以下是我正在尝试做的简化版本。基本上我有 2 个注册表，每个表单的内容将写入不同的表。当我加载表单时，我在第 xx 行收到许多未定义索引错误，并且我仍然在第 75 行和第 84 行的这个示例中收到错误，我认为错误归因于某些字段是“必需的”，但该页面尚未加载为它使用一些 JS 飘走了。... 请帮忙！<?phpob_start(); //Turns on output bufferingsession_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>
1 回答
HUWWW
TA贡献1874条经验 获得超12个赞
由于您一次只提交一种格式，因此您不能插入到两个表中。您需要检查提交的是哪个表单，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，因此您可以分辨出在 PHP 中使用了哪一个。
您还应该使用准备好的语句而不是将变量替换到 SQL 中，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);
}
?>
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