我正在使用PHP和MySQL进行调查。我无法弄清楚用什么方法来显示随机的10个问题及其答案。每个问题可以有2到5个答案。
以下是数据库的设置方法:
questions:
ID | quID | question
answers:
ID | an_quID | anID | answer
quID和an_quID将表链接在一起。
如果我执行以下操作(不是确切的代码;为简洁起见而编辑)。我最终做了20次数据库调用。
$sql = "SELECT * FROM questions ORDER BY RAND() LIMIT 10";
while($row = mysql_fetch_array($result)) {
$quID = $row[quID];
echo "HTML code here";
"SELECT * FROM answers WHERE an_quID = '$quID'"
while($inside_row = mysql_fetch_array($inside_result)) {
$answer = $inside_row[answer];
echo "HTML code here";
}
}
我还尝试使用array_push(),只需2次数据库调用,将所有结果推送到数组中。但是array_push的问题在于,据我所知,你不能在你指定密钥的地方建立关联数组。
显示随机的10个问题及其答案的理想方法是什么?
答案 0 :(得分:1)
而不是每个问题1次调用以获取答案,您可以将所有问题ID拉出来并使用MySQL中的IN()语法立即获取所有答案。然后循环以根据需要展平数据
SELECT * FROM answers WHERE an_quID IN (1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
答案 1 :(得分:1)
joshtonic非常接近。我会做这样的事情:
SELECT * FROM answers WHERE an_quID IN (SELECT quID FROM questions ORDER BY RAND() LIMIT 10);
正如Kuchen所说,您需要对数据库进行一些规范化。你不应该使用id和question_id(id应该是你所需要的全部);
答案 2 :(得分:0)
这样的事情:
SELECT * FROM questions AS q INNER JOIN answers AS a ON q.quID = a.an_quID
ORDER BY RAND() LIMIT 10
然后在您的代码中,您可以根据使用相同 quID
返回的行来过滤结果答案 3 :(得分:0)
您不需要那么多ID,每个表一个,并且可以有很多问题的答案,所以我们需要一个非唯一字段,在答案中指向问题ID:
questions:
id | question
answers:
id | id_questions | answer | correct
循环中的数据库查询(99%的时间)都是坏事 获取问题,并使用WHERE IN和问题ID来获取所需的条目。
基本的工作示例:
/** Tables
CREATE TABLE `questions` (
`id` INT NOT NULL AUTO_INCREMENT,
`question` VARCHAR(255) NOT NULL,
PRIMARY KEY (`id`)
)
ENGINE = MyISAM;
CREATE TABLE `test`.`answers` (
`id` INT NOT NULL AUTO_INCREMENT,
`id_questions` INT NOT NULL,
`answer` VARCHAR(255) NOT NULL,
`correct` BOOL NOT NULL,
PRIMARY KEY (`id`)
)
ENGINE = MyISAM;
*/
/** Test Data
INSERT INTO `questions` (id,question) VALUES
(null,'Pick A!'),(null,'Pick B!'),(null,'Pick C!'),(null,'Pick D!');
INSERT INTO `answers` (id,id_questions,answer,correct) VALUES
(null,1,'Answer A',1),(null,1,'Answer B',0),(null,1,'Answer C',0),(null,1,'Answer D',0),
(null,2,'Answer A',0),(null,2,'Answer B',1),(null,2,'Answer C',0),(null,2,'Answer D',0),
(null,3,'Answer A',0),(null,3,'Answer B',0),(null,3,'Answer C',1),(null,3,'Answer D',0),
(null,4,'Answer A',0),(null,4,'Answer B',0),(null,4,'Answer C',0),(null,4,'Answer D',1);
*/
define('NUM_OF_QUESTIONS_TO_DISPLAY',3);
// DB login data
$db_login = array('host' => 'localhost',
'user' => 'test',
'pass' => 'tset',
'base' => 'test'
);
// DB connect
$db = new mysqli($db_login['host'],$db_login['user'],$db_login['pass'],$db_login['base']);
if (mysqli_connect_errno())
trigger_error('DB connection failed: '.mysqli_connect_errno());
// Get questions
if ($res = $db->query('SELECT q.id, q.question FROM questions q
ORDER BY RAND() LIMIT '.(int)NUM_OF_QUESTIONS_TO_DISPLAY.';')) {
$questions = array();
$questions_ids = array();
while ($r = $res->fetch_assoc()) {
$questions[] = $r;
$questions_ids[] = $r['id'];
}
}
// Get the answers
if ($res = $db->query('SELECT a.id_questions, a.answer, a.correct FROM answers a
WHERE (a.id_questions IN ('.implode(',',$questions_ids).'))
ORDER BY RAND();')) {
$answers = array();
while ($r = $res->fetch_assoc())
$answers[$r['id_questions']][] = $r;
}
?>
<html>
<head>
<title>Foo!</title>
</head>
<body style="margin: 100px;">
<?php foreach ($questions as $question) : ?>
<strong><?php echo $question['question']; ?></strong>
<small>(Question id:<?php echo $question['id']; ?>)</small>
<ul>
<?php foreach ($answers[$question['id']] as $answer) : ?>
<li><?php echo $answer['answer'].($answer['correct'] ? '(*)' : ''); ?></li>
<?php endforeach; ?>
</ul>
<?php endforeach; ?>
</body>
</html>
(不是一个随时可用的脚本,但它可以帮助你解决你遇到的问题)