Update innerHTML by an email - javascript

I don't know if this is possible so I'm very open to trying other methods to achieve similar results.
Basically I have a website and a certain div
<div id="mainContent">Content here...</div>
What I want to do with this, is be able to send an email to a certain address or something similar and the body of the email will define the "innerHTML" of this div.
I can imagine it would look something like this in Javascript:
document.getElementbyId("mainConetent").innerHTML = emailBody;
With obviously "emailBody" being a pre-defined variable.
I am almost certain nothing like this could work but is there any way to achieve a similar thing?
I also know that 'innerHTML' in javascript becomes restored when a page is refreshed. I would like it to be there permenantly on the original document, until a new email is recieved with the new content. This method with javascript is probably not the way to do is. I just used it to explain bettter

Store email body in input hidden field and then using javascript you can retrieve them based on id and store them in respective div.Try this:
html:
<div id="mainContent">Content here...</div>
<input type="hidden" value="<?php echo $inputbody;?>" id="emailbody">
javascript:
var emailBody = document.getElementbyId("emailbody").value;
document.getElementbyId("mainContent").innerHTML = emailBody;
Updated Example:
<?php
$to = "someone#example.com";
$subject = "Test mail";
$inputbody = "Hello! This is a simple email message."; // insert this into hidden field
$from = "someonelse#example.com";
$headers = "From:" . $from;
?>

In your index.php file add the following piece of code for database connection
<?php
// Create connection
$con=mysqli_connect("localhost","username","password","db_name");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_fetch_array(mysqli_query($con,"SELECT * FROM table_name Where type='mail'"));
//Inside your mainContent div
<div id="mainContent"><?php echo $result['innerhtml'] ?></div>
?>
In your js file
If your email is being sent on button click perform this
$('#button').on('click',function(){
emailBody = document.getElementbyId("mainConetent").innerHTML;
$.ajax({
type: "POST",
url: "update_mail content.php",
data: "emailBody="+ emailBody,
success: function(){
}
});
})
//Updates the content in the databse table usinganother php file named update_mail content.php .This flow might help u mate.. :)

Related

PHP get information from database when echoed html code is clicked

In my code, I am retrieving data from the database using a while loop so i can have all the $FormData['fullname'] in the db.
What i want to do is, have each name display on the page and also have clickable so when someone clicks a name, it gets their user_id and pulls up information about them.
The problem i am having is that I can't figure out a way to make a it so where i can get the user_id when the user clicks a name. I tried putting a "name" in the button attribute and I checked if isset() but that didn't work.
If someone can properly figure out a way for me to basically, display all the fullnames in my database and when someone clicks a name, it pulls up information about them that is stored in the database. Here is my code
$stmtGet = $handler->prepare("SELECT * FROM formdata");
$stmtGet->execute();
while($formData = $stmtGet->fetch()){
echo "<button name='name'>$formData[fullname]</button>";
if($_SERVER['REQUEST_METHOD'] =="POST"){
if(isset($_POST['name'])){
echo "ok";
}else{
echo "bad";
}
}
}
As far i can see you are trying hit a button inside the while loop , i would not say its a bad approach , but i will suggest you not to do that . and from your code i can see you have lack of understanding post and get request learn from here . and other than this you need to know the transition of web url . how its actually works . anyway , i have given a sample code without . i hope it will help you understanding this concept.
$stmtGet = $handler->prepare("SELECT * FROM formdata");
$stmtGet->execute();
while($formData = $stmtGet->fetch(PDO::FETCH_ASSOC)){
$id = $formData['formdataid'];
echo "<a href='somepagename.php?infoid={$id}'>". $formData['fullname']."</a></br>";
}
now in the somepagename.php file or in the same page you can actually show the details information for instance
if(isset($_GET['infoid'])){
$stmt = $handler->prepare("select * from formdata where formdataid='"$_GET['infoid']"'");
$qry = $stmt->execute();
$row = $qry->fetch(PDO::FETCH_ASSOC);
echo "<p>id =".$row['formdataid']."</p></br>";
echo "<p>id =".$row['name']."</p></br>";
echo "<p>id =".$row['email']."</p></br>";
echo "<p>id =".$row['address']."</p></br>";
code is not executed , it may have semicolon or comma error warning . you have to fix those on your own . this example above shown you only the way it works .
if still you have problem ask , or see the documentation
I should stress that this is not production code and you should totally validate the data input coming in before posting queries to your DB. You can do something like this.
<?php
// Connect
$connection = mysqli_connect('localhost', 'username', 'password', 'database','port');
// Grab all users
$sql = 'SELECT * FROM users';
$users = mysqli_query($connection, $sql);
if (($_SERVER['REQUEST_METHOD'] == 'POST') && !empty($_POST['user_id'])) {
$query = "SELECT * FROM users WHERE user_id = {$_POST['user_id']};";
$user = mysqli_fetch_assoc(mysqli_query($connection, $query));
}
?>
// This only runs if our $user variable is set.
<?php if (isset($user)) : ?>
<?php foreach ($user as $key => $value) : ?>
<span><?= print_r($value) ?></span>
<?php endforeach; ?>
<?php endif; ?>
// Display all users in a dropdown and when the button is clicked
// submit it via post to this page.
<form action="<?= $_SERVER['PHP_SELF'] ?>" method="post">
<select name="user_id">
<?php foreach ($users as $user) : ?>
<option value="<?= $user['user_id'] ?>"><?= $user['name'] ?></option>
<?php endforeach; ?>
</select>
<button type="submit">Submit</button>
</form>
This is going to refresh your page every time. If you want to have an interactive page you are going to need to use JavaScript/AJAX to update the page elements without reloading the page. This example just demonstrates how you can achieve this with PHP and HTML.
you need to know about server-side language(php) and client-side language(javascript).
php runs before page loaded. it cannot runs when click something by itself(with ajax, it can).
most interactions without page move runs with javascript.
in your case, there are two methods.
store all information into hidden input or button's attribute. and get them by user_id via javascript.
use ajax to call another php that can select informations you need by user_id.
both use javascript or jquery. so you must learn about them.

