I'm currently doing a Web Programming module at university and have been having trouble with some of the homework set. We are meant to insert code that updates our current mysql table with new information (gender, age, email, comment). This information needs to be inserted into the row of each persons session generated ID (currID). How do we code for the updated information to be inserted into a session-specific row?
<?php
session_start();
include('muqHeader.html');
include('commonSrc.php');
include('../shareCode/mysqlLink.php');
if ($_SERVER['REQUEST_METHOD'] == 'POST'):
// update the mf record
if (filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)){
}else{
echo "Not a valid email address";
}
if(filter_var($_POST['comment'], FILTER_SANITIZE_STRING)){
}else{
echo "Text includes invalid characters";
}
$gender = $_POST['gender'];
$age = $_POST['age'];
$email = $_POST['email'];
$comment = $_POST['comment'];
$currID = $_SESSION['currID'];
if ($_POST['submit']){
$sql = "UPDATE muq
SET (gender='$gender', age = '$age', email = '$email', comment = '$comment')
WHERE (muqID = '$currID')";
}
if (#mysqli_query($link, $sql)) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . #mysqli_error($link);
}
else:
$useTime = implode(',', $_SESSION['useTime'] );
$usedM = implode( ',', $_SESSION['usedM'] );
$tmp = array();
for($i=0; $i < count($_SESSION['freqRate']); $i++) {
$tmp[$i] = implode( '', $_SESSION['freqRate'][$i] ); // empty string as 'glue'
}
$freqRate = implode( ',', $tmp );
$dateTime = $_SESSION['dateTime'];
$taskTime = (time() - $_SESSION['startTime']) / 60; //in minutes
$sql = "INSERT INTO muq
(dateTime, taskTime, useTime, usedM, freqRate)
VALUES ('$dateTime', '$taskTime', '$useTime', '$usedM', 'freqRate')";
$link = connectDB();
#mysqli_query( $link, $sql );
$_SESSION['currID'] = #mysqli_insert_id($link);
#mysqli_close($link);
?>
Well, before answearing your question, here is some coding rules you need to respect:
1- You don't have to use more lines than what you need. This means you don't have to do an an empty "if" using 4 lines if you can do it in 2 lines.
Example:
Instead of:
if (filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)){
}else{
echo "Not a valid email address";
}
You can do:
if (!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL))
echo "Not a valid email address";
Second thing, to update a row in a database, you need an ID. This the key you are going to use to tell your db engine which row you are going to update because if not "he" will not know which row "he" should update (I'm considering the db engine as a person like me and you :D )
So, you need to inject that key (account ID or whatever) in your session so that you can use later when updating your database by telling you db engine that "he" needs to update that row identified by that key.
Related
I'm trying to make a system where an administrator can add multiple people at the same time into a database. I want this system to prevent the administrator from adding people with email addresses already existing in the database.
IF one of the emails in the _POST["emailaddress"] matches with one of the emailaddresses in the db, the user should get a message saying one of the emails already exists in the database. To achieve this, I've tried using the function array_intersect(). However, upon doing so I get a warning saying:
Warning: array_intersect(): Argument #2 is not an array in ... addingusers.php on line 41
At first i thought it had something to do with the fact my second argument was an associative array, so I tried the function array_intersect_assoc, which returns the same warning. How can I solve this?
