I'm following a tutorial by mmtuts on youtube to show how to post comments to a myphpadmin database. All of my code is exactly the same as his, but I'm working from a different starting point becuase I already had a website I was working on and I just wanted to add the new code.
Basically, the video showed the code working flawlessly and my posts do not show up in the database like his did.
https://www.youtube.com/watch?v=4pPGOF5MI4U
".setComments($conn)." on the second document of code is blue instead of white like in the video.
<?php
require 'includes/dbh.inc.php';
session_start();
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="description" content="This is an example of a meta description. This will often show up in search results.">
<meta name=viewport content="width=device-width, initial-scale=1">
<title>TAG</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<div id="headerContainer">
<?php
if (isset($_SESSION['userID'])) {
$id = $_SESSION['userID'];
$sqlImg = "SELECT * FROM profileimg WHERE userid='$id'";
$resultImg = mysqli_query($conn, $sqlImg);
while ($rowImg = mysqli_fetch_assoc($resultImg)) {
if ($rowImg['status'] == 0) {
$filename = "profilepics/profile".$id."*";
$fileinfo = glob($filename);
$fileext = explode(".", $fileinfo[0]);
$fileactualext = $fileext[1];
echo "<div class=userPicture><img src='profilepics/profile".$id.".".$fileactualext."?".mt_rand()."'></div>";
}
else {
echo "<div class='userPicture'><img src='profilepics/noUser.png'></div>";
}
}
echo '
<div class="userName">'. $_SESSION['userUserName'] .'</div>
<div id="logoutForm">
<form action="includes/logout.inc.php" method="post">
<button type="Submit" name="logout-submit">Logout</button>
</form>
</div>
<div class="upload">
<form action="upload.php" method="POST" enctype="multipart/form-data">
<input type="file" name="file">
<button type="submit" name="submit">Profile Image</button>
</form>
</div>
';
}
else {
echo '
<div class="userPicture"><img src="profilepics/noUser.png"></div>
<div class="userName">You are not logged in!</div>
<div id="loginForm">
<form action="includes/login.inc.php" method="post">
<input type="text" name="mailuid" placeholder="Username/E-mail">
<input type="password" name="password" placeholder="Password">
<button type="Submit" name="login-submit">Login</button>
</form>
</div>
<div id="signupForm">
or Signup
</div>
';
}
?>
</div>
<?php
require "header.php";
date_default_timezone_set('America/Chicago');
include 'includes/comments.inc.php';
?>
<div class="homeBody">
<p>Starting Filler</p>
<p>-</p>
<p>-</p>
<p>-</p>
<video width="320" height="240" controls>
<source src="videos/sample.mp4" type="video/mp4">
Your browser does not support the video tag.
</video>
<?php
echo "<form method='POST' action='".setComments($conn)."'>
<input type='hidden' name='uid' value='Anonymous'>
<input type='hidden' name='date' value='".date('Y-m-d H:i:s')."'>
<textarea name='message'></textarea><br>
<button type='submit' name='commentSubmit'>Comment</button>
</form>";
?>
<?php
function setComments($conn) {
if (isset($POST['commentSubmit'])) {
$uid = $_POST['uid'];
$date = $_POST['date'];
$message = $_POST['message'];
$sql = "INSERT INTO comments (uid, date, message) VALUES ('$uid', '$date', '$message')";
$result = $conn->query($sql);
}
}
<?php
$servername = "localhost";
$dBUsername = "root";
$dBPassword = "thisisnotmyactualpassword";
$dBName = "tagloginsystem";
$conn = mysqli_connect($servername, $dBUsername, $dBPassword, $dBName);
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
}
All I want is for the posts to make it into the database.
Try making a separate PHP-file and have your action attribute inside the form tag point to it. Right now it looks like you are running the function in the action attribute. In the PHP-file you can run your PHP function and write the PHP you need.
