【发布时间】：2021-01-10 03:41:12
【问题描述】：
我想将 HTML 表单中的值发送到 php 脚本，该脚本应该能够读取这些值。
HTML：
<form id="details" method="POST" onsubmit="return submit(n,i,p,e,img)">
Full name: <strong name="name_1"></strong><br><br>
ID No:<strong name="org_number_1"></strong><br><br>
Mobile No:<strong name="ph_number_1"></strong><br><br>
E-mail: <strong name="email_1"></strong><br><br>
ID Card: <img src="" alt="preview" name="image" style="width: 100px; height: 100px;"><br><br>
<button id="go" type="submit">It's correct</button>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script src="preview_back_end.js" async></script>
Javascript：
function submit(n,i,p,e,img){
console.log("Going into submit()")
$.ajax({
type : "POST", //type of method
url : "database_registration.php", //your page
data : { name_1 : n, email_1 : e, image : img, org_number_1: i, ph_number_1: p },// passing the values
success: function(){
//do what you want here...
alert(data);
}
});
return false;
}
var arr=document.cookie.split(';')
for(var i=0; i<arr.length; i++){
var c=arr[i].split('=');
if (c[0].trim()=='name'){
document.getElementsByName("name_1")[0].innerHTML=decodeURIComponent(c[1]);
var nme=decodeURIComponent(c[1]);
}
else if(c[0].trim()=='ID No'){
document.getElementsByName("org_number_1")[0].innerHTML=decodeURIComponent(c[1]);
var id=decodeURIComponent(c[1]);
}
else if(c[0].trim()=='Mobile No'){
document.getElementsByName("ph_number_1")[0].innerHTML=decodeURIComponent(c[1]);
var phone=decodeURIComponent(c[1]);
}
else if(c[0].trim()=='Email'){
document.getElementsByName("email_1")[0].innerHTML=decodeURIComponent(c[1]);
var email=decodeURIComponent(c[1]);
}
}
const image = localStorage.getItem("Image");
document.getElementsByName("image")[0].src=image;
var img=image
submit(nme,id,phone,email,img);
php：
<?php
// Server name is localhost
$servername = "localhost";
// In my case, user name will be root
$username = "root";
// Password is empty
$password = "";
//current date
$date = date("Y-m-d");
// get values
$name=$_POST['name_1'];
$org_number=$_POST['org_number_1'];
$ph_number=$_POST['ph_number_1'];
$email=$_POST['email_1'];
$image=$_POST['image'];
echo $name ."<BR>";
echo $org_number ."<BR>";
echo $ph_number ."<BR>";
echo $email ."<BR>";
echo $image ."<BR>";
// Creating a connection
$conn = new mysqli($servername,
$username, $password, "Employee_information");
// Check connection
if ($conn->connect_error) {
die("Connection failure: "
. $conn->connect_error);
}
// Creating a table Employees
$sql="CREATE TABLE IF NOT EXISTS Employees(Sl_no int AUTO_INCREMENT PRIMARY KEY, Full_name varchar(30) NOT NULL,
ID_no INT(2) NOT NULL, Contact INT(10) NOT NULL, Email varchar(30) NOT NULL, registration_date DATE,
ID_preview blob(10M))";
$conn -> query($sql);
// Inserting records
$stmt = $conn->prepare("INSERT INTO Employees (Full_name, ID_no, Contact, Email,
registration_date, ID_preview)
VALUES (?, ?, ?, ?, ?, ?)");
$stmt->bind_param("siissb", $name, $org_number, $ph_number, $email, $date, $image);
$stmt->execute();
if($stmt->execute())
echo "records inserted";
else
echo $stmt->error;
// Closing connection
$stmt->close();
$conn->close();
?>
当我运行这个时，我在 php 脚本中得到一个错误：
Warning: Undefined array key "name_1"
Warning: Undefined array key "org_number_1"
Warning: Undefined array key "ph_number_1"
Warning: Undefined array key "email_1"
除此之外，我在控制台中收到此错误：
POST http://127.0.0.1:5500/database_registration.php 405 (Method Not Allowed)
为什么数组键未定义？控制台错误是什么意思？我该如何解决这个问题？
编辑： 尝试打印$_POST 数组，使用：
<?php
echo 'Post variables:<br />';
print_r($_POST);
?>
数组完全为空：
Post variables:
Array ( )
【问题讨论】：
-
你可以使用 $.ajax 通过 java 脚本而不是表单来发布，你可以在 php 中使用 $_get 来获取元素值
-
我尝试在 php 中使用
$_POST['name_value']，但收到Undefined array key "name_value"警告。 -
你已经在你的表单中使用了 post 方法，所以你必须使用
$_POST['name_1']使用输入名称将结果接收到 php 文件中 -
上面写着
Undefined array key "name_1"
标签： javascript php html forms http-post