修改
我已经实施了建议的更改，但我仍然无法让它工作：
表单页面跟随（login.php）
<?php
$mac=$_POST['mac'];
$ip=$_POST['ip'];
$username=$_POST['username'];
$linklogin=$_POST['link-login'];
$linkorig=$_POST['link-orig'];
$error=$_POST['error'];
$chapid=$_POST['chap-id'];
$chapchallenge=$_POST['chap-challenge'];
$linkloginonly=$_POST['link-login-only'];
$linkorigesc=$_POST['link-orig-esc'];
$macesc=$_POST['mac-esc'];
if (isset($_POST['postcode'])) {
$postcode = $_POST['postcode'];
}
if (isset($_POST['email'])) {
$email = $_POST['email'];
}
?>
**SOME HTML HERE**
<script src="jquery-3.2.1.min.js"></script>
<script>
var js-postcode = document.login.getElementsByName("postcode").value;
var js-email = document.login.getElementsByName("email").value;
var formdata = {postcode:js-postcode,email:js-email};
$("button").click(function(){
$.ajax(
{
type: "POST",
url: "database.php", //Should probably echo true or false depending if it could do it
data : formdata,
success: function(feed) {
if (feed!="true") {
// DO STUFF
} else {
console.log(feed);
// WARNING THAT IT WASN'T DONE
}
}}}
</script>
</head>
<body>
<table width="100%" style="margin-top: 10%;">
<tr>
<td align="center" valign="middle">
<table width="240" height="240" style="border: 1px solid #cccccc; padding: 0px;" cellpadding="0" cellspacing="0">
<tr>
<td align="center" valign="bottom" height="175" colspan="2">
<!-- removed $(if chap-id) $(endif) around OnSubmit -->
<form name="login" action="<?php echo $linkloginonly; ?>" method="post" onSubmit="return doLogin()" >
<input type="hidden" name="dst" value="<?php echo $linkorig; ?>" />
<input type="hidden" name="popup" value="true" />
<table width="100" style="background-color: #ffffff">
<tr><td align="right">login</td>
<td><input style="width: 80px" name="username" type="text" value="<?php echo $username; ?>"/></td>
</tr>
<tr><td align="right">password</td>
<td><input style="width: 80px" name="password" type="password"/></td>
</tr>
<tr><td align="right">Postcode</td>
<td><input style="width: 80px" name="postcode" type="text" /></td>
</tr>
<tr><td align="right">Email</td>
<td><input style="width: 80px" name="email" type="text" /></td>
</tr>
<td><button><input type="submit" value="OK" /></button></td>
</tr>
</table>
</form>
</td>
</tr>
</table>
</td>
</tr>
</table>
<script type="text/javascript">
<!--
document.login.username.focus();
//-->
</script>
</body>
</html>
并调用文件database.php如下：
<?php
if ((isset($_POST['postcode'])) && (isset($_POST['email']))) {
$postcode = $_POST['postcode'];
$email = $_POST['email'];
$connect= new mysqli_connect('xx','xx','xx','xx');
if ($conn->connect_errno) {
echo "There was a problem connecting to MySQL: (" . $conn->connect_errno . ") " . $conn->connect_error;
}
if (!($sql = $conn->prepare("INSERT INTO visitors(postcode,email) VALUES(postcode,email)"))) {
echo "Prepare failed: (" . $conn->errno . ") " . $conn->error;
}
//NOTE: the "ss" part means that $postcode and $email are strings (mysql is expecting datatypes of strings). For example, if $postcode is an integer, you would do "is" instead.
if (!$sql->bind_param("ss", $postcode, $email)) {
echo "Binding parameters failed: (" . $sql->errno . ") " . $sql->error;
}
if (!$sql->execute()) {
echo "Execute failed: (" . $sql->errno . ") " . $sql->error;
}
} else {
echo 'Variables did not send through ajax.'; // any echoed values would be sent back to javascript and stored in the 'response' variable of your success or fail functions for testing.