Clickable HTML list to query db and output to <div>

Bear with me; this is my first StackOverflow question. I'm having trouble writing a proper algorithm. Perhaps doing all this to "force" it means I'm over-complicating a simple concept, but it's also very likely I'm just unaware of the fix.
I'm building an experimental cookbook that reads from a database and displays in the browser. I created a list (note: NOT a <ul> or <ol> element) that is populated by <span> items generated by a PDO query to the database. These spans reference the name of each recipe in the database.
<p>
<?php
$recList = $pdo->query('SELECT name FROM Recipe');
$rowCount = $recList->rowCount();
echo 'There are ' . $rowCount . ' recipes currently in our database!';
echo '<br /><br />';
while ($row = $recList->fetch()) {
echo '<span class="recName"';
echo '>' . $row['name'] . "</span><br />";
}
?>
</p>
I then created a scrolling div element:
<div id="recWindow">
<!-- Display recipe queried from the database here -->
<?php require("$DOCUMENT_ROOT/$rec_source"); ?>
</div>
I would like the user to be able to click on the recipe names generated by php and the chosen recipe to display within the <div>. Choosing different recipes should not cause the page to reload.
I feel like the answer lies in an AJAX request to a php file listening for a variable containing the recipe to display, but that's where I'm stuck. Somehow I need to pass the php list items a unique identifier that is recognized by javascript, which in turn handles the onclick change in the div by passing that identifier BACK to php to query the database. While typing that out, I'm almost certain that I've over-complicated this entire process.
I thought of using a dropdown select menu and a GET request, but I'd like to retain the clickable names function if possible.
Answers that conclude my proposed method is too "dirty" and point me in a better direction are completely acceptable. I'm happy to provide any necessary information I left out. Thank you so much in advance.
Environment: Virtual LAMP (CentOS7, MariaDB)
Try something like this
<p>
<?php
$recList = $pdo->query('SELECT name FROM Recipe');
$rowCount = $recList->rowCount();
echo 'There are ' . $rowCount . ' recipes currently in our database!';
echo '<br /><br />';
while ($row = $recList->fetch()) {
echo '<span class="recName" data-id="' . $row['id'] . '"';
echo '>' . $row['name'] . "</span><br />";
}
?>
</p>
<div id="recWindow">
<!-- Display recipe queried from the database here -->
<?php require("$DOCUMENT_ROOT/$rec_source"); ?>
</div>
<script type="text/javascript">
$(document).ready(function() {
$("body").on("click", "recName", function() {
//* get id of required recipe
var recId = $(this).attr("data-id");
//* send ajax-request to back-end
$.ajax({
url: "/get-recipe.php",
method: "GET",
data: {
id: recId
},
success: function(respond) {
//* put recipe-data into container
$("#recWindow").html(respond);
}
});
});
});
</script>
I hope, it shows you the main idea

