我试图将数据从表单发送到数据库。这是我使用的形式:

<form name="foo" action="form.php" method="POST" id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />
    <input type="submit" value="Send" />
</form>

典型的方法是提交表单,但这会导致浏览器重定向。使用jQuery和Ajax,是否有可能捕获所有表单的数据并将其提交给PHP脚本(例如,form. PHP)?


当前回答

这是使用ajax在HTML中填充选择选项标签的代码,而XMLHttpRequest和API是用PHP和PDO编写的

conn.php

<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "db_event";
try {
    $conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
    echo "Connection failed: " . $e->getMessage();
}
?>

category.php

<?php
 include 'conn.php';
try {
    $data = json_decode(file_get_contents("php://input"));
    $stmt = $conn->prepare("SELECT *  FROM events ");
    http_response_code(200);
    $stmt->execute();
    
    header('Content-Type: application/json');

    $arr=[];
    while($value=$stmt->fetch(PDO::FETCH_ASSOC)){
        array_push($arr,$value);
    }
    echo json_encode($arr);
   
  } catch(PDOException $e) {
    echo "Error: " . $e->getMessage();
  }

script.js

var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
    if (this.readyState == 4 && this.status == 200) {
        data = JSON.parse(this.responseText);

        for (let i in data) {


            $("#cars").append(
                '<option value="' + data[i].category + '">' + data[i].category + '</option>'

            )
        }
    }
};
xhttp.open("GET", "http://127.0.0.1:8000/category.php", true);
xhttp.send();

index . html


<!DOCTYPE html>
<html lang="en">

<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="style.css">
    <script src="https://code.jquery.com/jquery-3.6.0.min.js"
        integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
    <title>Document</title>
</head>

<body>
    <label for="cars">Choose a Category:</label>

    <select name="option" id="option">
        
    </select>
    
    <script src="script.js"></script>
</body>

</html>

其他回答

在你的php文件中输入:

$content_raw = file_get_contents("php://input"); // THIS IS WHAT YOU NEED
$decoded_data = json_decode($content_raw, true); // THIS IS WHAT YOU NEED
$bar = $decoded_data['bar']; // THIS IS WHAT YOU NEED
$time = $decoded_data['time'];
$hash = $decoded_data['hash'];
echo "You have sent a POST request containing the bar variable with the value $bar";

在你的js文件中发送一个带有数据对象的ajax

var data = { 
    bar : 'bar value',
    time: calculatedTimeStamp,
    hash: calculatedHash,
    uid: userID,
    sid: sessionID,
    iid: itemID
};

$.ajax({
    method: 'POST',
    crossDomain: true,
    dataType: 'json',
    crossOrigin: true,
    async: true,
    contentType: 'application/json',
    data: data,
    headers: {
        'Access-Control-Allow-Methods': '*',
        "Access-Control-Allow-Credentials": true,
        "Access-Control-Allow-Headers" : "Access-Control-Allow-Headers, Origin, X-Requested-With, Content-Type, Accept, Authorization",
        "Access-Control-Allow-Origin": "*",
        "Control-Allow-Origin": "*",
        "cache-control": "no-cache",
        'Content-Type': 'application/json'
    },
    url: 'https://yoururl.com/somephpfile.php',
    success: function(response){
        console.log("Respond was: ", response);
    },
    error: function (request, status, error) {
        console.log("There was an error: ", request.responseText);
    }
  })

或者保持表单提交的原样。只有当您希望发送带有计算附加内容的修改过的请求,而不仅仅是一些由客户机输入的表单数据时,才需要这种方法。例如哈希、时间戳、用户id、会话id等等。

自从引入了Fetch API,就没有理由再用jQuery Ajax或xmlhttprequest来实现了。要POST表单数据到php脚本在香草JavaScript你可以做以下:

异步函数postData() { 尝试{ Const res = await fetch('../php/contact.php', { 方法:“文章”, new FormData(document.getElementById('form')) }) if (!res.ok)抛出新的错误('网络响应不正常'); } catch (err) { console.log (err) } } <form id="form" action="javascript:postData()" > <input id="name" name="name" placeholder=" name" type="text" required> <input type="submit" value=" submit" > > < /形式

下面是一个非常基本的php脚本示例,它获取数据并发送电子邮件:

<?php
    header('Content-type: text/html; charset=utf-8');

    if (isset($_POST['name'])) {
        $name = $_POST['name'];
    }

    $to = "test@example.com";
    $subject = "New name submitted";
    $body = "You received the following name: $name";
    
    mail($to, $subject, $body);

