我正在尝试让我的HTML表单通过Javascript,然后将其传递给将发送给MySQL的PHP。
但是我要么在浏览器中加载JS文件,要么在浏览器中加载PHP文件。
这是我的HTML表单:
<div class="form" id="form" onclick="submitForm()">
<form id='contactform' action="js/contactform.js" method="post" onSubmit="submitForm()">
<input type="text" id="name" placeholder="Name" autofocus required><br>
<input type="text" id="email" placeholder="Email" required><br>
<input type="tel" id="telephone" placeholder="Telephone"><br>
Enquiry : <input type="radio" id="subject" value="enquiry" required>
Booking : <input type="radio" id="subject" value="booking" required><br>
<textarea id="message" required rows="20" cols="20" placeholder="Enter your message and I will try to get back to you within 2 days."></textarea><br>
<input type="submit" name="submit" value="Submit" class="submitbutton"/>
<input type="reset" name="clearbutton" value="Clear" class="clearbutton"/>
</form>
<div id="outcome"></div>
我希望将表单提交的结果放入&#34;结果&#34;格
我的JS代码:
function getOutput() {
getRequest(
'php/getinfo.php',
drawOutput,
drawError
);
return false;
}
// handles drawing an error message
function drawError () {
var container = document.getElementById("content");
container.innerHTML = 'Bummer: there was an error!';
}
// handles the response, adds the html
function drawOutput(responseText) {
var container = document.getElementById("content");
container.innerHTML = responseText;
}
// helper function for cross-browser request object
function getRequest(url, success, error) {
var req = false;
try{
// most browsers
req = new XMLHttpRequest();
} catch (e){
// IE
try{
req = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
// try an older version
try{
req = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
return false;
}
}
}
if (!req) return false;
if (typeof success != 'function') success = function () {};
if (typeof error!= 'function') error = function () {};
req.onreadystatechange = function(){
if(req .readyState == 4){
return req.status === 200 ?
success(req.responseText) : error(req.status)
;
}
}
req.open("GET", url, true);
req.send(null);
return req;
}
function submitForm(){
var name = document.getElementById('name').value;
var booking = document.getElementById('subject').value;
var enquiry = document.getElementById('subject').value;
var email = document.getElementById('email').value;
var telephone = document.getElementById('telephone').value;
var message = document.getElementById('message').value;
getRequest(
'php/form.php?' + params, //URL for the PHP file
procesOutput,
processError
);
return false;
}
function processOutput(){
var container = document.getElementById('outcome');
container.innerHTML = responseText;
}
function processError(){
alert("There has been an error, please try again");
}
和我的PHP代码:
$con=mysqli_connect("DBLocation","Username","Password",'DBName');
if (mysqli_connect_errno()){
die("Error: " . mysqli_connect_error());
}
$result = mysqli_query($con,"INSERT INTO `Contact`(`Name`, `Email`, `Telephone`, `Enquiry`, `Booking`, `Message`) VALUES ([value-2],[value-3],[value-4],[value-5],[value-6],[value-7])");
if ($conn->query($sql) === TRUE){
echo "Thank you for contacting us, I will replay to you soon!";
}
else {
echo "I'm sorry but an Error has occured. Please try again shortly";
}
mysql_close($conn);
?>
我已经看过w3schools页面以及其他一些问题,但我似乎无法理解它。
答案 0 :(得分:1)
有几件事,首先是getRequest
是什么?第二关,responseText
定义在哪里?第三,我会检查你的控制台,因为我很确定submitForm
中有错误。我看到很多getElementById
s,但你的输入都没有与它们相关的ID:
<input type="text" name="name" placeholder="Name" autofocus required>
应该是:
<input type="text" id="name" placeholder="Name" autofocus required>
答案 1 :(得分:0)
删除onClick
,onSubmit
,action="js/contactform.js" method="post"
属性,因为您不需要它们。在js中捕获表单提交。 How to? - link
在表单元素上添加ID,以便您可以选择它们:
var name = document.getElementById('name').value;
答案 2 :(得分:0)
我相信你想使用Ajax,你可以轻松地使用jQuery。
我相信您的问题的答案就在这个页面上:https://learn.jquery.com/ajax/ajax-and-forms/
如果jQuery对你而言是新的,你可能想要阅读一些关于jQuery的内容:https://learn.jquery.com/about-jquery/
如果您不想使用jQuery,可以直接在javascript中使用XMLHttpRequest。