PHP/MySQL-计算具有给定前缀的字段数

是否可以计算具有给定前缀的已填充字段的数量

e、 g.我有一张问答表，上面有多项选择答案。有时一个问题有3个答案选项，有时有4或5个。每个答案都在一个字段中，如

answer1\u a

，

answer1\u b

，

answer1\u c

，等等，是否有办法确定这些字段中有多少字段具有数据（对于给定记录）？我需要这个来确定HTML中要显示多少个多选选项

也许以某种方式使用mysql\u num\u字段

谢谢，

---

Geoff要在SQL中解决这个问题，必须明确列出所有字段名。。假设在未提供数据时字段值设置为

NULL

，SQL解决方案可能如下所示：

SELECT *, 
    ( answer1_a IS NOT NULL +
      answer1_b IS NOT NULL +
      answer1_c IS NOT NULL +
      answer1_d IS NOT NULL +
      answer1_e IS NOT NULL ) AS number_of_answers
FROM answers

CREATE TABLE QUESTION(
  ID INTEGER NOT NULL,
  QUESTION VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID)
);

INSERT INTO QUESTION VALUES(1, 'My first question');
INSERT INTO QUESTION VALUES(2, 'My second question');
INSERT INTO QUESTION VALUES(3, 'My third question');

CREATE TABLE ANSWER(
  ID INTEGER NOT NULL,
  ID_QUESTION INTEGER NOT NULL,
  ANSWER VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID),
  FOREIGN KEY (ID_QUESTION) REFERENCES QUESTION(ID)
);

INSERT INTO ANSWER VALUES (1, 1, 'First possible answer for question 1');
INSERT INTO ANSWER VALUES (2, 1, 'Second possible answer for question 1');
INSERT INTO ANSWER VALUES (3, 1, 'Third possible answer for question 1');
INSERT INTO ANSWER VALUES (4, 2, 'First possible answer for question 2');
INSERT INTO ANSWER VALUES (5, 2, 'Second possible answer for question 2');
INSERT INTO ANSWER VALUES (6, 3, 'First possible answer for question 3');
INSERT INTO ANSWER VALUES (7, 3, 'Second possible answer for question 3');
INSERT INTO ANSWER VALUES (8, 3, 'Third possible answer for question 3');
INSERT INTO ANSWER VALUES (9, 3, 'Fourth possible answer for question 3');

SELECT
  QUESTION.ID,
  COUNT(*) as NB_ANSWER
FROM QUESTION
INNER JOIN ANSWER ON QUESTION.ID = ANSWER.ID_QUESTION
GROUP BY QUESTION.ID
ORDER BY QUESTION.ID;

一种可扩展的方法（当您更改表结构时会自动增长，例如插入一个新的答案“f”）是在PHP中进行这种计算：

<?php
$res = mysql_query("SELECT * FROM answers");
while ( $row = mysql_fetch_assoc($res) ) {
    $count = 0;
    foreach ( $row as $fieldname => $value ) {
        if ( strncmp($fieldname, 'answer1_', 8) === 0 &&
             $value !== NULL ) {
            $count++;
        }
    }
    $row['number_of_answers'] = $count;
    $resultset[] = $row;
}
print_r($resultset);

---

要在SQL中解决此问题，必须明确列出所有字段名。。假设在未提供数据时字段值设置为
NULL
，SQL解决方案可能如下所示：

SELECT *, 
    ( answer1_a IS NOT NULL +
      answer1_b IS NOT NULL +
      answer1_c IS NOT NULL +
      answer1_d IS NOT NULL +
      answer1_e IS NOT NULL ) AS number_of_answers
FROM answers

CREATE TABLE QUESTION(
  ID INTEGER NOT NULL,
  QUESTION VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID)
);

INSERT INTO QUESTION VALUES(1, 'My first question');
INSERT INTO QUESTION VALUES(2, 'My second question');
INSERT INTO QUESTION VALUES(3, 'My third question');

CREATE TABLE ANSWER(
  ID INTEGER NOT NULL,
  ID_QUESTION INTEGER NOT NULL,
  ANSWER VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID),
  FOREIGN KEY (ID_QUESTION) REFERENCES QUESTION(ID)
);

INSERT INTO ANSWER VALUES (1, 1, 'First possible answer for question 1');
INSERT INTO ANSWER VALUES (2, 1, 'Second possible answer for question 1');
INSERT INTO ANSWER VALUES (3, 1, 'Third possible answer for question 1');
INSERT INTO ANSWER VALUES (4, 2, 'First possible answer for question 2');
INSERT INTO ANSWER VALUES (5, 2, 'Second possible answer for question 2');
INSERT INTO ANSWER VALUES (6, 3, 'First possible answer for question 3');
INSERT INTO ANSWER VALUES (7, 3, 'Second possible answer for question 3');
INSERT INTO ANSWER VALUES (8, 3, 'Third possible answer for question 3');
INSERT INTO ANSWER VALUES (9, 3, 'Fourth possible answer for question 3');

SELECT
  QUESTION.ID,
  COUNT(*) as NB_ANSWER
FROM QUESTION
INNER JOIN ANSWER ON QUESTION.ID = ANSWER.ID_QUESTION
GROUP BY QUESTION.ID
ORDER BY QUESTION.ID;

