Php 插入到多个表中的帮助
我正在建立一个学习计划 我已经设计了一个草稿界面和数据库，目前我正在尝试将测试注册细节插入数据库，但我已经做了三天了，没有成功。我将非常感谢你能帮我指出我做错了什么 我收到的信息/错误如下：Php 插入到多个表中的帮助,php,mysql,Php,Mysql,我正在建立一个学习计划 我已经设计了一个草稿界面和数据库，目前我正在尝试将测试注册细节插入数据库，但我已经做了三天了，没有成功。我将非常感谢你能帮我指出我做错了什么 我收到的信息/错误如下： Connected successfully Error details for Result 2: Cannot add or update a child row: a foreign key constraint fails (`p00702`.`universityreferences`, CON
Connected successfully
Error details for Result 2: Cannot add or update a child row: a foreign key constraint fails (`p00702`.`universityreferences`, CONSTRAINT `universityreferences_ibfk_1` FOREIGN KEY (`userid`) REFERENCES `users` (`userid`) ON UPDATE CASCADE).
Error details for Result 3: Cannot add or update a child row: a foreign key constraint fails (`p00702`.`mobiles`, CONSTRAINT `mobiles_ibfk_1` FOREIGN KEY (`userid`) REFERENCES `users` (`userid`) ON UPDATE CASCADE).
Error details for Result 4: Cannot add or update a child row: a foreign key constraint fails (`p00702`.`logins`, CONSTRAINT `logins_ibfk_1` FOREIGN KEY (`userid`) REFERENCES `users` (`userid`) ON UPDATE CASCADE).
All queries were rolled back
$id=$mysqli->insert\u id在第一次查询之后也不起作用，我不知道为什么，但我怀疑这可能是根本原因
我的PHP代码是：
<?php
if (isset($_POST['submitted'])) {
include 'serverSide/connectToServer.php';
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$studentId = $_POST['studentId'];
$mobile = $_POST['mobile'];
$username = $_POST['username'];
$password = $_POST['password'];
//turn off autocommit
$mysqli->autocommit(false);
$flag = true;
// attempt insert query executions
$query1 = "INSERT INTO users (firstname, lastname) VALUES ('$firstName', '$lastName');";
$id = $mysqli->insert_id;
$query2 = "INSERT INTO universityreferences (userid, universityreference) VALUES ('$id', '$studentId');";
$id2 = $mysqli->insert_id;
$query3 = "INSERT INTO mobiles (userid, mobile) VALUES ('$id2', '$mobile');";
$id3 = $mysqli->insert_id;
$query4 = "INSERT INTO logins (userid, username, password) VALUES ('$id3', '$username', '$password');";
$result1 = mysqli_query($mysqli, $query1);
if (!$result1) {
$flag = false;
echo "Error details for Result 1: " . mysqli_error($mysqli) . ".";
}
$result2 = mysqli_query($mysqli, $query2);
if (!$result2) {
$flag = false;
echo "Error details for Result 2: " . mysqli_error($mysqli) . ".";
}
$result3 = mysqli_query($mysqli, $query3);
if (!$result3) {
$flag = false;
echo "Error details for Result 3: " . mysqli_error($mysqli) . ".";
}
$result4 = mysqli_query($mysqli, $query4);
if (!$result4) {
$flag = false;
echo "Error details for Result 4: " . mysqli_error($mysqli) . ".";
}
if ($flag) {
mysqli_commit($mysqli);
echo "All queries were executed successfully";
} else {
mysqli_rollback($mysqli);
echo "All queries were rolled back";
}
mysqli_close($mysqli);
}
?>
<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
<head>
<meta charset="UTF-8">
<link rel="shortcut icon" type="image/png" href="/images/favicon.png"/>
<title>Just-Read</title>
<link rel="stylesheet" type="text/css" href="css/styles.css">
</head>
<body>
<!-- Link to external JavaScript file -->
<script src="javascript/validator.js"></script>
<div id="container">
<div id="header">
<!-- Web site Logo -->
<div class="logo">
<img src="images/logo.png" width="128" height="93.5" alt="Logo"/><br><br>
</div>
<div id="logoText">
<h1>Just Read</h1>
</div>
</div>
<div id="leftColumn">
<h4>The ultimate study planner</h4>
</div>
<div id="rightColumn">
<!-- Registration Form -->
<h3>Please fill out the form below</h3>
<form name="Register" action="registration.php" onsubmit="return registrationValidator()" autocomplete="on" method="POST">
<!--According to YouTuber Ralph Philips, this makes sure a blank form cannot be submitted to the database-->
<input type="hidden" name="submitted" value="true"/>
<div class="register">
<label><b>First Name*</b></label>
<input type="text" id="firstName" name="firstName" placeholder="Enter your first name" autofocus/>
<label><b>Last Name*</b></label>
<input type="text" id="lastName" name="lastName" placeholder="Enter your last name" />
<label><b>Student ID*</b></label>
<input type="text" id="studentId" name="studentId" placeholder="Enter your university ID" />
<label><b>Mobile</b></label>
<input type="text" id="mobile" name="mobile" placeholder="Enter your phone number" />
<label><b>Email Address (Username)*</b></label>
<input type="email" id="username" name="username" placeholder="Enter your email address" />
<label><b>Password*</b></label>
<input type="password" id="password" name="password" placeholder="Enter your password" />
<button type="submit">Register</button>
</div>
<div id="back">
<a href="index.php">Back</a>
</div>
<div id="mandatoryFields">
<h4>* Mandatory Fields</h4>
</div>
</form>
</div>
<div id="footer">
Copyright © 2017, Chizzy Meka.
</div>
</div>
</body>
</html>
提前感谢您的帮助。问题是您在执行查询之前正在使用mysqli->insert\u id
。定义第一个查询时，数据库尚未生成id，因为尚未执行插入
正确的做法是
$query1 = "INSERT INTO users (firstname, lastname) VALUES ('$firstName', '$lastName');";
$result1 = mysqli_query($mysqli, $query1);
if (!$result1) {
$flag = false;
echo "Error details for Result 1: " . mysqli_error($mysqli) . ".";
}
$id = $mysqli->insert_id;
//go on with the next insert
//defining the query, executing it and then using the generated id
所有这些错误消息都是由以下事实引起的：当您运行query2时，您违反了您在user\u id
上定义的外键，因为您插入了一个空值的行。问题是，您试图在执行任何查询之前获取最后插入的id。例如：
$query1 = "INSERT INTO users (firstname, lastname) VALUES ('$firstName', '$lastName');";
$id = $mysqli->insert_id;
在上面的代码中，您只定义sql语句，但不执行它。因此，最后一个插入id不会返回任何内容
仅在执行查询后执行insert\u id（）
方法：
$result1 = mysqli_query($mysqli, $query1);
可能是重复的你好，@asteriskTheServer，谢谢你提出类似的问题，但是，我已经看到了那篇帖子，在我决定发布自己的帖子之前，它没有告诉我足够的信息。虽然我同意有相似之处，但我相信我的PHP是错误的，建议的答案不能帮助我解决这个问题。此外，我正在处理更多的表格，这使得事情变得更加棘手，因此，如果有人能调查我自己的情况并提供一些建议，我将不胜感激。非常感谢，@Stefano Zanini，我会让你知道它是如何工作的。它现在正在工作。非常感谢您，希望您在需要时也能找到帮助！非常感谢，@Shadow，我会让你知道它是如何工作的。它现在正在工作。非常感谢您，希望您在需要时也能找到帮助！
$result1 = mysqli_query($mysqli, $query1);