将数据插入到 2 个单独的 mySQL 表中

Question

下面是我正在尝试做的简化版本。 基本上我有 2 个注册 forms 并且每个表单的内容都将写入不同的表。 当我加载表单时，我在第 xx 行收到许多错误未定义索引，并且在第 75 行和第 84 行的此示例中仍然出现错误，我认为错误归结为某些“必需”字段，但该页面尚未加载为它使用一些 JS 飘走了。 ... 请帮忙！

<?php
ob_start(); //Turns on output buffering
session_start(); //This starts a session variable where it store the input so the user doesnt have to start over if they get an error
$con = mysqli_connect("localhost","root","","test-sql"); //connection variables
  
  $timezone = date_default_timezone_set("Europe/London");
  
   if(mysqli_connect_errno()){
        echo "failed to connect:" . mysqli_connect_errno();
    }
?>

<html>
    <head>
            <! here we add the links and the jquery>
    <title>Register Project</title>
    </head>
    <body>
        
        <div id="first">
                    <form action="test-sql.php" method="POST">
                        <input type="text" name="Inf_fname" placeholder="First Name" required>
                        <br>
                        <input type="text" name="Inf_lname" placeholder="Last Name" required>
                        </form>
                    </div>
        
        <div id="second">
                
                    <form action="test-sql.php" method="POST">
                        
                       <br>
                    <input type="text" name="Cust_fname" placeholder="First Name" required>
                    <br>
                                               
                    <input type="text" name="Cust_lname" placeholder="Last Name" required>
                    <br>
    
                    <input type="submit" name="register_button" value="Register">
                        
                       </form>
                    </div>
    </body>
    
</html>

<?php

$inf_fname = "";
$cust_fname = "";
$inf_lname = "";
$cust_lname = "";

if(isset($_POST['register_button'])) {
    
    $inf_fname = strip_tags($_POST['Inf_fname']);
    $inf_fname = str_replace(' ', '', $inf_fname);
    $_SESSION['Inf_fname'] = $inf_fname;
    
    $cust_fname = strip_tags($_POST['Cust_fname']);
    $cust_fname = str_replace(' ', '', $cust_fname);
    $_SESSION['Cust_fname'] = $cust_fname;
    
    $inf_lname = strip_tags($_POST['Inf_lname']);
    $inf_lname = str_replace(' ', '', $inf_lname);
    $_SESSION['Inf_lname'] = $inf_lname;
    
    $cust_lname = strip_tags($_POST['Cust_lname']);
    $cust_lname = str_replace(' ', '', $cust_lname);
    $_SESSION['Cust_lname'] = $cust_lname;
    
    $query = mysqli_query($con, "INSERT INTO inf VALUES('', '$inf_fname', '$inf_lname')");
    $query = mysqli_query($con, "INSERT INTO customer VALUES('', '$cust_fname', '$cust_lname')");
    
}
?>

Answer 1

由于您一次只提交一种格式，因此您不能同时插入两个表格。 您需要检查提交了哪个表单，并插入到相应的表中。 每个表单都需要一个具有不同名称或值的提交按钮，因此您可以知道 PHP 中使用了哪个。

您还应该使用准备好的语句而不是将变量替换为 SQL，以防止 SQL 注入。 我已经展示了如何做到这一点。

<?php
ob_start(); //Turns on output buffering
session_start(); //This starts a session variable where it store the input so the user doesnt have to start over if they get an error
$con = mysqli_connect("localhost","root","","test-sql"); //connection variables
  
$timezone = date_default_timezone_set("Europe/London");
  
if(mysqli_connect_errno()){
    echo "failed to connect:" . mysqli_connect_errno();
}
?>

<html>
<head>
<! here we add the links and the jquery>
<title>Register Project</title>
</head>
<body>
        
<div id="first">
    <form action="test-sql.php" method="POST">
    <input type="text" name="Inf_fname" placeholder="First Name" required>
    <br>
    <input type="text" name="Inf_lname" placeholder="Last Name" required>
    <input type="submit" name="inf_submit" value="Register">
    </form>
</div>

<div id="second">
                
    <form action="test-sql.php" method="POST">
                        
    <br>
    <input type="text" name="Cust_fname" placeholder="First Name" required>
    <br>
                                               
    <input type="text" name="Cust_lname" placeholder="Last Name" required>
    <br>
    
    <input type="submit" name="cust_submit" value="Register">
                        
    </form>
</div>
</body>

</html>

 <?php

 $inf_fname = "";
 $cust_fname = "";
 $inf_lname = "";
 $cust_lname = "";

 if(isset($_POST['inf_submit'])) {

     $inf_fname = strip_tags($_POST['Inf_fname']);
     $inf_fname = str_replace(' ', '', $inf_fname);
     $_SESSION['Inf_fname'] = $inf_fname;
     $inf_lname = strip_tags($_POST['Inf_lname']);
     $inf_lname = str_replace(' ', '', $inf_lname);
     $_SESSION['Inf_lname'] = $inf_lname;

     $stmt = mysqli_prepare($con, "INSERT INTO inf VALUES (?, ?)");
     mysqli_stmt_bind_param($stmt, "ss", $inf_fname, $inf_lname);
     mysqli_stmt_execute($stmt);
 } elseif (isset($_POST['cust_submit'])) {

     $cust_fname = strip_tags($_POST['Cust_fname']);
     $cust_fname = str_replace(' ', '', $cust_fname);
     $_SESSION['Cust_fname'] = $cust_fname;
     $cust_lname = strip_tags($_POST['Cust_lname']);
     $cust_lname = str_replace(' ', '', $cust_lname);
     $_SESSION['Cust_lname'] = $cust_lname;

     $stmt = mysqli_prepare($con, "INSERT INTO customer VALUES (?, ?)");
     mysqli_stmt_bind_param($stmt, "ss", $cust_fname, $cust_lname);
     mysqli_stmt_execute($stmt);
 }
 ?>

将数据插入到 2 个单独的 mySQL 表中

问题描述

1 个解决方案

解决方案1
1 已采纳 2020-08-10 20:39:36

将数据插入到 2 个单独的 mySQL 表中

问题描述

1 个解决方案

解决方案1 1 已采纳 2020-08-10 20:39:36

解决方案1
1 已采纳 2020-08-10 20:39:36