php - MySql PHP 从逗号分隔数据(标签)中选择不同值的计数

Question

如何从存储为 MySql 中的逗号分隔值的数据中选择不同值的计数？最后我将使用 PHP 从 MySql 中输出数据。

里面有什么，是每个帖子的标签。所以最后，我尝试像 stackoverflow 处理它的标签一样输出数据，如下所示:

tag-name x 5

这就是表格中数据的样子(对内容感到抱歉，但它是一个食谱网站)。

"postId"    "tags"                                  "category-code"
"1"         "pho,pork"                              "1"
"2"         "fried-rice,chicken"                    "1"
"3"         "fried-rice,pork"                       "1"
"4"         "chicken-calzone,chicken"               "1"
"5"         "fettuccine,chicken"                    "1"
"6"         "spaghetti,chicken"                     "1"
"7"         "spaghetti,chorizo"                     "1"
"8"         "spaghetti,meat-balls"                  "1"
"9"         "miso-soup"                             "1"
"10"        "chanko-nabe"                           "1"
"11"        "chicken-manchurian,chicken,manchurain" "1"
"12"        "pork-manchurian,pork,manchurain"       "1"
"13"        "sweet-and-sour-pork,pork"              "1"
"14"        "peking-duck,duck"                      "1"

输出

chicken             5 // occurs 5 time in the data above
pork                4 // occurs 4 time in the data above
spaghetti           3 // an so on
fried-rice          2
manchurian          2
pho                 1
chicken-calzone     1
fettuccine          1
chorizo             1
meat-balls          1
miso-soup           1
chanko-nabe         1
chicken-manchurian  1
pork-manchurian     1
sweet-n-sour-pork   1
peking-duck         1
duck                1

我正在尝试选择其中所有不同值的计数，但由于它是逗号分隔的数据，因此似乎没有办法做到这一点。 select distinct 将不起作用。

你能想出一个好方法，无论是在 mysql 中还是使用 php 来获取输出，就像我所做的那样？

Answer 1

解决方案

我真的不知道如何在不创建包含数字的表格的情况下将逗号分隔值的水平列表转换为行列表，数字的数量与逗号分隔的值一样多。如果您可以创建此表，这是我的答案:

SELECT 
  SUBSTRING_INDEX(SUBSTRING_INDEX(all_tags, ',', num), ',', -1) AS one_tag,
  COUNT(*) AS cnt
FROM (
  SELECT
    GROUP_CONCAT(tags separator ',') AS all_tags,
    LENGTH(GROUP_CONCAT(tags SEPARATOR ',')) - LENGTH(REPLACE(GROUP_CONCAT(tags SEPARATOR ','), ',', '')) + 1 AS count_tags
  FROM test
) t
JOIN numbers n
ON n.num <= t.count_tags
GROUP BY one_tag
ORDER BY cnt DESC;

返回:

+---------------------+-----+
| one_tag             | cnt |
+---------------------+-----+
| chicken             |   5 |
| pork                |   4 |
| spaghetti           |   3 |
| fried-rice          |   2 |
| manchurain          |   2 |
| pho                 |   1 |
| chicken-calzone     |   1 |
| fettuccine          |   1 |
| chorizo             |   1 |
| meat-balls          |   1 |
| miso-soup           |   1 |
| chanko-nabe         |   1 |
| chicken-manchurian  |   1 |
| pork-manchurian     |   1 |
| sweet-and-sour-pork |   1 |
| peking-duck         |   1 |
| duck                |   1 |
+---------------------+-----+
17 rows in set (0.01 sec)

见 sqlfiddle

---

说明

场景

1. 我们使用逗号连接所有标签以仅创建一个标签列表，而不是每行一个
2. 我们计算列表中有多少标签
3. 我们找到了如何在这个列表中获得一个值
4. 我们找到了如何将所有值作为不同的行来获取
5. 我们按值(value)分组计算标签

上下文

让我们构建您的架构:

CREATE TABLE test (
    id INT PRIMARY KEY,
    tags VARCHAR(255)
);

INSERT INTO test VALUES
    ("1",         "pho,pork"),
    ("2",         "fried-rice,chicken"),
    ("3",         "fried-rice,pork"),
    ("4",         "chicken-calzone,chicken"),
    ("5",         "fettuccine,chicken"),
    ("6",         "spaghetti,chicken"),
    ("7",         "spaghetti,chorizo"),
    ("8",         "spaghetti,meat-balls"),
    ("9",         "miso-soup"),
    ("10",        "chanko-nabe"),
    ("11",        "chicken-manchurian,chicken,manchurain"),
    ("12",        "pork-manchurian,pork,manchurain"),
    ("13",        "sweet-and-sour-pork,pork"),
    ("14",        "peking-duck,duck");

连接所有标签列表

我们将在一行中处理所有标签，因此我们使用 GROUP_CONCAT 来完成这项工作:

SELECT GROUP_CONCAT(tags SEPARATOR ',') FROM test;

返回以逗号分隔的所有标签:

pho,pork,fried-rice,chicken,fried-rice,pork,chicken-calzone,chicken,fettuccine,chicken,spaghetti,chicken,spaghetti,chorizo,spaghetti,meat-balls,miso-soup,chanko-nabe,chicken-manchurian,chicken,manchurain,pork-manchurian,pork,manchurain,sweet-and-sour-pork,pork,peking-duck,duck

