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

HTML:

    <form name="foo" action="form.php" method="POST" id="foo">
        <label for="bar">A bar</label>
        <input id="bar" class="inputs" name="bar" type="text" value="" />
        <input type="submit" value="Send" onclick="submitform(); return false;" />
    </form>

JavaScript:

   function submitform()
   {
       var inputs = document.getElementsByClassName("inputs");
       var formdata = new FormData();
       for(var i=0; i<inputs.length; i++)
       {
           formdata.append(inputs[i].name, inputs[i].value);
       }
       var xmlhttp;
       if(window.XMLHttpRequest)
       {
           xmlhttp = new XMLHttpRequest;
       }
       else
       {
           xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
       }
       xmlhttp.onreadystatechange = function()
       {
          if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
          {

          }
       }
       xmlhttp.open("POST", "insert.php");
       xmlhttp.send(formdata);
   }

其他回答

我使用如下所示的方法。它像文件一样提交所有东西。

$(document).on("submit", "form", function(event)
{
    event.preventDefault();

    var url  = $(this).attr("action");
    $.ajax({
        url: url,
        type: 'POST',
        dataType: "JSON",
        data: new FormData(this),
        processData: false,
        contentType: false,
        success: function (data, status)
        {

        },
        error: function (xhr, desc, err)
        {
            console.log("error");
        }
    });
});

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。

<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<form method="post" id="form_content" action="Javascript:void(0);">
    <button id="desc" name="desc" value="desc" style="display:none;">desc</button>
    <button id="asc" name="asc"  value="asc">asc</button>
    <input type='hidden' id='check' value=''/>
</form>

<div id="demoajax"></div>

<script>
    numbers = '';
    $('#form_content button').click(function(){
        $('#form_content button').toggle();
        numbers = this.id;
        function_two(numbers);
    });

    function function_two(numbers){
        if (numbers === '')
        {
            $('#check').val("asc");
        }
        else
        {
            $('#check').val(numbers);
        }
        //alert(sort_var);

        $.ajax({
            url: 'test.php',
            type: 'POST',
            data: $('#form_content').serialize(),
            success: function(data){
                $('#demoajax').show();
                $('#demoajax').html(data);
                }
        });

        return false;
    }
    $(document).ready(function_two());
</script>

如果你想用jQuery Ajax发送数据,那么就不需要表单标签和提交按钮

例子:

<script>
    $(document).ready(function () {
        $("#btnSend").click(function () {
            $.ajax({
                url: 'process.php',
                type: 'POST',
                data: {bar: $("#bar").val()},
                success: function (result) {
                    alert('success');
                }
            });
        });
    });
</script>

<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input id="btnSend" type="button" value="Send" />