我在这里有问题表我有584行，即584个问题，本章节表中的章节表有类型栏，我为不同类型添加了章节，即类型1的18章和类型2和18的18章类型3的章节，因此章节的id是不同的。
现在我在类型1的问题表中添加了584个问题，现在这些584个问题我想在同一个问题表中添加但是对于类型2和问题表有chapter_id列，所以chapter_id是80到97.所以对于每章如果我们用章节号划分584，即18则结果为32，所以对于每章32个问题将添加章节id，如前32章章节id将为80，另一章章节id将为81等等
但是18 * 32 = 576和584-576 = 08，所以根据我的逻辑，8个问题不会在表格中插入，但如果我使用foreach章节id变量并且我没有得到增量。
<?php
require 'Database.php';
$dbh = new PDO('mysql:host=localhost;dbname=airman_handbook','siddhi', 'siddhi');
$sth = $dbh->prepare("SELECT * FROM `questions`");
$sth->execute();
$result = $sth->fetchAll(PDO::FETCH_ASSOC);
$count = $sth->rowCount();
$chapterId = 80;
$i = 0;
foreach ($result as $chapter) {
while ($i <= 32)
{
$stmt = $dbh->prepare("INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer,type,chapterId) VALUES (?, ?, ?, ?, ?, ?, ?,?)");
$stmt->execute(array($result['question'], $result['answer_a'], $result['answer_b'], $result['answer_c'], $result['answer_d'], $result['answer'], 2, $chapterId));
}
$i++;
$chapterId ++;
}
?>
Chapter table :
id
title
type
33
Ch 1: Air Force heritage
1
34
Ch 2: Enlisted History
1
35
Ch 3: Organization
1
36
Ch 4: Air Force doctrine, AEF, Joint force
1
37
Ch 5: Emergency management
1
38
Ch 6: Standards of conduct
1
39
Ch 7: Enforcing standards and legal issues
1
40
Ch 8: Military customs, courtesies, and protocol f...
1
41
Ch 9: Enlisted force development
1
42
Ch 10: Leadership
1
43
Ch 12: Training and education
1
44
Ch 13: Resource management
1
45
Ch 14: Communicating in today's Air Force
1
46
Ch 15: Personnel programs
1
47
Ch 16: Wing support
1
48
Ch 17: Dress and appearance
1
49
Ch 18: Fit force
1
50
Ch 19: Security
1
80
Ch 1: Air Force heritage
2
81
Ch 2: Enlisted History
2
82
Ch 3: Organization
2
83
Ch 4: Air Force doctrine, AEF, Joint force
2
84
Ch 5: Emergency management
2
85
Ch 6: Standards of conduct
2
86
Ch 7: Enforcing standards and legal issues
2
87
Ch 8: Military customs, courtesies, and protocol f...
2
88
Ch 9: Enlisted force development
2
89
Ch 10: Leadership
2
90
Ch 12: Training and education
2
91
Ch 13: Resource management
2
92
Ch 14: Communicating in today's Air Force
2
93
Ch 15: Personnel programs
2
94
Ch 16: Wing support
2
95
Ch 17: Dress and appearance
2
96
Ch 18: Fit force
2
97
Ch 19: Security
2
98
Ch 1: Air Force heritage
3
99
Ch 2: Enlisted History
3
100
Ch 3: Organization
3
101
Ch 4: Air Force doctrine, AEF, Joint force
3
102
Ch 5: Emergency management
3
103
Ch 6: Standards of conduct
3
104
Ch 7: Enforcing standards and legal issues
3
105
Ch 8: Military customs, courtesies, and protocol f...
3
106
Ch 9: Enlisted force development
3
107
Ch 10: Leadership
3
108
Ch 12: Training and education
3
109
Ch 13: Resource management
3
110
Ch 14: Communicating in today's Air Force
3
111
Ch 15: Personnel programs
3
112
Ch 16: Wing support
3
113
Ch 17: Dress and appearance
3
114
Ch 18: Fit force
3
115
Ch 19: Security
3
Questions :
id
question
answer_a
answer_b
answer_c
answer_d
answer
type
chapterId
56
_______ achieved the first powered, sustained, con...
Otto Lilienthal
Orville and Wilbur Wright
The Tuskegee Airmen
Lieutenant Thomas E. Selfridge
B
1
24
57
_______ achieved the first powered, sustained, con...
Otto Lilienthal
Orville and Wilbur Wright
The Tuskegee Airmen
Lieutenant Thomas E. Selfridge
B
1
33
58
The U.S. and ___________ signed the Anti-Ballistic...