Trying to make php login work with Phonegap

I understand that for a login / register system to work within Phonegap, you have to use aJax with your php. I've got a sucessful php login and register page working but I'm unsure where to begin with jQuery / aJax a.k.a where I'm meant to put it, and what exactly I should be putting in. I was wondering if someone would know how to point me into the right direction.
jQuery
jQuery is a framework built on Javascript. Javascript is a client-side (browser) language . It runs on your device, unlike PHP that gets executed on the server.
You need to include jQuery in the HTML of your login page using script tags:
<script src="https://code.jquery.com/jquery-2.2.3.min.js"></script>
jQuery provides a way to target any html element within your document and perform certain functions on that element. You specify what element by using the following syntax:
$(element).doSomething();
You can select classes or IDs:
<p id="myparagraph">A paragraph of text</p>
<p class="myparagraphclass">A paragraph of text</p>
$('#myparagraph').doSomething();
$('.myparagraphclass').doSomething();
AJAX
AJAX is a method introduced with Javascript that allows a page to request another url along with the result of that request. You will need to use AJAX login with Cordova/Phonegap because the "app" you're building is based on Javascript.
Thankfully, jQuery provides some really nice and easy to use AJAX methods.
Putting it together
I notice from a previous question that you have already created a PHP script that checks the login credentials are correct. Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /public_html/access/login.php on line 15
I have edited slightly the code within that question (/access/login.php):
require_once($_SERVER['DOCUMENT_ROOT'] . "/html5up-aerial/access/functions.php");
$username = trim($_POST['username']);
$password = trim($_POST['password']);
if ($username&&$password) {
session_start();
require_once($_SERVER['DOCUMENT ROOT'] . "db_connect.php");
mysqli_select_db($db_server, $db_database) or
die("Couldn't find db");
$username = clean_string($db_server, $username);
$password = clean_string($db_server, $password);
$query = "SELECT * FROM users WHERE username='$username'";
$result = mysqli_query($db_server, $query);
*if($row = mysqli_fetch_array($result)){*
$db_username = $row['username'];
$db_password = $row['password'];
if($username==$db_username&&salt($password)==$db_password){
$_SESSION['username']=$username;
$_SESSION['logged']="logged";
//header('Location: home.php'); // Have commented this out
$message = "YOU ARE NOW LOGGED IN!"; // <- ADDED THIS
}else{
$message = "<h1>Incorrect password!</h1>";
}
}else{
$message = "<h1>That user does not exist!</h1>" .
"Please <a href='index.php'>try again</a>";
}
mysqli_free_result($result);
require_once("db_close.php");
}else{
$message = "<h1>Please enter a valid username/password</h1>";
}
//header/footer only required if submitting to a seperate page
echo $message; // ADDED THIS
die(); // ADDED THIS
This will be the PHP script that AJAX requests.
Now we create the HTML document with a login form and include jQuery and write our ajax code:
<script src="https://code.jquery.com/jquery-2.2.3.min.js"></script>
<form class="login-form" method="post">
Username: <input name="username" /><br />
Password: <input name="password" type="password" /><br />
<input type="submit" value="Login" />
</form>
<script>
$('.login-form').on('submit', function(e) { // Listen for submit
e.preventDefault(); // Don't actually submit the form
var data = $(this); // Put the form in a variable
$.ajax({
type: "POST",
url: '/access/login.php',
data: $(data).serialize(), // Make form data into correct format
success: function(response) {
alert(response); // Alert with the response from /access/login.php
}
});
});
</script>
To debug this code you will need to use Chrome development toolbar or Firefox Firebug. Hope this helps.