这是使用ajax在HTML中填充选择选项标签的代码,而XMLHttpRequest和API是用PHP和PDO编写的

conn.php

<?php
$servername = "localhost";
$username = "root";
$password = "root";
$database = "db_event";
try {
    $conn = new PDO("mysql:host=$servername;dbname=$database", $username, $password);
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (PDOException $e) {
    echo "Connection failed: " . $e->getMessage();
}
?>

category.php

<?php
 include 'conn.php';
try {
    $data = json_decode(file_get_contents("php://input"));
    $stmt = $conn->prepare("SELECT *  FROM events ");
    http_response_code(200);
    $stmt->execute();
    
    header('Content-Type: application/json');

    $arr=[];
    while($value=$stmt->fetch(PDO::FETCH_ASSOC)){
        array_push($arr,$value);
    }
    echo json_encode($arr);
   
  } catch(PDOException $e) {
    echo "Error: " . $e->getMessage();
  }

script.js

var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function () {
    if (this.readyState == 4 && this.status == 200) {
        data = JSON.parse(this.responseText);

        for (let i in data) {


            $("#cars").append(
                '<option value="' + data[i].category + '">' + data[i].category + '</option>'

            )
        }
    }
};
xhttp.open("GET", "http://127.0.0.1:8000/category.php", true);
xhttp.send();

index . html


<!DOCTYPE html>
<html lang="en">

<head>
    <meta charset="UTF-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <link rel="stylesheet" href="style.css">
    <script src="https://code.jquery.com/jquery-3.6.0.min.js"
        integrity="sha256-/xUj+3OJU5yExlq6GSYGSHk7tPXikynS7ogEvDej/m4=" crossorigin="anonymous"></script>
    <title>Document</title>
</head>

<body>
    <label for="cars">Choose a Category:</label>

    <select name="option" id="option">
        
    </select>
    
    <script src="script.js"></script>
</body>

</html>

这是一篇非常好的文章,包含了关于jQuery表单提交你需要知道的一切。

文章摘要:

简单HTML表单提交

HTML:

<form action="path/to/server/script" method="post" id="my_form">
    <label>Name</label>
    <input type="text" name="name" />
    <label>Email</label>
    <input type="email" name="email" />
    <label>Website</label>
    <input type="url" name="website" />
    <input type="submit" name="submit" value="Submit Form" />
    <div id="server-results"><!-- For server results --></div>
</form>

JavaScript:

$("#my_form").submit(function(event){
    event.preventDefault(); // Prevent default action
    var post_url = $(this).attr("action"); // Get the form action URL
    var request_method = $(this).attr("method"); // Get form GET/POST method
    var form_data = $(this).serialize(); // Encode form elements for submission

    $.ajax({
        url : post_url,
        type: request_method,
        data : form_data
    }).done(function(response){ //
        $("#server-results").html(response);
    });
});

HTML Multipart/ Form -data表单提交

要将文件上传到服务器,我们可以使用XMLHttpRequest2提供的FormData接口,该接口构造一个FormData对象,可以使用jQuery Ajax轻松地将其发送到服务器。

HTML:

<form action="path/to/server/script" method="post" id="my_form">
    <label>Name</label>
    <input type="text" name="name" />
    <label>Email</label>
    <input type="email" name="email" />
    <label>Website</label>
    <input type="url" name="website" />
    <input type="file" name="my_file[]" /> <!-- File Field Added -->
    <input type="submit" name="submit" value="Submit Form" />
    <div id="server-results"><!-- For server results --></div>
</form>

JavaScript:

$("#my_form").submit(function(event){
    event.preventDefault(); // Prevent default action
    var post_url = $(this).attr("action"); // Get form action URL
    var request_method = $(this).attr("method"); // Get form GET/POST method
    var form_data = new FormData(this); // Creates new FormData object
    $.ajax({
        url : post_url,
        type: request_method,
        data : form_data,
        contentType: false,
        cache: false,
        processData: false
    }).done(function(response){ //
        $("#server-results").html(response);
    });
});

我希望这能有所帮助。

请核对一下。它是完整的Ajax请求代码。

$('#foo').submit(function(event) {
    // Get the form data
    // There are many ways to get this data using jQuery (you
    // can use the class or id also)
    var formData = $('#foo').serialize();
    var url = 'URL of the request';

    // Process the form.
    $.ajax({
        type        : 'POST',   // Define the type of HTTP verb we want to use
        url         : 'url/',   // The URL where we want to POST
        data        : formData, // Our data object
        dataType    : 'json',   // What type of data do we expect back.
        beforeSend : function() {

            // This will run before sending an Ajax request.
            // Do whatever activity you want, like show loaded.
        },
        success:function(response){
            var obj = eval(response);
            if(obj)
            {
                if(obj.error==0){
                    alert('success');
                }
                else{
                    alert('error');
                }
            }
        },
        complete : function() {
            // This will run after sending an Ajax complete
        },
        error:function (xhr, ajaxOptions, thrownError){
            alert('error occured');
            // If any error occurs in request
        }
    });

    // Stop the form from submitting the normal way
    // and refreshing the page
    event.preventDefault();
});