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
我还有一个主意。
提供下载文件的PHP文件的URL。 然后你必须通过ajax触发相同的URL，我检查了这第二个请求只在你的第一个请求完成下载文件后才给出响应。所以你可以得到它的事件。
它正在通过ajax处理同样的第二个请求。
其他回答
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>
你可以使用serialize。下面是一个例子。
$("#submit_btn").click(function(){
$('.error_status').html();
if($("form#frm_message_board").valid())
{
$.ajax({
type: "POST",
url: "<?php echo site_url('message_board/add');?>",
data: $('#frm_message_board').serialize(),
success: function(msg) {
var msg = $.parseJSON(msg);
if(msg.success=='yes')
{
return true;
}
else
{
alert('Server error');
return false;
}
}
});
}
return false;
});
要使用jQuery发出Ajax请求，可以通过以下代码实现。
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>
JavaScript:
方法1
/* Get from elements values */
var values = $(this).serialize();
$.ajax({
url: "test.php",
type: "post",
data: values ,
success: function (response) {
// You will get response from your PHP page (what you echo or print)
},
error: function(jqXHR, textStatus, errorThrown) {
console.log(textStatus, errorThrown);
}
});
方法2
/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
var ajaxRequest;
/* Stop form from submitting normally */
event.preventDefault();
/* Clear result div*/
$("#result").html('');
/* Get from elements values */
var values = $(this).serialize();
/* Send the data using post and put the results in a div. */
/* I am not aborting the previous request, because it's an
asynchronous request, meaning once it's sent it's out
there. But in case you want to abort it you can do it
by abort(). jQuery Ajax methods return an XMLHttpRequest
object, so you can just use abort(). */
ajaxRequest= $.ajax({
url: "test.php",
type: "post",
data: values
});
/* Request can be aborted by ajaxRequest.abort() */
ajaxRequest.done(function (response, textStatus, jqXHR){
// Show successfully for submit message
$("#result").html('Submitted successfully');
});
/* On failure of request this function will be called */
ajaxRequest.fail(function (){
// Show error
$("#result").html('There is error while submit');
});
从jQuery 1.8开始，.success()、.error()和.complete()回调函数已弃用。要为最终删除它们做好准备，请使用.done()、.fail()和.always()来代替。
MDN: abort()。如果请求已经发送，此方法将中止请求。
我们已经成功地发送了一个Ajax请求，现在是时候向服务器抓取数据了。
PHP
当我们在Ajax调用中发出POST请求(类型:" POST ")时，我们现在可以使用$_REQUEST或$_POST获取数据:
$bar = $_POST['bar']
您还可以通过以下两种方式查看您在POST请求中获得了什么。顺便说一句，确保设置了$_POST。否则您将得到一个错误。
var_dump($_POST);
// Or
print_r($_POST);
您正在向数据库中插入一个值。确保在执行查询之前正确地敏感或转义了所有请求(无论是GET还是POST)。最好的方法是使用准备好的语句。
如果你想将任何数据返回到页面，你可以通过如下所示的回显数据来实现。
// 1. Without JSON
echo "Hello, this is one"
// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));
然后你可以得到:
ajaxRequest.done(function (response){
alert(response);
});
有几种简便的方法。您可以使用下面的代码。它做同样的功。
var ajaxRequest= $.post("test.php", values, function(data) {
alert(data);
})
.fail(function() {
alert("error");
})
.always(function() {
alert("finished");
});
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。
ajax的基本用法是这样的:
HTML:
<form id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
jQuery:
// Variable to hold request
var request;
// Bind to the submit event of our form
$("#foo").submit(function(event){
// Prevent default posting of form - put here to work in case of errors
event.preventDefault();
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "/form.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
});
注意:从jQuery 1.8开始，.success()， .error()和.complete()已被弃用，取而代之的是.done()， .fail()和.always()。
注意:请记住，上面的代码段必须在DOM ready之后完成，因此应该将其放在$(document).ready()处理程序中(或使用$()简写)。
提示:你可以像这样链接回调处理程序:$.ajax().done().fail().always();
PHP(即form.php):
// You can access the values posted by jQuery.ajax
// through the global variable $_POST, like this:
$bar = isset($_POST['bar']) ? $_POST['bar'] : null;
注意:总是消毒发布的数据，以防止注入和其他恶意代码。
你也可以在上面的JavaScript代码中使用。post来代替。ajax:
$.post('/form.php', serializedData, function(response) {
// Log the response to the console
console.log("Response: "+response);
});
注意:上面的JavaScript代码适用于jQuery 1.8及更高版本，但它应该适用于jQuery 1.5之前的版本。