}
?>
我仍然没有从表单到数据库的任何东西。即使我将变量换成字符串，我也没有得到任何数据库，但是如果我单独运行database.php就可以了。当然，我现在接近让这个工作了......任何帮助表示感谢，并非常感谢迄今为止所提供的帮助。
*******************************原始问题如下***************** **
我有一个简单的表格如下：
<form name="login" action="somethingelse.php" method="post" onSubmit="return doLogin()" >
<input type="hidden" name="dst" value="<?php echo $linkorig; ?>" />
<input type="hidden" name="popup" value="true" />
<table width="100" style="background-color: #ffffff">
<tr><td align="right">login</td>
<td><input style="width: 80px" name="username" type="text" value="<?php e$
</tr>
<tr><td align="right">password</td>
<td><input style="width: 80px" name="password" type="password"/></td>
</tr>
<tr><td align="right">Postcode</td>
<td><input style="width: 80px" name="postcode" type="text" /></td>
</tr>
<tr><td align="right">Email</td>
<td><input style="width: 80px" name="email" type="text" /></td>
</tr>
<td><button><input type="submit" value="OK" /></button></td>
</tr>
</table>
</form>
因为我需要使用表单操作来执行其他操作，所以我需要在单击按钮时使用jQuery将数据发送到数据库。特别是从表单中获取的邮政编码和电子邮件地址。与jQuery相关的代码部分如下所示：
<script language="JavaScript" >
$(document).ready(function(){
$("button").click(function(){
mysqli_query();
});
});
</script>
调用函数mysqli_query是通过include语句声明的，因此它位于不同的文件中。调用的函数如下所示：
mysqli_query( $connect, "INSERT INTO visitors(postcode,email) VALUES(postcode,email)");
我已经绕圈了好几天了。我知道我接近使其工作，但不能跨越终点线。有人可以指出我在这里做错了吗？
答案 0 :(得分：3)
警告： 从不信任用户输入，始终清理输入首先使用预准备语句否则，你让自己容易受到 SQL INJECTION ATTACKS
的攻击你正在混淆，Javascript是一种客户端语言，而mysqli是基于PHP的功能在服务器端。
你应该做的是ajax调用，使用不同PHP文件的值来建立数据库连接并插入数据。
var dataString = "postcode="+ postcode+"&email="+email;
$.ajax({
type: "POST",
url: "file_that_does_the_work.php", //Should probably echo true or false depending if it could do it
data: dataString,
success: function(feed) {
if (feed=="true") {
// DO STUFF
} else {
console.log(feed);
// WARNING THAT IT WASN'T DONE
}
}
file_that_does_the_work.php
<?
include("config.php"); // your thing that configures the connection
$postcode = sanitizationfunction($_POST["postcode"]);
$email = sanitizationfunction($_POST["email"]);
$query = $connection->prepare('INSERT INTO visitors(postcode,email) VALUES(?,?)');
$query->bindParam(1, $postcode);
$query->bindParam(2, $email);
if ($query->execute()) {
echo "true";
} else {
echo "false";
}
?>
答案 1 :(得分：1)
<强> form.php的
<table width="100" style="background-color: #ffffff">
<tr><td align="right">login</td>
<td><input style="width: 80px" name="username" type="text" value="<?php echo $username?>"/>
</tr>
<tr><td align="right">password</td>
<td><input style="width: 80px" name="password" type="password"/></td>
</tr>
<tr><td align="right">Postcode</td>
<td><input style="width: 80px" name="postcode" type="text" /></td>
</tr>
<tr><td align="right">Email</td>
<td><input style="width: 80px" name="email" type="text" /></td>
</tr>
<td><input type="submit" value="OK" /></td>
</tr>
</table>
</form>
`
<强> somethingelse.php
<?php
foreach ($_POST as $key => $value) {
echo $key."=".$value."<br/>";
}
?>
我将连接部分留给您：D
答案 2 :(得分：1)
因此，正如其他人所指出的那样，您正在混淆客户端代码和服务器端代码。您需要将所有表单数据发送到php文件。 jquery ajax将数据发送到脚本，并确定此调用是否成功。如果调用不成功，则可以运行测试逻辑。如果是，则可以执行其他逻辑，例如提醒用户提交成功表单。
以下是该过程的一个示例：
AJAX：
<script>
var formData = 'some data' // Get your form values and save here - postcode and email
$("button").click(function(){
$.ajax ({
method: 'POST',// you can do either post or get...
url: "page_to_handle_mysql_code.php",
data: formData
success: function( response ) {
//do something like alert("Submitted Successfully!");
}
fail: function( response) {
//Do testing such as console.log(response); NOTE: Response will be what ever your php page sends back.
}
});
)};
</script>
在您的php页面上：page_to_handle_mysql_code.php
<?php
if ((isset($_POST['postcode'])) && (isset($_POST['email']))) {
$postcode = $_POST['postcode'];
$email = $_POST['email'];
//connect to mysql - I prefer prepared statements as the variables are prepared for safety when sent to MySQL
$conn = new mysqli($servername, $username, $password, $dbname);//you can either put the actually values in, or I include another php page in this one that sets my variables so I can resuse my code easily.
if ($conn->connect_errno) {
echo "There was a problem connecting to MySQL: (" . $conn->connect_errno . ") " . $conn->connect_error;
}
if (!($sql = $conn->prepare("INSERT INTO visitors(postcode,email) VALUES(?,?)"))) {
echo "Prepare failed: (" . $conn->errno . ") " . $conn->error;
}
//NOTE: the "ss" part means that $postcode and $email are strings (mysql is expecting datatypes of strings). For example, if $postcode is an integer, you would do "is" instead.
if (!$sql->bind_param("ss", $postcode, $email)) {
echo "Binding parameters failed: (" . $sql->errno . ") " . $sql->error;
}
if (!$sql->execute()) {
echo "Execute failed: (" . $sql->errno . ") " . $sql->error;
}
} else {
echo 'Variables did not send through ajax.'; // any echoed values would be sent back to javascript and stored in the 'response' variable of your success or fail functions for testing.
}
?>
这可以帮助您将值输入MySQL。我希望它有所帮助！
答案 3 :(得分：-1)
您可以使用jquery
提交表单 nodes是PHP中的一个函数，您的javascript无法访问该函数。你必须从你的PHP发出一个http调用，你的PHP将收到它并在其末尾运行mysqli_query