After redirect, how to display error messages coming from original file?

Here's what I'm trying to achieve: I want to redirect the user if any errors I check for are found to a html/php form (that the user see's first where inputs are previously created) with custom error messages.
Details: The User see's the HTML/PHP form first where they enter names in a csv format. After they click create, the names are processed in another file of just php where the names are checked for errors and other such things. If an error is found I want the User to be redirected to the HTML/PHP form where they can fix the errors and whatever corresponding error messages are displayed. Once they fix the names the User can click the 'create user' button and processed again (without errors hopefully) and upon completion, redirect user to a page where names and such things are displayed. The redirect happens after the headers are sent. From what I've read this isn't the best thing but, for now, it'll do for me.
Code For HTML/PHP form:
<!DOCTYPE HTML>
<HTML>
<head>
<title>PHP FORM</title>
</head>
<body>
<form method="post" action="processForm.php">
Name: <input type="text" name="names" required = "required"><br>
<input type="submit" value="Create Users" onclick="formInputNames"><br>
Activate: <input type="checkbox" name="activate">
</form>
<?php
// include 'processForm.php';
// errorCheck($fullname,$nameSplit,$formInputNames);
?>
</body>
</html>
I tried messing around with 'include' but it doesn't seem to do anything, however, I kept it here to help illustrate what I'm trying to achieve.
Code For Process:
$formInputNames = $_POST['names'];
$active = (isset($_POST['activate'])) ? $_POST['activate'] : false;
//checks if activate checkbox is being used
$email = '#grabby.com';
echo "<br>";
echo "<br>";
$fullnames = explode(", ", $_POST['names']);
if ($active == true) {
$active = '1';
//sets activate checkbox to '1' if it has been selected
}
/*----------------------Function to Insert User---------------------------*/
A Function is here to place names and other fields in database.
/*-------------------------End Function to Insert User--------------------*/
/*-----------------------Function for Errors---------------------*/
function errorCheck($fullname,$nameSplit,$formInputNames){
if ($formInputNames == empty($fullname)){
echo 'Error: Name Missing Here: '.$fullname.'<br><br>';
redirect('form.php');
}
elseif ($formInputNames == empty($nameSplit[0])) {
echo 'Error: First Name Missing in: '.$fullname.'<br><br>';
redirect('form.php');
}
elseif ($formInputNames == empty($nameSplit[1])) {
echo 'Error: Last Name Missing in: '.$fullname.'<br><br>';
redirect('form.php');
}
elseif (preg_match('/[^A-Za-z, ]/', $fullname)) {
echo 'Error: Found Illegal Character in: '.$fullname.'<br><br>';
redirect('form.php');
}
}
/*-----------------------------End Function for Errors------------------------*/
/*--------------------------Function for Redirect-------------------------*/
function redirect($url){
$string = '<script type="text/javascript">';
$string .= 'window.location = "' .$url. '"';
$string .= '</script>';
echo $string;
}
/*-------------------------End Function for Redirect-----------------------*/
// Connect to database
I connect to the database here
foreach ($fullnames as $fullname) {
$nameSplit = explode(" ", $fullname);
//opens the database
I Open the database here
errorCheck($fullname,$nameSplit,$formInputNames);
$firstName = $nameSplit[0];//sets first part of name to first name
$lastName = $nameSplit[1];//sets second part of name to last name
$emailUser = $nameSplit[0].$email;//sets first part and adds email extension
newUser($firstName,$lastName,$emailUser,$active,$conn);
redirect('viewAll.php');
//echo '<META HTTP-EQUIV="Refresh" Content="0; URL=viewAll.php">';
//if you try this code out, you can see my redirect to viewAll doesn't work when errors are found...I would appreciate help fixing this as well. My immediate fix is using the line under it but I don't like it.
}
Any help is certainly appreciated.Thank You
Also it's worth noting I'm new to php. I would like to have an answer in php as well (if possible).
There's multiple ways of doing so. I personally would use AJAX. On a 'form submit', run a javascript function calling an AJAX request to a .php file to check the form information, all using post method. Calculate all the $_POST['variables'] checking for your defined errors. You would have an html element print the errors via AJAX request.
If there are 0 errors then in the request back return a string as so that your javascript function can look for if its ready to go. If ready to go, redirect the user to where ever you please.
AJAX is not hard and I only suggested the idea sense you put javascript in your tags.
Another method:
Having all your code on one .php file. When you submit the form to the same .php file check for the errors (at the top of the file). If $_POST['variables'] exist, which they do after you submit the form, you echo your errors in the needed places. If zero errors then you redirect the page.