统计所有标签

为了计算所有标签，我们得到标签的完整列表的长度，并且在将 , 替换为空之后，我们删除了完整的标签列表的长度。我们加 1，因为分隔符在两个值之间。

SELECT LENGTH(GROUP_CONCAT(tags SEPARATOR ',')) - LENGTH(REPLACE(GROUP_CONCAT(tags SEPARATOR ','), ',', '')) + 1 AS count_tags
FROM test;

返回:

+------------+
| count_tags |
+------------+
|         28 |
+------------+
1 row in set (0.00 sec)

获取标签列表中的第N个标签

我们使用SUBSTRING_INDEX函数来获取

-- returns the string until the 2nd delimiter\'s occurrence from left to right: a,b
SELECT SUBSTRING_INDEX('a,b,c', ',', 2);

-- return the string until the 1st delimiter, from right to left: c
SELECT SUBSTRING_INDEX('a,b,c', ',', -1);

-- we need both to get: b (with 2 being the tag number)
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX('a,b,c', ',', 2), ',', -1);

按照这样的逻辑，要获取列表中的第三个标签，我们使用:

SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(tags SEPARATOR ','), ',', 3), ',', -1)
FROM test;

返回:

+-------------------------------------------------------------------------------------+
| SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(tags SEPARATOR ','), ',', 3), ',', -1) |
+-------------------------------------------------------------------------------------+
| fried-rice                                                                          |
+-------------------------------------------------------------------------------------+
1 row in set (0.00 sec)

获取所有值作为不同的行

我的想法有点棘手:

1. 我知道我们可以通过连接表来创建行
2. 我需要使用上面的请求获取列表中的第 N 个标签

因此，我们将创建一个表格，其中包含从 1 到列表中可能拥有的最大标签数的所有数字。如果您可以有 1M 值，请创建从 1 到 1,000,000 的 1M 条目。对于 100 个标签，这将是:

CREATE TABLE numbers (
  num INT PRIMARY KEY
);

INSERT INTO numbers VALUES
    ( 1 ), ( 2 ), ( 3 ), ( 4 ), ( 5 ), ( 6 ), ( 7 ), ( 8 ), ( 9 ), ( 10 ), 
    ( 11 ), ( 12 ), ( 13 ), ( 14 ), ( 15 ), ( 16 ), ( 17 ), ( 18 ), ( 19 ), ( 20 ), 
    ( 21 ), ( 22 ), ( 23 ), ( 24 ), ( 25 ), ( 26 ), ( 27 ), ( 28 ), ( 29 ), ( 30 ), 
    ( 31 ), ( 32 ), ( 33 ), ( 34 ), ( 35 ), ( 36 ), ( 37 ), ( 38 ), ( 39 ), ( 40 ), 
    ( 41 ), ( 42 ), ( 43 ), ( 44 ), ( 45 ), ( 46 ), ( 47 ), ( 48 ), ( 49 ), ( 50 ), 
    ( 51 ), ( 52 ), ( 53 ), ( 54 ), ( 55 ), ( 56 ), ( 57 ), ( 58 ), ( 59 ), ( 60 ), 
    ( 61 ), ( 62 ), ( 63 ), ( 64 ), ( 65 ), ( 66 ), ( 67 ), ( 68 ), ( 69 ), ( 70 ), 
    ( 71 ), ( 72 ), ( 73 ), ( 74 ), ( 75 ), ( 76 ), ( 77 ), ( 78 ), ( 79 ), ( 80 ), 
    ( 81 ), ( 82 ), ( 83 ), ( 84 ), ( 85 ), ( 86 ), ( 87 ), ( 88 ), ( 89 ), ( 90 ), 
    ( 91 ), ( 92 ), ( 93 ), ( 94 ), ( 95 ), ( 96 ), ( 97 ), ( 98 ), ( 99 ), ( 100 );

现在，我们使用以下查询获得第 numth(num 是 number 中的一行):

SELECT n.num, SUBSTRING_INDEX(SUBSTRING_INDEX(all_tags, ',', num), ',', -1) as one_tag
FROM (
  SELECT
    GROUP_CONCAT(tags SEPARATOR ',') AS all_tags,
    LENGTH(GROUP_CONCAT(tags SEPARATOR ',')) - LENGTH(REPLACE(GROUP_CONCAT(tags SEPARATOR ','), ',', '')) + 1 AS count_tags
  FROM test
) t
JOIN numbers n
ON n.num <= t.count_tags

返回:

+-----+---------------------+
| num | one_tag             |
+-----+---------------------+
|   1 | pho                 |
|   2 | pork                |
|   3 | fried-rice          |
|   4 | chicken             |
|   5 | fried-rice          |
|   6 | pork                |
|   7 | chicken-calzone     |
|   8 | chicken             |
|   9 | fettuccine          |
|  10 | chicken             |
|  11 | spaghetti           |
|  12 | chicken             |
|  13 | spaghetti           |
|  14 | chorizo             |
|  15 | spaghetti           |
|  16 | meat-balls          |
|  17 | miso-soup           |
|  18 | chanko-nabe         |
|  19 | chicken-manchurian  |
|  20 | chicken             |
|  21 | manchurain          |
|  22 | pork-manchurian     |
|  23 | pork                |
|  24 | manchurain          |
|  25 | sweet-and-sour-pork |
|  26 | pork                |
|  27 | peking-duck         |
|  28 | duck                |
+-----+---------------------+
28 rows in set (0.01 sec)

统计标 checkout 现次数

只要我们现在有了 classic 行，我们就可以轻松计算每个标签的出现次数。

见 top of this answer查看请求。