The code on addingusers.php
<?php
session_start();
error_reporting(E_ALL);
ini_set('display_errors',1);
$conn = mysqli_connect('localhost','*','*','*');
$condition = false; // this is for the other part of my code which involves inserting the output into db
$name = $_POST["name"];
$affix = $_POST["affix"];
$surname = $_POST["surname"];
$emailaddress = $_POST["emailaddress"];
$password = $_POST["password"];
//amount of emailaddresses in db
$checkquery2 = mysqli_query($conn, "
SELECT COUNT(emailaddress)
FROM users
");
$result2 = mysqli_fetch_array($checkquery2);
// the previously mentioned amount is used here below
for($i=0; $i<$result2[0]; $i++){
// the actual emails in the db itself
$q1 = mysqli_query($conn, "
SELECT
emailaddress
FROM
users
");
// goes through all the emails
$result_array1 = array();
while ($row1 = mysqli_fetch_assoc($q1)) {
$result_array1[] = $row1;
}
$query1 = $result_array1[$i]["emailaddress"];
}
// HERE LIES THE ISSUE
for($i=0; $i<count($emailaddress); $i++){
if (count(array_intersect_assoc($emailaddress, $query1)) > 0) {
echo "One of the entered emails already exists in the database...";
echo '<br><button onclick="goBack()">Go Back</button>
<script>
function goBack() {
window.history.back();
}
</script><br>';
$condition = false;
}
else{
$condition = true;
}
}
EDIT
as the comments point out, $query1 is indeed not an array it is a string. However, the problem remains even IF i remove the index and "[emailaddress]", as in, the code always opts to the else-statement and never to if.
$query1 is not an array, it's just one email address. You should be pushing onto it in the loop, not overwriting it.
You also have more loops than you need. You don't need to perform SELECT emailaddress FROM users query in a loop, and you don't need to check the intersection in a loop. And since you don't need those loops, you don't need to get the count first.
<?php
session_start();
error_reporting(E_ALL);
ini_set('display_errors',1);
$conn = mysqli_connect('localhost','*','*','*');
$condition = false; // this is for the other part of my code which involves inserting the output into db
$name = $_POST["name"];
$affix = $_POST["affix"];
$surname = $_POST["surname"];
$emailaddress = $_POST["emailaddress"];
$password = $_POST["password"];
$q1 = mysqli_query($conn, "
SELECT
emailaddress
FROM
users
");
// goes through all the emails
$result_array1 = array();
while ($row1 = mysqli_fetch_assoc($q1)) {
$result_array1[] = $row1['emailaddress'];
}
$existing_addresses = array_intersect($emailaddress, $result_array1);
if (count($existing_addresses) > 0) {
echo "Some of the entered emails already exists in the database: <br>" . implode(', ', $existing_addresses);
echo '<br><button onclick="goBack()">Go Back</button>
<script>
function goBack() {
window.history.back();
}
</script><br>';
$condition = false;
}
else{
$condition = true;
}
It's working, but when I add the data in to my database, the data will be twice. I don't know if my syntax is wrong or my code is wrong.
Here's the structure:
//if submit is clicked
$checkin = $_POST['text_checkin'];
while ($row = mysqli_fetch_array($reservation)) {
if (isset($_POST['submitBtn'])) {
if ($row['reservefrom'] == $checkin) {
echo "Same Date";
return;
}
else
{
$lastname = $_POST['text_lastname'];
$firstname = $_POST['text_firstname'];
$address = $_POST['text_address'];
$tnumber = $_POST['text_tnumber'];
$cnumber = $_POST['text_cnumber'];
$email = $_POST['text_email'];
$checkin = $_POST['text_checkin'];
$checkout = $_POST['text_checkout'];
$room = $_POST['text_room'];
$tour = $_POST['text_tour'];
$guest = $_POST['text_guest'];
$query = "INSERT INTO reservation
(lastname, firstname, homeaddress,
telephonenumber, cellphonenumber, email,
reservefrom, reserveto, room, tour,
guestnumber)
values ('$lastname', '$firstname', '$address',
'$tnumber', '$cnumber', '$email', '$checkin',
'$checkout', '$room', '$tour', '$guest')";
mysqli_query($db, $query);
echo "Data Submitted!";
}
}
}
You're getting multiple inserts because you are looping for each record in $reservations. You should first look into why you are getting multiple records if you expected just a single record reservation.
