1、要把以下的一段sql轉換成為java語句

SELECT DISTINCT

????c.company_code AS companyCode,

????c.user_name AS userName

FROM user AS c

WHERE CHAR_LENGTH(c.user_name) = 11 AND c.user_name LIKE '1%'

????AND NOT EXISTS (SELECT 1 FROM temp_phone AS p WHERE p.phone = c.user_name)

2、起初我用的查詢如下，發(fā)現(xiàn)結果與SQL執(zhí)行結果數據量出現(xiàn)了很大的差異，結果發(fā)現(xiàn)likeLeft("user_name", "1")并不能替換c.user_name LIKE '1%'

public List<User> getUserList(List<String> phoneList) {

???? return? this.list(new QueryWrapper<User>()

????.likeLeft("user_name", "1")? ? ? ? ? ? ? ?

????.eq("CHAR_LENGTH(user_name)", 11)? ? ? ? ? ? ? ?

????.notIn("user_name", phoneList))；

}

3、于是最后修改為

public List<User> getUserList(List<String> phoneList) {

????List<User> userList = this.list(new QueryWrapper<User>()? ? ? ? ? ? ? ? ????????.eq("CHAR_LENGTH(user_name)", 11)? ? ? ? ? ? ? ?

????????.notIn("user_name", phoneList));

????return userList.stream().distinct().filter(user -> user.getUserName().startsWith("1")).collect(Collectors.toList());

}

4、至于為什么likeLeft("user_name", "1")不能替換c.user_name LIKE '1%'，likeLeft到底是個什么情況還在研究中，記錄一下，研究好了再來補上