一种可扩展的方法（当您更改表结构时会自动增长，例如插入一个新的答案“f”）是在PHP中进行这种计算：

<?php
$res = mysql_query("SELECT * FROM answers");
while ( $row = mysql_fetch_assoc($res) ) {
    $count = 0;
    foreach ( $row as $fieldname => $value ) {
        if ( strncmp($fieldname, 'answer1_', 8) === 0 &&
             $value !== NULL ) {
            $count++;
        }
    }
    $row['number_of_answers'] = $count;
    $resultset[] = $row;
}
print_r($resultset);

鉴于您的表结构，Kaii的答案是正确的

对于未来，我建议使用如下真正的关系模型：

SELECT *, 
    ( answer1_a IS NOT NULL +
      answer1_b IS NOT NULL +
      answer1_c IS NOT NULL +
      answer1_d IS NOT NULL +
      answer1_e IS NOT NULL ) AS number_of_answers
FROM answers

CREATE TABLE QUESTION(
  ID INTEGER NOT NULL,
  QUESTION VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID)
);

INSERT INTO QUESTION VALUES(1, 'My first question');
INSERT INTO QUESTION VALUES(2, 'My second question');
INSERT INTO QUESTION VALUES(3, 'My third question');

CREATE TABLE ANSWER(
  ID INTEGER NOT NULL,
  ID_QUESTION INTEGER NOT NULL,
  ANSWER VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID),
  FOREIGN KEY (ID_QUESTION) REFERENCES QUESTION(ID)
);

INSERT INTO ANSWER VALUES (1, 1, 'First possible answer for question 1');
INSERT INTO ANSWER VALUES (2, 1, 'Second possible answer for question 1');
INSERT INTO ANSWER VALUES (3, 1, 'Third possible answer for question 1');
INSERT INTO ANSWER VALUES (4, 2, 'First possible answer for question 2');
INSERT INTO ANSWER VALUES (5, 2, 'Second possible answer for question 2');
INSERT INTO ANSWER VALUES (6, 3, 'First possible answer for question 3');
INSERT INTO ANSWER VALUES (7, 3, 'Second possible answer for question 3');
INSERT INTO ANSWER VALUES (8, 3, 'Third possible answer for question 3');
INSERT INTO ANSWER VALUES (9, 3, 'Fourth possible answer for question 3');

SELECT
  QUESTION.ID,
  COUNT(*) as NB_ANSWER
FROM QUESTION
INNER JOIN ANSWER ON QUESTION.ID = ANSWER.ID_QUESTION
GROUP BY QUESTION.ID
ORDER BY QUESTION.ID;

在这里查看它的工作情况：
鉴于您的表结构，Kaii的答案是正确的

对于未来，我建议使用如下真正的关系模型：

SELECT *, 
    ( answer1_a IS NOT NULL +
      answer1_b IS NOT NULL +
      answer1_c IS NOT NULL +
      answer1_d IS NOT NULL +
      answer1_e IS NOT NULL ) AS number_of_answers
FROM answers

CREATE TABLE QUESTION(
  ID INTEGER NOT NULL,
  QUESTION VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID)
);

INSERT INTO QUESTION VALUES(1, 'My first question');
INSERT INTO QUESTION VALUES(2, 'My second question');
INSERT INTO QUESTION VALUES(3, 'My third question');

CREATE TABLE ANSWER(
  ID INTEGER NOT NULL,
  ID_QUESTION INTEGER NOT NULL,
  ANSWER VARCHAR(100) NOT NULL,

  PRIMARY KEY (ID),
  FOREIGN KEY (ID_QUESTION) REFERENCES QUESTION(ID)
);

INSERT INTO ANSWER VALUES (1, 1, 'First possible answer for question 1');
INSERT INTO ANSWER VALUES (2, 1, 'Second possible answer for question 1');
INSERT INTO ANSWER VALUES (3, 1, 'Third possible answer for question 1');
INSERT INTO ANSWER VALUES (4, 2, 'First possible answer for question 2');
INSERT INTO ANSWER VALUES (5, 2, 'Second possible answer for question 2');
INSERT INTO ANSWER VALUES (6, 3, 'First possible answer for question 3');
INSERT INTO ANSWER VALUES (7, 3, 'Second possible answer for question 3');
INSERT INTO ANSWER VALUES (8, 3, 'Third possible answer for question 3');
INSERT INTO ANSWER VALUES (9, 3, 'Fourth possible answer for question 3');

SELECT
  QUESTION.ID,
  COUNT(*) as NB_ANSWER
FROM QUESTION
INNER JOIN ANSWER ON QUESTION.ID = ANSWER.ID_QUESTION
GROUP BY QUESTION.ID
ORDER BY QUESTION.ID;

请在此处查看它的工作情况：
使用循环。无论哪种方式，您都需要获取此字段的值，对吗？检查循环，如果为null或为空，则不显示它。是否可以发布表结构？使用循环。无论哪种方式，您都需要获取此字段的值，对吗？检查循环，如果为null或为空，则不显示。可以发布表结构吗？我认为您不需要
SUM（）
，如果每行都是唯一的问题，只需将它们添加在一起就可以了。我认为您不需要
SUM（）
，如果每一行都是一个唯一的问题，仅仅把它们加在一起就可以了。+1-我还可以建议
ALTER TABLE answer ADD COLUMN sequence INT NOT NULL；更改表格答案添加唯一性（id_问题，顺序）（语法不正确，但你明白了）。+1-我还可以建议
改变表格答案添加列序列INT not NULL；更改表格答案添加唯一性（id_问题，顺序）（语法不正确，但你明白了）。