That aside, alter your code by replacing your while loop with:
if(isset($_POST['submitBtn']) && $row = mysqli_fetch_array($reservation)){
if($row['reservefrom'] == $checkin) die("Same Date");
$lastname = $_POST['text_lastname'];
$firstname = $_POST['text_firstname'];
// ... other values, then execute your query
}else{
// either submitBtn was not posted or no result were found in $reservation
}
I noticed also that you use return in your code, but the code doesn't seem to be within a function so that's confusing. If it is within a function, it's probably a bad idea to echo from within unless the function is specifically meant to send data directly to the browser.
end web developer, i was given a CMS done from another team and i have to link with my front-end. I have made some modifications, but due to my lack of php knowledge i have some issue here.
My users are able to fill up a form, where 1 text field is asking for their photo link. I want to check for if the value entered is not equal to what i want, then i will query insert a default avatar photo link to mysql to process.
code that i tried on php
// check if the variable $photo is empty, if it is, insert the default image link
if($photo = ""){
$photo="images/avatarDefault.png";
}
doesn't seem to work
<?php
if($_SERVER["REQUEST_METHOD"] === "POST")
{
//Used to establish connection with the database
include 'dbAuthen.php';
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{
//Used to Validate User input
$valid = true;
//Getting Data from the POST
$username = sanitizeInput($_POST['username']);
$displayname = sanitizeInput($_POST['displayname']);
$password = sanitizeInput($_POST['password']);
//hash the password using Bcrypt - this is to prevent
//incompatibility from using PASSWORD_DEFAULT when the default PHP hashing algorithm is changed from bcrypt
$hashed_password = password_hash($password, PASSWORD_BCRYPT);
//Determining Type of the User
//if B - User is student
//if A - User is adin
if($_POST['type'] == 'true')
$type = 'B';
else
$type = 'A';
$email = sanitizeInput($_POST['email']);
$tutorGroup = sanitizeInput($_POST['tutorGroup']);
$courseID = sanitizeInput($_POST['courseID']);
$description = sanitizeInput($_POST['desc']);
$courseYear = date("Y");
$website = sanitizeInput($_POST['website']);
$skillSets = sanitizeInput($_POST['skillSets']);
$specialisation = sanitizeInput($_POST['specialisation']);
$photo = sanitizeInput($_POST['photo']);
// this is what i tried, checking if the value entered is empty, but doesn't work
if($photo = ""){
$photo="images/avatarDefault.png";
}
$resume = sanitizeInput($_POST['resume']);
//Validation for Username
$sql = "SELECT * FROM Users WHERE UserID= '$username'";
if (mysqli_num_rows(mysqli_query($con,$sql)) > 0){
echo 'User already exists! Please Change the Username!<br>';
$valid = false;
}
if($valid){
//Incomplete SQL Query
$sql = "INSERT INTO Users
VALUES ('$username','$displayname','$hashed_password','$type','$email', '$tutorGroup', ";
//Conditionally Concatenate Values
if(empty($courseID))
{
$sql = $sql . "NULL";
}
else
{
$sql = $sql . " '$courseID' ";
}
//Completed SQL Query
$sql = $sql . ", '$description', '$skillSets', '$specialisation', '$website', '$courseYear', '$photo', '$resume', DEFAULT)";
//retval from the SQL Query
if (!mysqli_query($con,$sql))
{
echo '*Error*: '. mysqli_error($con);
}
else
{
echo "*Success*: User Added!";
}
}
//if student create folder for them
if ($type == 'B')
{
//Store current reporting error
$oldErrorReporting = error_reporting();
//Remove E_WARNING from current error reporting level to prevent users from seeing code
error_reporting($oldErrorReporting ^ E_WARNING);
//Set current reporting error();
error_reporting($oldErrorReporting);
}
mysqli_close($con);
}
}
function sanitizeInput($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
i've tried finding a way on mysql to insert default values but it seem impossible, so i have no choice but to query insert through php.
I have the logic but i'm not sure how to implement on the php with my lack of knowledge, i was thinking of checking either
1) if the photo link does not have the word .png/.jpg, $photo != ".png"
2) if the photo link length is too low $.photo.length < 10
can someone help me look into the code and tell me what i'm doing wrong? Thanks!
