Question

如何在两对＃或多对＃之间找到字符串。

要搜索的示例文字：这是＃important＃，需要进一步阐述。记得在回家之前购买#milk＃。

我希望结果如下：

重要

回家之前喝牛奶

Answer 1

您可以尝试此代码

<?php
    $s =  'This is #important# and needs to elaborated further. Remember to buy #milk before coming home#';
    for($i=0;$i<strlen($s);$i++)
        if($s[$i] == "#") {
        $i++;
        $str="";
        while($s[$i]!='#') {
            echo $s[$i];
            $i++;
        }
        echo '<br>';
    };
    ?>

然后你会得到你的结果

重要

回家之前喝牛奶

Answer 2

编辑1

create table t91
(   id int auto_increment primary key,
    thing varchar(1000) not null
);
insert t91(thing) values
('This is #important# and needs to elaborated further. Remember to buy #milk before coming home#'),
('This is #important# and needs to elaborated further. Remember to buy #milk home#'),
('This is #');


select id,thing,theCount
from
(   SELECT id,thing,
        ROUND (   
            (
                LENGTH(thing)
                - LENGTH( REPLACE ( thing, "#", "") ) 
            ) / LENGTH("#")        
        ) AS theCount    
    FROM t91
) d
where d.theCount>1
-- 2 rows returned (id 1 and 2)

以上编辑使用了从This Answer

挖走的策略

EDIT2

-- truncate table t91;
create table t91
(   id int auto_increment primary key,
    thing varchar(1000) not null
);
insert t91(thing) values
('This is #important# and needs to elaborated further. Remember to buy #milk before coming home#'),
('This is #important# and needs to elaborated further. Remember to buy #milk home#'),
('This is #'),
('This is #1# ... #2# ... #');

功能：

DROP FUNCTION IF EXISTS findInsideHashMarks;
DELIMITER $$
CREATE FUNCTION findInsideHashMarks(s VARCHAR(200),segments INT)
RETURNS VARCHAR(200)
BEGIN
    DECLARE i,nextPos,i1,i2 INT;
    DECLARE sOut VARCHAR(200);
    SET i=0;
    SET nextPos=1;
    SET sOut='';
    WHILE i<segments DO
        -- LOCATE(substr,str,pos)
        SET i1=LOCATE('#',s,nextPos);
        IF i1>0 THEN
            SET i1=i1+1;
            SET nextPos=i1+1;
            SET i2=LOCATE('#',s,nextPos);
            IF i2>0 THEN
                SET nextPos=i2+1;
                SET i2=i2-1;
                SET sOut=CONCAT(sOut,SUBSTRING(s,i1,i2-i1+1));
            END IF;
        END IF;
        SET i=i+1;
        IF i<segments THEN
            SET sOut=CONCAT(sOut,',');
        END IF;
    END WHILE;
    RETURN sOut;
END$$
DELIMITER ;

查询：

SELECT id,
SUBSTRING_INDEX(SUBSTRING_INDEX(segString, ',', n.digit+1), ',', -1) dummy
FROM
(   select id,thing,theCount,cast(floor(theCount / 2) as unsigned) as segments,
    findInsideHashMarks(thing,cast(floor(theCount / 2) as unsigned)) as segString
    FROM
    (   SELECT id,thing,
            ROUND (   
                (   LENGTH(thing)
                    - LENGTH( REPLACE ( thing, "#", "") ) 
                ) / LENGTH("#")        
            ) AS theCount    
        FROM t91
    ) d1
    where d1.theCount>1
) d2
INNER JOIN
(SELECT 0 digit UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3) n
ON LENGTH(REPLACE(d2.segString, ',' , '')) <= LENGTH(d2.segString)-n.digit
ORDER BY d2.id,n.digit;

输出：

+----+-------------------------+
| id | dummy                   |
+----+-------------------------+
|  1 | important               |
|  1 | milk before coming home |
|  2 | important               |
|  2 | milk home               |
|  4 | 1                       |
|  4 | 2                       |
+----+-------------------------+

集合（a，b，c）到行的数字突破受到了来自用户Answer的fthiella的启发（低估）。正如fthiella的答案中所指出的那样（或者没有明确表示），UNION策略可以扩展为在#个标记之间拾取10个或更多数据块。

另请注意Edit1底部的先前归因。

EDIT3

Edit2的函数和查询的分支，但使用永久帮助程序表来避免查询派生表UNION中的n。它最多支持100个#段。使用先前的（Edit2）t91表和函数。

架构：

CREATE TABLE 4kTable
(   -- a helper table of about 4k consecutive ints
    id int auto_increment primary key,
    thing int null
)engine=MyISAM;

insert 4kTable (thing) values (null),(null),(null),(null),(null),(null),(null),(null),(null);
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
insert 4kTable (thing) select thing from 4kTable;
-- verify:
-- select min(id),max(id),count(*) from 4kTable;
-- 1 4608 4608

ALTER TABLE 4kTable ENGINE = InnoDB; -- *********** it is now InnoDB

-- verify:
-- select min(id),max(id),count(*) from 4kTable;
-- 1 4608 4608
-- no innodb auto_increment gaps (consecutive block)

查询：

SELECT id, 
SUBSTRING_INDEX(SUBSTRING_INDEX(segString, ',', n.digit+1), ',', -1) dummy 
FROM 
(   select d1.id,thing,theCount,cast(floor(theCount / 2) as unsigned) as segments, 
    findInsideHashMarks(thing,cast(floor(theCount / 2) as unsigned)) as segString 
    FROM 
    (   SELECT id,thing, 
            ROUND (   
                (   LENGTH(thing) 
                    - LENGTH( REPLACE ( thing, "#", "") )  
                ) / LENGTH("#")         
            ) AS theCount     
        FROM t91 
    ) d1 
    where d1.theCount>1 
) d2 
INNER JOIN
(select id-1 as digit from 4kTable where id<=100) n
ON LENGTH(REPLACE(d2.segString, ',' , '')) <= LENGTH(d2.segString)-n.digit
ORDER BY d2.id,n.digit;

与Edit2 Output部分相同的输出。我通常在一个系统中有一些永久的助手表，尤其是带有日期的left joins。

MySQL在两个＃/多对＃之间获取字符串＃

2 个答案:

编辑1

EDIT2

EDIT3