PHP post请求未识别的索引错误
PHP post请求未识别的索引错误
提问于 2016-08-15 16:14:17
我正在尝试编写一个页面来向php脚本发出POST请求，我觉得我做得很好，它在其他地方都可以工作，所以看起来似乎是这样，但我一直收到一个“不明错误”，它不工作，我怎么才能让它工作呢？
Javascript：
$(document).ready(function() {
$("#x").click(function() {
var email = $("email").val();
var pass = $("password").val();
var confirmPass = $("confirmPassword").val();
var name = $("name").val();
var question = $("question").val();
var answer = $("answer").val();
if(pass != confirmPass) {
alert("Passwords do not match!");
return;
}
var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
$.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
alert(result);
window.location.href = "../Dashboard";
}});
});
});PHP：
<?php
$servername = "localhost";
$username = "root";
$password = "*********";
$dbname = "myDB";
$conn = new mysqli($servername, $username, $password, $dbname);
$email = $_POST["email"];
$pass = $_POST["pass"];
$name = $_POST["name"];
$question = $_POST["question"];
$answer = $_POST["answer"];
$sql = "INSERT INTO accounts (accountEmail, accountPassword, accountName, accountQuestion, accountRecover) VALUES ('$email', '$pass', '$name', '$question', '$answer')";
$conn->close();
if(mysql_affected_rows() > 0) {
$response = "Account added successfully!";
}
else {
$response = "Couldn't add account!";
}
$pre = array("Response" => $response);
echo json_encode($pre);
?>回答 3
Stack Overflow用户
发布于 2016-08-15 16:21:06
您需要正确使用jquery。
例如，var email = $("email").val(); //IS WRONG应该是(如果您输入了“id=”) var email = $("#email").val();，如果您只有姓名，则可以使用var email = $("[name='email']").val();
有点离题:如果您正在使用form ajax submit，可以考虑使用jquery方法serialize https://api.jquery.com/serialize/来获取所有的表单值(或者一些jquery ajaxform插件)。
还有，请！不要使用不安全的mysql语句。看在上帝的份上，使用事先准备好的语句。如果需要非常基本的东西，只需使用准备好的语句或考虑使用https://phpdelusions.net/pdo/pdo_wrapper
还有一个小技巧:在回显json之前，将json头部设为<?php header('Content-type:application/json;charset=utf-8');
Stack Overflow用户
发布于 2016-08-15 16:30:04
你的代码无法工作的原因有很多。@AucT和@ so已经解决了你的Javascript方面的问题，所以我将重点介绍PHP。您的查询代码为：
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "...";
$conn->close();请注意：
- 你永远不会执行你的查询。
$sql只是一个保存在内存中的字符串。 - 您正在将
mysqli函数与mysql_函数(mysql_affected_rows)混合在一起；这将不起作用您正在将POST数据直接插入到查询中，因此您很容易受到SQL injection - At的攻击。最后，您回显了JSON，但是您并没有告诉浏览器要使用这种格式的
改为执行以下操作：
$conn = new mysqli(...);
//SQL with ? in place of values is safe against SQL injection attacks
$sql = "INSERT INTO accounts (accountEmail, accountPassword,
accountName, accountQuestion, accountRecover) VALUES (?, ?, ?, ?, ?)";
$error = null;
//prepare query and bind params. save any error
$stmt = $conn->prepare($sql);
$stmt->bind_param('sssss',$email,$pass,$name,$question,$answer)
or $error = $stmt->error;
//run query. save any error
if(!$error) $stmt->execute() or $error = $stmt->error;
//error details are in $error
if($error) $response = "Error creating new account";
else $response = "Successfully created new account";
//set content-type header to tell the browser to expect JSON
header('Content-type: application/json');
$pre = ['Response' => $response];
echo json_encode($pre);Stack Overflow用户
发布于 2016-08-15 16:32:05
我认为你搞错了你的jquery数据，他们应该有标识符，像id用'#‘表示，类用'.’表示，这是你在输入参数中有“id=name of id=”的输入参数：
$(document).ready(function() {
$("#x").click(function() {
var email = $("#email").val();
var pass = $("#password").val();
var confirmPass = $("#confirmPassword").val();
var name = $("#name").val();
var question = $("#question").val();
var answer = $("#answer").val();
if(pass != confirmPass) {
alert("Passwords do not match!");
return;
}
var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
$.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
alert(result);
window.location.href = "../Dashboard";
}});
});
});或者像这样，在输入参数中有“name of class= field”：
$(document).ready(function() {
$("#x").click(function() {
var email = $(".email").val();
var pass = $(".password").val();
var confirmPass = $(".confirmPassword").val();
var name = $(".name").val();
var question = $(".question").val();
var answer = $(".answer").val();
if(pass != confirmPass) {
alert("Passwords do not match!");
return;
}
var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
$.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
alert(result);
window.location.href = "../Dashboard";
}});
});
});或者，如果您想直接使用该名称，请遵循以下内容：
$(document).ready(function() {
$("#x").click(function() {
var email = $("input[name='email']").val();
var pass = $("input[name='pasword']").val();
var confirmPass = $("input[name='confirmPassword']").val();
var name = $("input[name='name']").val();
var question = $("input[name='question']").val();
var answer = $("input[name='answer']").val();
if(pass != confirmPass) {
alert("Passwords do not match!");
return;
}
var stuff = {email: email, pass: pass, name: name, question: question, answer: answer};
$.ajax({method: "POST", url: "addAccount.php", data: stuff, success: function(result) {
alert(result);
window.location.href = "../Dashboard";
}});
});
});我希望这对你有帮助
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原文链接：
https://stackoverflow.com/questions/38951550
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