Question

我在MySQL数据库中有一组数据。我正在检索由名为login的字段排序的结果列表。当我检索这个集合时，其中两行按以下顺序排列。

cco1@blah.com
cco10.test@blah.com

然而，当我在Java中比较它们时（由于合并对象应用程序端需要进行后处理而必需），cco10.test@blah.com的值小于cco1@blah.com。换句话说，String比较期望它们按以下顺序排列。

cco10.test@blah.com
cco1@blah.com

作为一个整体，其他所有内容都以正确的顺序返回。我认为这里的差异很可能是Java和MySQL字符串比较处理某些值的方式的差异。如何以一致的顺序返回这些内容（我不管他们是哪种方式，只需要一致性）。

查询我正在运行：

select t0.id as envUserId , t0.environment_id as envId, t0.environment_name as envName,   t0.customer_name as customerName, t0.version version, t0.user_id as userId, t0.login as userLogin,   t0.sso_granted_roles as sso_granted_roles,   t1z_.role_name as defaultRole, t3.id as customRoleId, t3.name as customRoleName
  from environment_user t0   
  left join ( 
    select distinct eu.id, eu.login       
    from  environment_user eu                 
    left join environment_user_role eur on eu.id = eur.environment_user_id   
    left join environment_user_custom_role eucr on eu.id = eucr.environment_user_id   
    left join custom_role cr on eucr.custom_role_id = cr.id        
    where eu.environment_id = '5a83069a-70d2-4d0e-9847-c709725281c5'             
    and (eur.role_name in ('Role1','Role2') 
        or cr.name in ('Role1','Role2'))       
    order by eu.login limit 0, 200) f on t0.id = f.id   
  left outer join environment_user_role t1z_ on t1z_.environment_user_id = t0.id   
  left outer join environment_user_custom_role ct1z_ on ct1z_.environment_user_id = t0.id   
  left outer join custom_role t3 on t3.id = ct1z_.custom_role_id   
  where t0.environment_id = '5a83069a-70d2-4d0e-9847-c709725281c5'     
  and t0.id = f.id
  order by userLogin asc

我得到了什么（上面和下面的额外线条为了清晰起见而被删除了）

'c2ad9f82-e0d5-4f8d-a5fe-a2d72d901b98', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '649ea0bc-dab7-4ad2-a534-546f9817e252', 'c9ca3e83-ccc6-4108-aee4-1bc41e6294ff@searchtestdomain.com', '0', 'Role1', NULL, NULL
'83313002-49a3-45f2-9013-e8dab15789d5', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '40d5c22a-33f8-4a37-a4db-63e3709cfae7', 'ccc@searchtestdomain.com', '0', 'Role1', NULL, NULL
'5ba69c88-a773-4d5b-835d-c88688867d6a', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '91a7609a-4809-4e27-9d6f-448ff62b38b3', 'cccc@searchtestdomain.com', '0', 'Role1', NULL, NULL
'6833a699-b5ca-46aa-8a53-23a6ef41e1f8', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '718808fa-3799-457f-9cdb-88ef887e0492', 'cco1@searchtestdomain.com', '0', 'Role1', NULL, NULL
'c466c478-8a32-4926-9cde-06a40071ac85', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '6282739d-76ea-4dbb-be5e-b7d64d3b3f3f', 'cco10.test@searchtestdomain.com', '0', 'Role1', NULL, NULL
'5b04d561-6c20-4703-aa96-f17eda0405b6', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', 'fb644427-46ab-42e4-8295-65a397409c0d', 'cd67848c-62fc-4cb3-ab7d-8b2d49709973@searchtestdomain.com', '0', 'Role1', NULL, NULL
'27116bed-a1a6-483c-9e7b-97158786245c', '5a83069a-70d2-4d0e-9847-c709725281c5', 'SearchTestDomainEnv', 'SearchTestDomainCustomer', '1', '246f392d-6d27-402e-837f-98384da0abb6', 'd1072b14-2956-432e-9243-72b68275fbe6@searchtestdomain.com', '0', 'Role1', NULL, NULL

Answer 1

我发现的是：

1）当我在MySQL中进行查询时，我得到如下结果：

2）在Test.java尝试查找两个字符串的顺序如下：

import java.util.Arrays;

public class Test {
  public static void main(String[] args) {   
      String[] arr={"cco1@blah.com", "cco10.test@blah.com"};
      Arrays.sort(arr);   
      System.out.println(Arrays.toString(arr));
  }
}

，输出为：

3）要查看排序规则，请执行以下查询：

SELECT table_catalog,
       table_schema,
       table_name,
       column_name,
       collation_name
FROM   information_schema.columns
WHERE  table_schema = 'test'
       AND column_name = 'email';

输出是：

因此，当MySQL排序规则为utf8_general_ci时，我们发现MySQL和Java都以相同的顺序对两个字符串进行排序。

Answer 2

请参阅<TableLayout android:divider="?android:attr/dividerHorizontal" android:showDividers="middle" ...

上的comment

String.compareTo()

MySql以类似的方式执行，但依赖于DB的/** * Compares two strings lexicographically. * The comparison is based on the Unicode value of each character in * the strings. The character sequence represented by this * {@code String} object is compared lexicographically to the * character sequence represented by the argument string. The result is * a negative integer if this {@code String} object * lexicographically precedes the argument string. The result is a * positive integer if this {@code String} object lexicographically * follows the argument string. The result is zero if the strings * are equal; {@code compareTo} returns {@code 0} exactly when * the {@link #equals(Object)} method would return {@code true}. * * This is the definition of lexicographic ordering. If two strings are * different, then either they have different characters at some index * that is a valid index for both strings, or their lengths are different, * or both. If they have different characters at one or more index * positions, let k be the smallest such index; then the string * whose character at position k has the smaller value, as * determined by using the < operator, lexicographically precedes the * other string. In this case, {@code compareTo} returns the * difference of the two character values at position {@code k} in * the two string -- that is, the value: * <blockquote><pre> * this.charAt(k)-anotherString.charAt(k) * </pre></blockquote> * If there is no index position at which they differ, then the shorter * string lexicographically precedes the longer string. In this case, * {@code compareTo} returns the difference of the lengths of the * strings -- that is, the value: * <blockquote><pre> * this.length()-anotherString.length() * </pre></blockquote> * * @param anotherString the {@code String} to be compared. * @return the value {@code 0} if the argument string is equal to * this string; a value less than {@code 0} if this string * is lexicographically less than the string argument; and a * value greater than {@code 0} if this string is * lexicographically greater than the string argument. */ public int compareTo(String anotherString) { int len1 = value.length; int len2 = anotherString.value.length; int lim = Math.min(len1, len2); char v1[] = value; char v2[] = anotherString.value; int k = 0; while (k < lim) { char c1 = v1[k]; char c2 = v2[k]; if (c1 != c2) { return c1 - c2; } k++; } return len1 - len2; }。更多信息请访问：this

Answer 3

在this Stack Overflow问题中找到了解决方案。看起来Java的词典排序不是基于utf8_unicode_ci中实现的自然语言排序。这里的解决方案是创建一个Collator并使用Collator的compareTo方法来执行我们的排序。

MySQL ORDER BY比较与Java compareTo比较

3 个答案: