So I'm working on an assignment for my class in which I am supposed to take a username and password and check it against a list contained in a table on a database I am connecting too.
Problem is when I am clicking the submit button nothing is happening I think this is likely to be some sort of error in syntax. Since I am new to PHP there is a good possibility it is something obvious, but not so much to me.
I have my database data stored in two PHP arrays (one for each field). I then converted the arrays to json which I will use in my JavaScript function that will be checked against the user inputted data.
I am including a form, a PHP script, and a JavaScript script in one document could this cause the issue?
Here is my code and thank you for any help!
<html>
<body>
<?php
/*config is included in order to protect my login info*/
require('config.php');
Echo "Project 4";
/*SQL connection*/
$conn = new mysqli(DB_HOST,DB_USER,DB_PASS,DB_NAME);
/*Checking Connection*/
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "SELECT * FROM p4Data";
$data2 = mysqli_query($conn, $sql);
/*Display Data*/
echo "<table border = 1 style='float:left'>
<tr>
<th>Username</th>
<th>Password</th>
</tr>";
//Array Declarations
$usernameArr = [];
$passwordArr = [];
while($records = mysqli_fetch_array($data2)){
array_push($usernameArr,$records["username"]);
array_push($passwordArr,$records["password"]);
}
echo "</table>";
//JSON Conversion
$usernameJson = json_encode($usernameArr);
$passwordJson = json_encode($passwordArr);
mysqli_close($conn);
?>
<!-- JAVA SECTION -->
<script type="text/javascript">
var obj = JSON.parse('<?= $usernameJson; ?>');
var obj2 = JSON.parse('<?= $passwordJson; ?>');
function verifUser(){
var usernameData = document.getElementById("username").value;
var passwordData = document.getElementById("password").value;
for (i = 0; i < 30; i++){
if(usernameData == obj[i]){
alert("Username verfied at " + i);
indexLocated = i;
break;
}
}
}
</script>
<form name='form-main'>
Username: <input type="text" id="username"><br>
Password: <input type="password" id="password"><br>
<input type="button" value="Login >>" id="submitButton"
onclick="verifUser()">
</form>
</body>
</html>
You can use post method to get the value of user input like this
<form method="post">
<input type="text" name="username">
<input type="password" name="password">
<input type="submit" name="submit" value="login">
</form>
and use this php code to get value when form is submitted
if(isset($_POST['submit'])){
$username_input = $_POST['username'];
$password_input = $_POST['password'];
}
Then make a query to sql where username = $username and password = $password. Like below
$sql query = " SELECT * FROM TABLE WHERE username = $username and password = $password";
And use
$num_rows = mysqli_num_rows($sql_query);
Now do a check of $num_rows = 1 that means input username and password is valid else echo Not valid
if($num_rows = 1){
**some code **
}else{
echo "Invalid information provided";
};
Related
I have never worked with $_COOKIES, and now I've been given the task to make it work.
I have been following a couple of tutorials online.
Found here: http://www.phpnerds.com/article/using-cookies-in-php/2
And then here:https://www.youtube.com/watch?v=Dsem42810H4
Neither of which worked for me.
Here is how my code ended up. I shortened it as much as I could.
Starting with the index.php page, which contains the initial login form:
<form role="form" action="index.php" method="post" id="loginForm" name="loginForm">
<input type="text" class="form-control" id="username" name="username"
value="<?php if(isset($_COOKIE['username'])) echo $_COOKIE['username']; ?>" />
<input type="password" class="form-control" id="password" name="password"
value="<?php if(isset($_COOKIE['password'])) echo $_COOKIE['password']; ?>"/>
<button type="button" id="loginSubmit" name="loginSubmit" class="btn btn-primary btn-block btn-flat">Sign In</button>
<input type="checkbox" id="rememberme"
<?php if(isset($_COOKIE['username'])){echo "checked='checked'";} ?> value="1" />
</form>
Here is the JavaScript used to send the form values:
$('#loginSubmit').on('click', function()
{
var username = $('#username').val();
var password = $('#password').val();
var rememberme = $('#rememberme').val();
// skipping the form validation
$.post('api/checkLogin.php', {username: username, password: password, rememberme:rememberme}, function(data)
{
// the data returned from the processing script
// determines which page the user is sent to
if(data == '0')
{
console.log('Username/Password does not match any records.');
}
if(data == 'reg-user")
{
window.location.href = "Home.php";
}
else
{
window.location.href = "adminHome.php";
}
});
});
Here is the processing script, called checkLogin.php. This is where I attempt to set the $_COOKIE:
<?php
include ("../include/sessions.php");
if(isset($_POST['username']) && isset($_POST['password']))
{
$username = strip_tags(mysqli_real_escape_string($dbc, trim($_POST['username'])));
$password = strip_tags(mysqli_real_escape_string($dbc, trim($_POST['password'])));
$rememberme = $_POST['rememberme'];
$select = "SELECT username, fullname, password FROM users WHERE username = '".$username."'";
$query = mysqli_query($dbc, $select);
$row = mysqli_fetch_array($query);
$dbusername = htmlentities(stripslashes($row['username']));
$dbfullname = htmlentities(stripslashes($row['fullname']));
$dbpassword = htmlentities(stripslashes($row['password']));
if(password_verify($password, $dbpassword))
{
// setting sessions here
$_SESSION['username'] = $username;
$_SESSION['fullname'] = $dbfullname;
// here is where I attempt to set the $_COOKIE
if(isset($remember))
{
setcookie('username', $_POST['username'], time()+60*60*24*365);
setcookie('password', $_POST['password'], time()+60*60*24*365);
}
else
{
setcookie('username', $_POST['username'], false);
setcookie('password', $_POST['password'], false);
}
echo $username; // this gets sent back to the JavaScript
mysqli_free_result($query);
}
else
{
// username/password does not match any records
$out = 0;
echo $out;
}
}
?>
So now that I have attempted to set the $_COOKIE, I can try to print it to the home page, like so:
<?php echo 'cookie ' . $_COOKIE["username"]; ?>
To which does not work, because all I see is the word 'cookie'.
