如何替换MySQL字符串中特定字符的所有其他实例? [英] How to replace every other instance of a particular character in a MySQL string?

问题描述

如何通过查询替换mysql列中的值，例如，列为options及其类型为varchar(255)

How to replace value in mysql column by query like, Column is options and its of type varchar(255)

来自

id   options
1    A|10|B|20|C|30
2    A|Positive|B|Negative

收件人

id   options
1    A|10,B|20,C|30
2    A|Positive,B|Negative

我是用php这样的.

<?php
    $str =  "A|10|B|20|C|30";
    $arr = explode("|",$str);
    $newArr = array();
    for($i=0;$i<count($arr);$i+=2){
      if($arr[$i] && $arr[$i+1]){
        $newArr[] = $arr[$i]."|".$arr[$i+1];
      }
    }
    echo "Before:".$str."\n";
    echo "After :".implode(",",$newArr);
?>

https://eval.in/841007

因此，我想在MySQL中做到这一点，而不是PHP.

So instead of PHP, I want to do this in MySQL.

推荐答案

您应考虑将数据存储在规范化架构中.在您的情况下，表格应如下所示:

You should consider to store your data in a normalized schema. In your case the table should look like:

| id | k |        v |
|----|---|----------|
|  1 | A |       10 |
|  1 | B |       20 |
|  1 | C |       30 |
|  2 | A | Positive |
|  2 | B | Negative |

此架构更加灵活，您将了解原因.

This schema is more flexible and you will see why.

那么如何将给定的数据转换为新的架构?您将需要一个包含序列号的帮助程序表.由于您的列是varchar(255)，因此您只能在其中存储128个值(+ 127个定界符).但是，我们只创建1000个数字.您可以使用具有足够行的任何表.但是，由于任何MySQL服务器都有information_schema.columns表，因此我将使用它.

So how to convert the given data into the new schema? You will need a helper table containing sequence numbers. Since your column is varchar(255) you can only store 128 values (+ 127 delimiters) in it. But let's just create 1000 numbers. You can use any table with enough rows. But since any MySQL server has the information_schema.columns table, I will use it.

drop table if exists helper_sequence;
create table helper_sequence (i int auto_increment primary key)
    select null as i
    from information_schema.columns c1
    join information_schema.columns c2
    limit 1000;

通过将两个表连接起来，我们将使用此数字作为字符串中值的位置.

We will use this numbers as position of the values in your string by joining the two tables.

要从定界字符串中提取值，可以使用substring_index()函数. i位置的值将是

To extract a value from a delimited string you can use the substring_index() function. The value at position i will be

substring_index(substring_index(t.options, '|', i  ), '|', -1)

在您的字符串中，有一个键序列及其后的值.键的位置是一个奇数.因此，如果键的位置为i，则对应值的位置将为i+1

In your string you have a sequence of keys followed by its values. The position of a key is an odd number. So if the position of the key is i, the position of the corresponding value will be i+1

要获取字符串中定界符的数量并限制我们的联接，我们可以使用

To get the number of the delimiters in the string and limit our join we can use

char_length(t.options) - char_length(replace(t.options, '|', ''))

以规范化形式存储数据的查询为:

The query to store the data in a normalized form would be:

create table normalized_table
    select t.id
        , substring_index(substring_index(t.options, '|', i  ), '|', -1) as k
        , substring_index(substring_index(t.options, '|', i+1), '|', -1) as v
    from old_table t
    join helper_sequence s
      on s.i <= char_length(t.options) - char_length(replace(t.options, '|', ''))
    where s.i % 2 = 1

现在运行select * from normalized_table，您将获得以下信息:

Now run select * from normalized_table and you will get this:

| id | k |        v |
|----|---|----------|
|  1 | A |       10 |
|  1 | B |       20 |
|  1 | C |       30 |
|  2 | A | Positive |
|  2 | B | Negative |

那么为什么这种格式是更好的选择?除了许多其他原因之外，一个原因是您可以使用

So why is this format a better choice? Besides many other reasons, one is that you can easily convert it to your old schema with

select id, group_concat(concat(k, '|', v) order by k separator '|') as options
from normalized_table
group by id;

| id |               options |
|----|-----------------------|
|  1 |        A|10|B|20|C|30 |
|  2 | A|Positive|B|Negative |

或您想要的格式

select id, group_concat(concat(k, '|', v) order by k separator ',') as options
from normalized_table
group by id;

| id |               options |
|----|-----------------------|
|  1 |        A|10,B|20,C|30 |
|  2 | A|Positive,B|Negative |

如果您不关心规范化而只希望完成此任务，则可以使用以下方式更新表

If you don't care about normalization and just want this task to be done, you can update your table with

update old_table o
join (
    select id, group_concat(concat(k, '|', v) order by k separator ',') as options
    from normalized_table
    group by id
) n using (id)
set o.options = n.options;

然后放下normalized_table.

但是那样您将无法使用简单的查询，例如

But then you won't be able to use simple queries like

select *
from normalized_table
where k = 'A'

请参阅rextester.com上的演示

See demo at rextester.com

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