EX:
<form method='POST' action="includes/comments.php">
I was missing the "_" in $_POST on the 3rd page of code
Related
I have a form that I am trying to use to reset passwords. I can get it to send the email out and I can click the link, but when I click the link all I get is the header. My form is not showing. I am extremely new to all of this. Any help will be much appreciated. I did search through some other posts, but none of them seemed to get me any closer to the right answer. Everything seems to work as it should, I just can't get my form to display so that I can enter the information.
<!DOCTYPE html>
<html lang="en">
<head>
<title>Reset Password Form</title>
</head>
<body>
<div class="container"><h2>Reset New Password Here</h2>
<?php
if($_GET['email'] && $_GET['token']) {
$conn = mysqli_connect("localhost", "root", "", "user");
// Check connection
if($conn === false){
die("ERROR: Could not connect. "
. mysqli_connect_error());
}
$email = $_GET['email'];
$token = $_GET['token'];
$query = mysqli_query("SELECT * FROM `customers` WHERE `reset_link_token`='".$token."' and `emailaddress`='".$email."';");
$current_date = date("Y-m-d H:i:s");
if (mysqli_num_rows($query) > 0) {
$row = mysqli_fetch_array($query);
if($row['expiry_date'] <= $current_date) { ?>
<form action="update-password.php" method="post">
<input name="email" type="hidden" value="<?php echo $email; ?>" />
<input name="reset_link_token" type="hidden" value="<?php echo $token; ?>" />
<div class="form-group"><label for="new-password">Password</label>
<input id="new-password" name="password" type="password" /></div>
<div class="form-group"><label for="confirm-password">Confirm Password</label>
<input id="confirm-password" name="confirm_password" type="password" /></div>
<input class="submit-btn" name="submit" type="submit" />
</form>
<?php }
} else {
echo "This forget password link has been expired";
}
}
?>
</div>
</body>
</html>
I'm creacting an admin zone for a blog I'm building. For some reason, the database is inserted with the last information I've submitted each time I refresh the page, instead of whenever I click "submit".
This is the page itself:
<?php
//send uploaded files to mysql
include_once 'conn.php';
include_once 'submit.php';
?>
<!DOCTYPE html>
<html>
<head>
<title>Your admin page</title>
</head>
<body>
<h1>Admin page</h1>
<h2>upload text</h2>
<form action="" method="POST">
<label>Header</label>
<input type="header" name="header" required><br>
<label>Text</label>
<input type="text" name="text" required><br>
<label>Image</label>
<input type="file" name="image">
<input type="submit" value="submit">
</form>
This is the submit.php file:
<?php
include_once 'conn.php';
$header = $_POST['header'];
$text = $_POST['text'];
$file_get = $_FILES['image']['name'];
$temp = $_FILES['image']['tmp_name'];
$file_to_saved = "uploads/".$file_get;
move_uploaded_file($temp, $file_to_saved);
if (isset($_POST['submit'])) {
$insert = ("INSERT INTO posts (postHeader,postText,picture) VALUES ('$header','$text','$file_to_saved')");
if ($link->query($insert) === TRUE) {
echo "record created succesfully";
} else{
echo "something went wrong with the file";
}
}
Hi for some reason my update statement is not working, most probably because of the id=%d, its not getting the id for the statement i think but its finding the id because it is listed in the url (shown Below). What is the problem please ?
This works when i Insert a number for the id, example : id=77, so most probably the problem is the %d how can i get it to find the id with the %d ?
