pg中replace和translate的用法说明(数据少的中文排序)

1.首先创建students表

CREATE TABLE students

(

id integer NOT NULL,

name character varying(255),

sex character varying(255),

class character varying(255),

"like" character varying(255),

school character varying(255),

phone character varying(255)

)

2.插入数据

INSERT INTO "public"."students" ("id", "name", "sex", "class", "like", "school", "phone")

VALUES ('1', '大猫', '女', '一年级', '绘画', '第三小学', '2345');

INSERT INTO "public"."students" ("id", "name", "sex", "class", "like", "school", "phone")

VALUES ('2', '小厌', '男', '三年级', '书法', '第四小学', '2346');

INSERT INTO "public"."students" ("id", "name", "sex", "class", "like", "school", "phone")

VALUES ('3', '库库', '女', '二年级', '绘画', '第三小学', '2342');

INSERT INTO "public"."students" ("id", "name", "sex", "class", "like", "school", "phone")

VALUES ('4', '艾琳', '女', '四年级', '书法,钢琴', '第四小学', '2349');

结果:

select * from students

如下图:

3.replace 的用法

replace(string text, from text, to text)

返回类型:text

解释:把字串string里出现地所有子字串from替换成子字串to

示例1:

select replace('一条黑色的狗','黑','黑白相间')

结果:一条黑色的狗 变成了 一条黑白相间色的狗

如下图:

示例2:

update students set name=replace(name,'大猫','小猫咪的姐姐')

结果:name为 ‘大猫'的这条数据name='小猫咪的姐姐'

示例3:

select * from students where school='第四小学' ORDER BY replace(name,'艾琳','1')

结果:

4.translate的用法

translate(string text, from text, to text)

返回类型:text

解释:把在string中包含的任何匹配from中的字符的字符转化为对应的在to中的字符。

示例1:

select translate('她真是好看', '好看','漂亮')

结果:

示例2:

select * from students where phone like '2%'

ORDER BY translate(class, '一二三四','1234')

结果:

示例3:

select * from students where phone like '2%'

ORDER BY translate(name, '库小厌猫咪艾','1234')

结果

结论:

有了translate再也不担心中文排序问题了(数据比较少的情况)

补充:pg中position、split_part、translate、strpos、length函数

我就废话不多说了,大家还是直接看代码吧~

select position('.' in '1.1.2.10');

select split_part('1.1.2.10','.',length('1.1.2.10') - length(translate('1.1.2.10','.',''))+1);

select split_part('1.1.2','.',length('1.1.2') - length(translate('1.1.2','.',''))+1);

select length(translate('1.1.2.10','.','a'))+1 as num

select translate('1.1.2.10','.','')

select strpos('1.1.2.10','.')

select instr('1.1.2.10','.',1,3)

select length('1.1.2.10') - length(translate('1.1.2.10','.',''))

以上为个人经验,希望能给大家一个参考,也希望大家多多支持。如有错误或未考虑完全的地方,望不吝赐教。

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