Iran
North Korea
the Union of Soviet Socialist Republics
Vietnam War
C
1
33
59
The purpose of Operation Desert Shield was to: (Ch...
defend Saudi Arabia and Persian Gulf states
respond to the attack on the U.S. Embassy in Iran
avoid a repetition of the Iranian hostage crisis
all of the above
A
1
24
编辑：
require 'Database.php';
ini_set('max_execution_time', 300);
$dbh = new PDO('mysql:host=localhost;dbname=airman_handbook','siddhi', 'siddhi');
$sth = $dbh->prepare("SELECT * FROM `questions`");
$sth->execute();
$result = $sth->fetchAll(PDO::FETCH_ASSOC);
$count = $sth->rowCount();
$chapterId = 80;
$i = 0;
foreach ($result as $chapter) {
while ($i <= 32)
{
$stmt = $dbh->prepare("INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer,type,chapterId) VALUES (?, ?, ?, ?, ?, ?, ?,?)");
$stmt->execute(array($result['question'], $result['answer_a'], $result['answer_b'], $result['answer_c'], $result['answer_d'], $result['answer'], 2, $chapterId));
// $count = $stmt->rowCount();
if($stmt)
{
echo 'row inserted';
}
}
$i++;
if($i > 32)
{
$i = 0;
}
$chapterId ++;
if($chapterId > 91)
{
$chapterId = 80;
}
}
?>
我能为此做些什么？请帮忙谢谢。
答案 0 :(得分：1)
经过讨论，以下内容将复制问题，并为每一章分配一个（大致）相同的数字。
require 'Database.php';
ini_set('max_execution_time', 300);
$dbh = new PDO('mysql:host=localhost;dbname=airman_handbook','siddhi', 'siddhi');
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sth = $dbh->prepare("SELECT * FROM `questions` WHERE `type` = 1 ORDER BY `id`;");
$sth->execute();
$result = $sth->fetchAll(PDO::FETCH_ASSOC);
$count = $sth->rowCount();
$StartChapterId = 79; // Will update first through foreach, so it should be 1 less than the 1st ID.
$NumChapters = 18;
$questionsPerChapter = ceil(584 / $NumChapters);
$chapterId = 0;
$num_inserted = 0;
foreach($result as $rowno => $chapter) {
if($rowno % $questionsPerChapter === 0) {
$chapterId++;
echo "<li>".$num_inserted." inserted in chapterId: ".$chapterId + $StartChapterId."</li>\n";
$num_inserted = 0;
}
$stmt = $dbh->prepare("INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer,type,chapterId) VALUES (?, ?, ?, ?, ?, ?, ?,?)");
$stmt->execute(array($chapter['question'], $chapter['answer_a'], $chapter['answer_b'], $chapter['answer_c'], $chapter['answer_d'], $chapter['answer'], 2, $chapterId + $StartChapterId));
$num_inserted++;
}
echo "<li>".$num_inserted." inserted in chapterId: ".$chapterId + $StartChapterId."</li>\n";
作为替代方案，这里是一个SQL语句，它将根据章节表中的内容进行复制。
这是一个SQL语句，应该复制问题，并根据您的要求更改类型和chapterId。
INSERT INTO `questions`
(`question`,
`answer_a`,
`answer_b`,
`answer_c`,
`answer_d`,
`answer`,
`type`,
`chapterId`)
SELECT `question`,
`answer_a`,
`answer_b`,
`answer_c`,
`answer_d`,
`answer`,
c2.`type` as `type`,
c2.`id` as `chapterId`
FROM `questions` q
JOIN `chapters` c1
ON c1.`id` = q.`chapterId` AND c1.`type` = 1
JOIN `chapters` c2
ON c2.`title` = c1.`title` AND c2.`id` <> c1.`id`
WHERE q.`type` = 1
ORDER BY c2.`type`, c2.`id`, q.`id`;