http://localhost/test/edit.php?id=77
<?php
ob_start();
session_start();
include_once 'logindb.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error);
if((isset($_POST['submit']))){
if((!isset($_POST['title'])) || (!isset($_POST['times'])) ){
echo "All values must be set";
}
else{
$title = $_POST['title'];
$times = $_POST['times'];
$film_id = $_GET['id'];
$upfile = 'Uploads/posters/'.$_FILES['userfile']['name'];
if(move_uploaded_file($_FILES['userfile']['tmp_name'],$upfile)){
echo "File moved into folder";
header( "Location: index.php" ) ;
}
else{
echo "Problem: could not move image file to destination directory";
}
$upfile2 = 'Uploads/trailers/'.$_FILES['userfile2']['name'];
$format1 = "UPDATE films SET titles = '$title', ftimes = '$times', poster = '$upfile', trailer = '$upfile2' WHERE id = %d";
$query = sprintf($format1, $film_id );
$result = mysqli_query($conn, $query)
or die("Error in query: ". mysqli_error($conn));
}
}
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="Empire.css"/>
<script type="text/javascript" src="script.js"> </script>
<title>Movie Form</title>
</head>
<body class="formola">
<div class="form-style-5">
<p class="top"> List Movies </p>
<form method = "post" action="edit.php" enctype="multipart/form-data">
<fieldset>
<legend><span class="number">1</span> Details </legend>
<input type="text" name="title" placeholder="Movie Title">
<input type="text" name="times" placeholder="Times">
</fieldset>
<fieldset>
<legend><span class="number">2</span> Attachments </legend>
<div class="form-group ">
<label> Film Poster <br> </label>
<input type="hidden" class="form-control" name="MAX_FILE_SIZE" value="100000000">
Upload this Film Poster: <br> <input name="userfile" id="userfile" type="file"> <br>
<input type="hidden" class="form-control" name="MAX_FILE_SIZE" value="10000000">
Upload this Trailer: <br> <input name="userfile2" id="userfile2" value="10000000" type="file"> <br>
</div>
</fieldset>
<input type="submit" name = "submit" value="Upload" />
</form>
</div>
</body>
</html>
Going to try to keep it short. I have a while loop in grid.php file to fill up a table as such...
<?php while($product = $products->fetch_assoc()) { ?>
<tr>
<td><?php echo $product['cd_id']?></td>
<td><?php echo $product['cd_title']?></td>
<td><?php echo $product['cd_musician_fname']?></td>
<td><?php echo $product['cd_musician_lname']?></td>
<td><?php echo $product['cd_price']?></td>
<td>Edit</td>
<td>Delete</td>
</tr>
<?php } ?>
If I click the first anchor tag takes me to a edit.php file and here is the head code for that file.
<?php include '_includes/db.php';
$cd_id = trim($_GET['id']);
$message = '';
include '_includes/connection.php';
if($db->connect_error){
$message = $db->connect_error;
}else{
$sql = "SELECT * FROM CD WHERE cd_id = $cd_id";
$result = $db->query($sql);
$row = $result->fetch_assoc();
if($db->error){
$message = $db->error;
}
}
?>
Now I will show the html of edit.php
<!-- Product Musician last name-->
<fieldset class="form-group">
<label for="cd_musician_lname">Musician's lirst name</label>
<input type="text" class="form-control" id="cd_musician_lname" name="cd_musician_lname" value="<?php echo $row['cd_musician_lname'];?>">
</fieldset>
<!-- End of Musician last name-->
<!-- Product price-->
<fieldset class="form-group">
<label for="cd_price">Product price</label>
<input type="text" class="form-control" id="cd_price" name="cd_price" value="<?php echo $row['cd_price'];?>">
</fieldset>
<!-- End of Product price-->
<!-- Form submit button-->
Update Record
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
I have the edit.php page working just fine but if I make changes in the fields and click the submit anchor tag I get all the fields of the row empty but the PK. Here is the code for the final edit_confirm.php file
<?php
include '_includes/db.php';
$cd_id = trim($_GET['id']);
$cd_title = $_POST['cd_title'];
$cd_musician_fname = $_POST['cd_musician_fname'];
$cd_musician_lname = $_POST['cd_musician_lname'];
$cd_price = $_POST['cd_price'];
$message = '';
include '_includes/connection.php';
if($db->connect_error){
die("Connection failed: ".$db->connect_error);
} else {
$sql = "UPDATE CD SET cd_title='".$cd_title."', cd_musician_fname='".
$cd_musician_fname."', cd_musician_lname='".
