können Sie Ajax verwenden
Beispiel -
$('#button').click(function() {//give id of submit button
var val1 = $('#text1').val();//get value you need to post
var val2 = $('#text2').val();//get value you need to post
$.ajax({
type: 'POST',
url: 'process.php',
data: { text1: val1, text2: val2 },
success: function(response) {
$('#result').html(response);//responce you recevied
}
}); });
beide Werte Get Post-Methode verwenden. Führen Sie Ihre Aktion aus und senden Sie im Ergebnis "True False".
--------------- explaning in tief mit neuem Beispiel ----------------
HTML Datei : ajaxsubmit.html
<!DOCTYPE html>
<html>
<head>
<title>Submit Form Using AJAX and jQuery</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<link href="css/refreshform.css" rel="stylesheet">
<script src="script.js"></script>
</head>
<body>
<div id="mainform">
<h2>Submit Form Using AJAX and jQuery</h2> <!-- Required div Starts Here -->
<div id="form">
<h3>Fill Your Information !</h3>
<div>
<label>Name :</label>
<input id="name" type="text">
<label>Email :</label>
<input id="email" type="text">
<label>Password :</label>
<input id="password" type="password">
<label>Contact No :</label>
<input id="contact" type="text">
<input id="submit" type="button" value="Submit">
</div>
</div>
</div>
</body>
</html>
PHP File: ajaxsubmit.php
<?php
$connection = mysql_connect("localhost", "root", ""); // Establishing Connection with Server..
$db = mysql_select_db("mydba", $connection); // Selecting Database
//Fetching Values from URL
$name2=$_POST['name1'];
$email2=$_POST['email1'];
$password2=$_POST['password1'];
$contact2=$_POST['contact1'];
//Insert query
$query = mysql_query("insert into form_element(name, email, password, contact) values ('$name2', '$email2', '$password2','$contact2')");
echo "Form Submitted Succesfully";
mysql_close($connection); // Connection Closed
?>
jQuery Datei: script.js
$(document).ready(function(){
$("#submit").click(function(){
var name = $("#name").val();
var email = $("#email").val();
var password = $("#password").val();
var contact = $("#contact").val();
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'name1='+ name + '&email1='+ email + '&password1='+ password + '&contact1='+ contact;
if(name==''||email==''||password==''||contact=='')
{
alert("Please Fill All Fields");
}
else
{
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "ajaxsubmit.php",
data: dataString,
cache: false,
success: function(result){
alert(result);
}
});
}
return false;
});
});
MY-SQL-Code Segment:
CREATE DATABASE mydba;
CREATE TABLE form_element(
id int(10) NOT NULL AUTO_INCREMENT,
name varchar(255) NOT NULL,
email varchar(255) NOT NULL,
password varchar(255) NOT NULL,
contact varchar(255) NOT NULL,
PRIMARY KEY (id)
)
Was für die process.php ist? – tash517
Ich habe dich nicht genau darüber informiert, welche process.php? Ich aktualisiere meine Antwort zu führen, was darin geschrieben ist nach Ihrer Frage –
Oh, ich habe wirklich keine PHP-Dateien atm, so dass ich verwirrt bin, was es für – tash517