PostgreSQL：检查数组中的每个项是否包含在较大的字符串中

我在PostgreSQL中有一个字符串数组：

SELECT ARRAY['dog', 'cat', 'mouse'];

我有一段很长的话：

Dogs and cats have a range of interactions. The natural instincts of each species lead towards antagonistic interactions, though individual animals can have non-aggressive relationships with each other, particularly under conditions where humans have socialized non-aggressive behaviors.

The generally aggressive interactions between the species have been noted in cultural expressions.

对于数组中的每个项目，我想检查它是否出现在我的大段落字符串中。我知道，对于任何一个字符串，我都可以执行以下操作：

SELECT paragraph_text ILIKE '%dog%';

---

但是有没有一种方法可以在不使用plpgsql的情况下同时检查数组中的每个字符串（任意数量的数组元素）

我相信您想要这样的内容（假设

段落_text

是表中名为

table

的列）：

函数

unnest（array）

从数组值创建记录表。您可以执行

交叉联接

，这意味着

表1

中的所有行都与该非最新表中的所有行合并

如果

段落\u text

是某种静态值（不是来自表），您可以只执行以下操作：

SELECT
    paragraph_text,
    sub.word,
    paragraph_text ILIKE '%' || sub.word || '%' as is_word_in_text
FROM (
         SELECT unnest(ARRAY['dog', 'cat', 'mouse']) as word
     ) as sub;

---

此解决方案仅适用于postgres 8.4及以上版本，因为早期版本不提供此解决方案

drop table if exists t;
create temp table t (col1 text, search_terms text[] );
insert into t values
   ('postgress is awesome', array['postgres', 'is', 'bad']), 
   ('i like open source', array['open', 'code', 'i']), 
   ('sql is easy', array['mysql']);

drop table if exists t1;
select *, unnest(search_terms) as search_term into temp t1 from t;

-- depending on how you like to do pattern matching. 
-- it will look for term not whole words
select *, position(search_term in col1) from t1;


-- This will match only whole words.
select *, string_to_array(col1, E' ')@>string_to_array(search_term, E' ') from t1;

基本上，您需要将搜索词数组展平为一列，然后将长字符串与每个搜索词逐行匹配。

谢谢--

交叉连接

是个好主意。

drop table if exists t;
create temp table t (col1 text, search_terms text[] );
insert into t values
   ('postgress is awesome', array['postgres', 'is', 'bad']), 
   ('i like open source', array['open', 'code', 'i']), 
   ('sql is easy', array['mysql']);

drop table if exists t1;
select *, unnest(search_terms) as search_term into temp t1 from t;

-- depending on how you like to do pattern matching. 
-- it will look for term not whole words
select *, position(search_term in col1) from t1;


-- This will match only whole words.
select *, string_to_array(col1, E' ')@>string_to_array(search_term, E' ') from t1;