$cd_musician_lname."', cd_price='".$cd_price."' WHERE cd_id = $cd_id ";
$db->query($sql);
var_dump($sql);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<?php include '_includes/main-head.php';?>
</head>
<body>
<?php include '_includes/main-navbar.php';?>
<div class="container">
<hr>
<?php
if($db->query($sql) === TRUE){ ?>
<h1>Record updated successfully.</h1>
<?php echo $cd_title; ?>
<?php echo $record->affected_rows ?>
<p> record was updated in the database.</p></br>
<?php } else { ?>
<p>Error updating the record: </p> <?php $db->error; ?>
<?php }; ?>
<hr>
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
</div>
<?php include '_includes/main-script.php';?>
</body>
</html>
If you notice in the edit_confirm.php I did a var_dump to see what are the values in the variables and it shows empty.
I need help with this.
Thank you in advance.
Man the better way to do this is make it simple to test if the record is updating or not
formsample.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<?php
include("connection.php");
$id = $_GET['id'];
$query= "select * from clients where id = '$id'";
$sql = mysqli_query($connect, $query);
$result = mysqli_fetch_assoc($sql);
?>
<form action="process.php" method="post">
<input type="text" name="name" id="name" value="<?php echo $result['name'] ?>" />
<input type="text" name="email" id="email" value="<?php echo $result['email'] ?>" />
<input type="hidden" name="id" id="id" value="<?php echo $id?>" />
<input type="submit" />
</form>
</body>
</html>
process.php
<?php
include("connection.php");
$id = $_POST['id'];
$name = $_POST['name'];
$email= $_POST['email'];
$query = "UPDATE clients set name= '$name', email ='$email' where id = '$id'";
$sql = mysqli_query($connect, $query);
?>
Update Record
This is not the proper way to submit a form - it won't work at all.
You need to have a form opening and closing tag, the target address is in the action attribute of the form element, and the method is on there too and should be post for this form (method="POST"). In your code you have a link where you should have a submit input so it won't submit the data, it will just redirect you to that URL. You should have something like this:
<input type="submit" value="Update Record" />
http://www.w3schools.com/html/html_forms.asp
I am trying to detect a form click using if(isset($_POST['appSelecter'])){ however it seems to not be returning true. This might be to do with the fact that my button click returns to the same page which would loose the form data i had just populated. Can someone confirm if my assumption is correct and if so - how would i need to change this?
Thanks
tried to only paste a sample piece of code to not confuse matters - seems i have made things worse - here is the full flow
<?php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<!--META-->
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Client Portal Login</title>
<!--STYLESHEETS-->
<link href="css/style.css" rel="stylesheet" type="text/css" />
<!--SCRIPTS-->
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.2.6/jquery.min.js"></script>
<!--Slider-in icons-->
<script type="text/javascript">
$(document).ready(function() {
$(".username").focus(function() {
$(".user-icon").css("left","-48px");
});
$(".username").blur(function() {
$(".user-icon").css("left","0px");
});
$(".password").focus(function() {
$(".pass-icon").css("left","-48px");
});
$(".password").blur(function() {
$(".pass-icon").css("left","0px");
});
});
</script>
</head>
<body>
<!--WRAPPER-->
<div id="wrapper">
<!--SLIDE-IN ICONS-->
<div class="user-icon"></div>
<div class="pass-icon"></div>
<!--END SLIDE-IN ICONS-->
<!--LOGIN FORM-->
<form name="login-form" class="login-form" action="index.php" method="post">
<!--HEADER-->
<div class="header">
<!--TITLE--><h1>Client Portal Login</h1><!--END TITLE-->
<!--DESCRIPTION--><span>Please login to your client portal</span><!--END DESCRIPTION-->
</div>
<!--END HEADER-->
<!--CONTENT-->
<div class="content">
<!--USERNAME--><input name="username" type="text" class="input username" value="Username" onfocus="this.value=''" /><!--END USERNAME-->