A very simple way with default values could be:
$photo = isset($photo) ? $photo : 'images/avatarDefault.png' ;
How it works is that it first it asks if the photo is set, if it is, use all ready inserted value, otherwise insert your default value,
Another (very alike) method to use:
$photo = !empty($photo) ? $photo : 'images/avatarDefault.png' ;
UPDATE
To check if it contains a certain "extension" would be a simple rewrite
$photo = preg_match('#\b(.jpg|.png)\b#', $photo ) ? $photo : "images/avatarDefault.png" ;
This way it checks wether the text / image link in $photo contains the .png file type, if it doesn't it inserts your default image
First thing that I notice is to use double =
if($photo == ""){
//...
}
i have this code to save note from text area
this is my post-note.php file
<?php
include('connect.php');
if(isset($_POST['note_title'])){
$note_title = $_POST['note_title'];
$note_description = $_POST['note_description'];
$login_user_id = $_SESSION['user_id'];
$errors = array();
if($note_title == ""){
$errors['note_title'] = 'fine';
}else{
$errors['note_title'] = 'fine';
}
if($note_description == ""){
$errors['note_description'] = '<span class="note_description">Please enter something</span>';
}elseif(strlen($note_description) < "3"){
$errors['note_description'] = '<span class="note_description">your note is too short</span>';
}else{
$errors['note_description'] = 'fine';
}
if($errors['note_title'] && $errors['note_description'] == 'fine'){
$Query = "INSERT INTO notes (note_title, login_user_id, note_description, is_private)
VALUE('$note_title', '".$login_user_id."', '".$note_description."','0')";
if (!mysql_query($Query)){
die('Error: ' . mysql_error());
}
$errors['done'] = 'done';
unset($_POST['note_title']);
unset($_POST['note_description']);
}
}
echo json_encode($errors);die;
?>`
i want to insert first time as new row then want to update that row in database
In notes table, add a column id (if its not already there).
On create note page use above code.
On update note page pass id field of that note in $_POST.
So if isset($_POST['id']) write an UPDATE note ... WHERE id=$_POST['id'].
This will update already created note or will insert if its a new note.
filter user inputs before inserting though
I want to update my data in mysql.
But, if i want update (ex. firstname), photo_profile will lost.
<?php
include 'function_page_user.php';
if(($_FILES['photo_profile']) and ($_POST['firstname']) and ($_POST['lastname']) and ($_POST['password']))
{
session_start();
include 'connect.php';
$foldername="assets/img/user/";
$firstname = mysql_real_escape_string($_POST["firstname"]);
$lastname = mysql_real_escape_string($_POST["lastname"]);
$pwd = mysql_real_escape_string($_POST["password"]);
if((!empty($firstname) and !empty($lastname) and !empty($pwd)) and($_FILES['photo_profile']))
{
$image = $foldername . basename ($_FILES['photo_profile'] ['name']);
mysql_query ("update user set firstname = '".$firstname."' , lastname = '".$lastname."' , password = '".$pwd."' , photo_profile='".$image."' where id_user ='".$_SESSION['id']."'");
move_uploaded_file($_FILES['photo_profile']['tmp_name'], $image);
echo "<script>alert ('File Succes To edit');</script>";
$page="formubahuser.php";
echo redirectPage($page);
}
else echo "variabel empty";
}
else
echo ("your data is not complete<a href=formubahuser.php>Fill it again</a>");
?>
You have a number of major problems in your code. Before you continue, you need to read about the following topics:
1) The php mysql_ functions have been deprecated. That means the functions will be removed in future versions of php. You should use pdo or mysqli instead
2) When you store passwords in your database, they should always, always, always be encrypted.
Regarding your question, I think you are asking how to change the metadata (such as a firstname) without unsetting the photo url. Try something like this:
$updatequery = "UPDATE user SET firstname = '".$firstname."' , lastname = '".$lastname."' , password = '".$pwd."'";
if( $_FILES['photo_profile'])
{
$image = $foldername . basename ($_FILES['photo_profile'] ['name']);
$updatequery .= ", photo_profile='".$image."'";
}
$updatequery .= " where id_user ='".$_SESSION['id']."'";