Upload an image without refreshing page in php

I want to upload an image without refreshing page. please help me for this purpose. I find many thing but ever
Complete Script :
you need ajax to do it and here some code to work for u :
ajaximage.php
Contains PHP code.
This script helps you to upload images into uploads folder.
Image file name rename into timestamp+session_id.extention
<?php
include('db.php');
session_start();
$session_id='1'; // User session id
$path = "uploads/";
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST") {
$name = $_FILES['photoimg']['name'];
$size = $_FILES['photoimg']['size'];
if(strlen($name)) {
list($txt, $ext) = explode(".", $name);
if(in_array($ext,$valid_formats)) {
if($size<(1024*1024)) // Image size max 1 MB
{
$actual_image_name = time().$session_id.".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name)) {
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE uid='$session_id'");
echo "<img src='uploads/".$actual_image_name."' class='preview'>";
}
else
echo "failed";
}
else
echo "Image file size max 1 MB";
}
else
echo "Invalid file format..";
}
else
echo "Please select image..!";
exit;
}
?>
index.php
Contains simple PHP and HTML code.
Here $session_id=1 means user id session value.
<?php
include('db.php');
session_start();
$session_id='1'; // User login session value
?>
<form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'>
Upload image <input type="file" name="photoimg" id="photoimg" />
</form>
<div id='preview'>
</div>
Sample database design for Users.
Users
Contains user details username, password, email, profile_image and profile_image_small etc.
CREATE TABLE `users` (
`uid` int(11) AUTO_INCREMENT PRIMARY KEY,
`username` varchar(255) UNIQUE KEY,
`password` varchar(100),
`email` varchar(255) UNIQUE KEY,
`profile_image` varchar(200),
`profile_image_small` varchar(200),
)
Javascript Code
$("#photoimg").on('change',function(){})
// photoimg is the ID name of INPUT FILE tag and
$('#imageform').ajaxForm()
//imageform is the ID name of FORM. While changing INPUT it calls FORM submit without refreshing page using ajaxForm() method.
<script type="text/javascript" src="http://ajax.googleapis.com/
ajax/libs/jquery/1.5/jquery.min.js"></script>
<script type="text/javascript" src="jquery.form.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$('#photoimg').on('change', function()
{
$("#preview").html('');
$("#preview").html('<img src="loader.gif" alt="Uploading...."/>');
$("#imageform").ajaxForm(
{
target: '#preview'
}).submit();
});
});
</script>
Uploading files to the server without a page refresh requires some additional client-side tools. These tools will then need to communicate with the PHP backend that you have written. Here are some popular solutions which offer what you are looking for:
Uploadify, my favorite of these solutions: http://www.uploadify.com/
SWFUpload, similar to Uploadify: http://swfupload.org/
jQuery Form Plugin, an AJAX-based uploader: http://jquery.malsup.com/form/#file-upload
Hope that helps.
Two good tutorials:
http://www.9lessons.info/2011/08/ajax-image-upload-without-refreshing.html
http://css-tricks.com/6522-ajax-image-uploading/
You need to use ajax to do that. Ajax will send the request to a PHP script that will do the work without refreshing the entire page.
Submit it via XMLHttpRequest. In a nutshell you would need to initialise a FormData() object and append your file to the object, then initiate an xhr connection, and send your object via xhr xhr.send.
This is all at a very basic level...
Or, better yet, use a pre-existing tool.
Lot of jquery plug ins are available, you can show progress bar too. refer this
http://www.phpletter.com/Demo/AjaxFileUpload-Demo/

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