<!--PASSWORD--><input name="password" type="password" class="input password" value="Password" onfocus="this.value=''" /><!--END PASSWORD-->
</div>
<!--END CONTENT-->
<!--FOOTER-->
<div class="footer">
<!--LOGIN BUTTON--><input type="submit" name="submit" value="Login" class="button" /><!--END LOGIN BUTTON-->
<!--REGISTER BUTTON--><input type="submit" name="submit" value="Register" class="register" /><!--END REGISTER BUTTON-->
</div>
<!--END FOOTER-->
</form>
<?php
include("application.php");
if(isset($_POST['submit'])){
$username=$_POST["username"];
$password=$_POST["password"];
$userid = logUserIn($username, $password);
if($userid > 0){
$applicationsForUser = getAppInformation($userid);
printUserApplicationSelectionForm($applicationsForUser);
if(isset($_POST['appSelecter'])) {
echo "this is a test message";
}
}
}
function printUserApplicationSelectionForm($applicationsForUser){
echo "<br/>";
echo "<br/>";
echo "<br/>";
echo "<br/>";
foreach ($applicationsForUser as $app) {
?>
<form action="index.php" method="post">
<input type="hidden" name="userid" value="<?php echo $app->getUserid(); ?>">
<input type="hidden" name="name" value="<?php echo $app->getName(); ?>">
<input type="hidden" name="created" value="<?php echo $app->getDateCreated(); ?>">
<input type="hidden" name="invoice" value="<?php echo $app->getInvoice(); ?>">
<input type="hidden" name="comment" value="<?php echo $app->getComment(); ?>">
<input type="submit" name="appSelecter" value="<?php echo $app->getName(); ?>">
</form>
<?php
}
}
function getAppInformation($userid){
$applicationsForUser = array();
$conn = new mysqli('localhost:3306', 'root', '', 'clientportal');
if ($conn->connect_errno > 0) {
die('Could not connect: ' . mysql_error());
}else{
//we have connected to the database
$sql = "SELECT * FROM application WHERE userid = '$userid'";
if(!$val = $conn->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}else{
$index = 0;
while($row = $val->fetch_assoc()){
$userid = $row['userid'];
$name = $row['name'];
$dateCreated = $row['date'];
$invoice = $row['invoiceid'];
$comment = $row['commentsid'];
$application = new Application($userid, $name, $dateCreated, $invoice, $comment);
$applicationsForUser[$index] = $application;
$index++;
}
}
}
$conn -> close();
return $applicationsForUser;
}
function logUserIn($username, $password) {
if(!isset($username) && !isset($password)){
return -1;
}
$result = -1;
//$conn = mysql_connect('localhost', 'web214-admin-ava', 'secondstory');
$conn = new mysqli('localhost:3306', 'root', '', 'clientportal');
if ($conn->connect_errno > 0) {
die('Could not connect: ' . mysql_error());
}else{
//we have connected to the database
$sql = "SELECT * FROM members WHERE username = '$username' AND password = '$password'";
if(!$val = $conn->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}else{
while($row = $val->fetch_assoc()){
$result = $row['id'];
break;
}
}
}
$conn -> close();
return $result;
}
?>
<!--END LOGIN FORM-->
</div>
<!--END WRAPPER-->
<!--GRADIENT--><div class="gradient"></div><!--END GRADIENT-->
</body>
</html>
You have used folowing in the form submit:
onClick="location.href='index.php'" // Making a GET request
This is not submitting the form using POST method. Remove this and it'll work.
Update: There is no submit button with name submit so this condion will not work:
if(isset($_POST['submit']))
Make it:
if(isset($_POST['appSelecter']))
You don't need if(isset($_POST['submit'])) instead use;
if(isset($_POST['appSelecter'])) {
$username=$_POST["username"];
$password=$_POST["password"];
$userid = logUserIn($username, $password);
if($userid > 0){
$applicationsForUser = getAppInformation($userid);
printUserApplicationSelectionForm($applicationsForUser);
}
}
You dont nee this
onClick="location.href='index.php'"
dont do anything , just apply value to button i, i think you have applied already ,
by location.href your request will be send by GET Method in thgis case no form elements sent to the server
if you allow native form submission then all form elements will be sent to server, in case of multiple forms , the only elements sent realted to that submit button form thats it