CHAR) = '1abcd') or any implicit convertion (for example,where CONCAT(id, '') = '1abcd').