在找到一些有趣的来源后，我最终得到了这个解决方案（添加了一个额外的函数来处理几个荷兰语字符）：

drop function if exists fn_remove_accents;
delimiter |
create function fn_remove_accents( textvalue varchar(20000) )
returns varchar(20000)
begin

set @textvalue = textvalue;

-- ACCENTS
set @withaccents = \'ŠšŽžÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÑÒÓÔÕÖØÙÚÛÜÝŸÞàáâãäåæçèéêëìíîïñòóôõöøùúûüýÿþƒ\';
set @withoutaccents = \'SsZzAAAAAAACEEEEIIIINOOOOOOUUUUYYBaaaaaaaceeeeiiiinoooooouuuuyybf\';
set @count = length(@withaccents);

while @count > 0 do
    set @textvalue = replace(@textvalue, substring(@withaccents, @count, 1), substring(@withoutaccents, @count, 1));
    set @count = @count - 1;
end while;

-- SPECIAL CHARS
set @special = \'!@#$%¨&*()_+=§¹²³£¢¬"`´{[^~}]<,>.:;?/°ºª+*|\\\\\'\'\';
set @count = length(@special);
while @count > 0 do
    set @textvalue = replace(@textvalue, substring(@special, @count, 1), \'\');
    set @count = @count - 1;
end while;

return @textvalue;

end
|

DROP FUNCTION IF EXISTS `slugify`;
DELIMITER ;;
CREATE DEFINER=`root`@`localhost` # I have no idea what this does
FUNCTION `slugify`(dirty_string varchar(200))
RETURNS varchar(200) CHARSET latin1
DETERMINISTIC
BEGIN
    DECLARE x, y , z Int;
    Declare temp_string, allowed_chars, new_string VarChar(200);
    Declare is_allowed Bool;
    Declare c, check_char VarChar(1);

    set allowed_chars = "abcdefghijklmnopqrstuvwxyz0123456789-";
    set temp_string = fn_remove_accents(LOWER(dirty_string));
    Select temp_string Regexp(\'&\') Into x;
    If x = 1 Then
        Set temp_string = replace(temp_string, \'&\', \' and \');
    End If;

    Select temp_string Regexp(\'[^a-z0-9]+\') into x;
    If x = 1 then
        set z = 1;
        While z <= Char_length(temp_string) Do
            Set c = Substring(temp_string, z, 1);
            Set is_allowed = False;
            Set y = 1;
            Inner_Check: While y <= Char_length(allowed_chars) Do
                If (strCmp(ascii(Substring(allowed_chars,y,1)), Ascii(c)) = 0) Then
                    Set is_allowed = True;
                    Leave Inner_Check;
                End If;
                Set y = y + 1;
            End While;
            If is_allowed = False Then
                Set temp_string = Replace(temp_string, c, \'-\');
            End If;

            set z = z + 1;
        End While;
    End If;

    Select temp_string Regexp("^-|-$|\'") into x;
    If x = 1 Then
        Set temp_string = Replace(temp_string, "\'", \'\');
        Set z = Char_length(temp_string);
        Set y = Char_length(temp_string);
        Dash_check: While z > 1 Do
            If Strcmp(SubString(temp_string, -1, 1), \'-\') = 0 Then
                Set temp_string = Substring(temp_string,1, y-1);
                Set y = y - 1;
            Else
                Leave Dash_check;
            End If;
            Set z = z - 1;
        End While;
    End If;

    Repeat
        Select temp_string Regexp("--") into x;
        If x = 1 Then
            Set temp_string = Replace(temp_string, "--", "-");
        End If;
    Until x <> 1 End Repeat;

    Return temp_string;
END;;
DELIMITER ;

UPDATE wp_posts SET post_name = slugify(post_title) WHERE post_name LIKE \'%copy%\' AND post_type = \'MY_POST_TYPE\';

最后，更新查询将匹配那些包含单词的重复帖子copy 在他们的slug中，但根据您的情况，您可能希望使此规则更加严格。我想对我的自定义帖子类型运行此查询，因此我在查询中添加了一个额外的帖子类型检查，这没什么大不了的。

匹配的帖子将根据标题（post\\u title）分配一个新生成的slug，或者您可以为slug生成指定其他来源。

资料来源：

[1] Replace non-latin characters selectively

[2] Regenerate a slug from a free-text name

[3] WebElaine\'s answer (on this page)