基本上我有2个简单的表...第一个叫做“user”,它是父表 . PK是索引自动递增 . 第二个表称为“useradvert” . “id”字段用作不自动递增的索引键 . 每当我尝试插入数据时,它都不会进入表格(useradvert) . 我的PHP页面上根本没有错误 . 我打开了错误报告 . 我设法创建一个没有错误的关系表 . 我试图解决这个问题好几天,并在互联网上搜索答案,但仍然无法找到并理解问题 . 问题是由于子表中的索引键(id)不是自动递增的吗?两个id键是否应自动递增?
谢谢..真的需要你的帮助tqs ..
下面的表定义为“users”-parent table和“useradvert”-child表;
- 表useradvert的表结构
CREATE TABLE IF NOT EXISTS useradvert
( id
int(10)unsigned NOT NULL, name2
varchar(60)COLLATE utf8_unicode_ci NOT NULL, color2
varchar(60)COLLATE utf8_unicode_ci NOT NULL, hobby2
varchar(60)COLLATE utf8_unicode_ci NOT NULL,KEY id
( id
))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_unicode_ci;
- 表用户的表结构
CREATE TABLE IF NOT EXISTS users
( id
int(10)unsigned NOT NULL AUTO_INCREMENT, name
varchar(60)COLLATE utf8_unicode_ci NOT NULL, telno
varchar(11)COLLATE utf8_unicode_ci NOT NULL, username
varchar(60)COLLATE utf8_unicode_ci NOT NULL, password
varchar( 60)COLLATE utf8_unicode_ci NOT NULL, date
timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,PRIMARY KEY( id
),UNIQUE KEY username
( username
),KEY id
( id
))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_unicode_ci AUTO_INCREMENT = 97;
- 转储表用户的数据
INSERT INTO users
( id
, name
, telno
, username
, password
, date
)VALUES(95,'Test Name','09999999999','test@test.com','$2y$12$fqdmAQk5c8qk8Eh2TWy2n.AdNO.lFjqmi2ruSzk8tsVXcK71OcPae','2015-12-24 05:00:13'),(96,'testtwo','10121212121','test2@mail.com','$2y$12$nHw0CjWCF5AS4VB3mjIBo.o7nxszxXh.t5FWGv3pFe5izWBOo0A0O','2015-12-24 05:20:19');
- 转储表的约束
- 表useradvert的约束
ALTER TABLE useradvert
ADD CONSTRAINT useradvert_ibfk_1
FOREIGN KEY( id
)REFERENCES users
( id
);
这是用户页面(useracc-test.php),用户希望将数据插入表“useradvert” . 页面显示以前注册的数据(来自表“user”,此页面还允许用户插入新数据(进入表“useradvert”) .
<?php
//useracc-test.php
/**
* Start the session.
*/
session_start();
ini_set('display_errors', 1);
error_reporting(E_ALL);
// require 'lib/password.php';
require 'connect-test.php';
$userName= isset($_POST['username']) ? $_POST['username'] : '';
$query = "SELECT id, name, username, telno FROM users WHERE username = ?";
$stmt = $conn->prepare($query);
$stmt->bind_param('s', $userName);
$stmt->execute();
$res = $stmt->get_result();
?>
<html>
<head>
<style type="text/css">
#apDiv2 {
position: absolute;
left: 51px;
top: 238px;
width: 237px;
height: 93px;
z-index: 1;
}
#apDiv1 {
position: absolute;
left: 134px;
top: 123px;
width: 234px;
height: 104px;
z-index: 2;
}
#apDiv3 {
position: absolute;
left: 58px;
top: 146px;
width: 219px;
height: 61px;
z-index: 2;
}
#apDiv4 {
position: absolute;
left: 302px;
top: 102px;
width: 365px;
height: 123px;
z-index: 3;
}
</style>
<link href="SpryAssets/SpryTabbedPanels.css" rel="stylesheet" type="text/css">
<script src="SpryAssets/SpryTabbedPanels.js" type="text/javascript"></script>
</head>
<body>
Your Personal details:</p>
<p><?php while($row = $res->fetch_array()): ?>
<p><?php echo $row['id']; ?></p>
<p><?php echo $row['name']; ?></p>
<p><?php echo $row['username']; ?></p>
<p><?php echo $row['telno']; ?>
<?php
// $userid = $_POST['id'];
$stmt=$conn->prepare("INSERT INTO useradvert (id,name2,color2,hobby2) VALUES (?,?,?,?)");
$stmt->bind_param("isss", $id, $name2, $color2, $hobby2);
$stmt->execute();
if (!$stmt)
{ printf("Errormessage: %s\n", $mysqli->error);}
else {
echo "New records created successfully";}
$stmt->close();
$conn->close();
?>
<form name="form2" method="post" action="useracc-test.php">
<p>INSERT YOUR INTEREST:</p>
<p>
</p>
ID:
<input name="id" type="hidden" id="id" value="<?php echo $row['id']; ?>">
<p>Name :
<input type="text" name="name2" id="name2">
</p>
<p>
<label for="warna2"></label>
Color :
<input type="text" name="color2" id="color2">
</p>
<p>
<label for="hobi2"></label>
Hobby:
<input type="text" name="hobby2" id="hobby2">
</p>
<p>
<input type="submit" name="submit" id="submit" value="submit">
</p>
<p> </p>
</form>
<?php endwhile; ?>
</body>
</html>
2 回答
问题解决了..我忘了if(isset($ _ POST ['submit'])).. LoL.I做了一些小改动只是为了添加上面的isset ..一切正常现在..
索引键不需要自动递增 . 相反,您应该将您的记录插入
users
表,抓取该记录的id
,然后在useradvert
表上手动插入id
.我怀疑你的错误是外键约束失败 . 如果在运行每个插入后检查错误,您可能会看到它 . 您给
useradvert
的id
必须首先在users
表上作为id
存在 .