Besides that, when I log out, I am hoping to see the login form already filled out, which is the overall task I have been trying to complete, but have been unsuccessful at doing so.
I the following code, I have a form that consists of three fields and two buttons. In the Review button, I would like to show any word in Arabic randomly and let the user show its translation in English by ticking the Show translation button.
<html>
<body>
<script>
function myFun1(var) {
document.getElementById("demo").innerHTML = "The translation in English is " + var;
}
</script>
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$english = $_POST["english"];
$arabic = $_POST["arabic"];
$example = $_POST["example"];
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<textarea name="english" rows="4" cols="70" placeholder="English">English</textarea>
<br>
<textarea name="arabic" rows="4" cols="70" placeholder="Arabic">Arabic</textarea>
<br>
<textarea name="example" rows="4" cols="70" placeholder="Example">Example</textarea>
<br><br>
<input type="submit" name="add" value="Add new">
<input type="submit" name="review" value="Review">
<br>
<p id="demo"></p>
</form>
<?php
$servername = "localhost";
$username = "xxx";
$password = "yyy";
$dbname = "vdb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['add'])) {
$sql = "INSERT INTO Vocabulary (English, Arabic, Example)
VALUES ('$english', '$arabic', '$example')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
elseif (isset($_POST['review'])) {
$sql = "SELECT COUNT(ID) as total FROM Vocabulary";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
#echo $row['total'];
$generated = rand(1,$row['total']);
$sql1 = "SELECT * FROM Vocabulary where ID = $generated";
$result1 = $conn->query($sql1);
$row1 = $result1->fetch_assoc();
echo "<br>";
echo $row1['Arabic'];
echo "<br><br>";
$eng = $row1['English'];
echo '<button onClick = "myFun('.$eng.')">Show translation</button>';
}
$conn->close();
?>
</body>
</html>
In the code, the following line creates the button and trigger the myFun1() function:
echo '<button onClick = "myFun('.$eng.')">Show translation</button>';
The problem is when the button is clicked, nothing happens (the message is not shown at all). Any ideas how to fix it?
Firstly change the argument var to some another argument name as var is a keyword in javascript
<script type="text/javascript">
function myFun(as) {
document.getElementById("demo").innerHTML = "The translation in English is " + as;
}
</script>
Secondly, you have to pass the string value in single or double quotes for that use inverted slash \ and rectify the function name from myFun() to myFun1()
echo '<button onClick = "myFun1(\''.$eng.'\')">Show translation</button>';
Rest your code is perfect.
You have definition of function myFun1(var), but you are calling myFun(). I guess this is the problem why there is nothing after clicking on button.
Add this to the top before
< html > tag
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$english = $_POST["english"];
$arabic = $_POST["arabic"];
$example = $_POST["example"];
}
?>
What I am trying to achieve is have the form appear if there was nothing submitted. So I add the form html to a variable and then echo the variable inside the html.
However ever since I implemented AJAX, inside the #results div. It also exports the jquery.js and my ajax script.
If I remove the PHP else code that displays the form, I won't see the form at all.
So how would someone make sure the js scripts aren't being inserted where they shouldn't be?
PHP:
if( isset($_POST["u_name"]) && isset($_POST["u_lastname"]) && isset($_POST["u_email"]) ){
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO testTable (Name, Lastname, Email)
VALUES ('".$_POST["u_name"]."','".$_POST["u_lastname"]."','".$_POST["u_email"]."')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error."";
}
$conn->close();
}else {
$final_content = '<form action="script.php" method="post" id="user_form">
<input type="text" name="u_name" placeholder="Name" id="user_name"> <br>
<input type="text" name="u_lastname" placeholder="Lastname" id="user_lastname"> <br>
<input type="email" name="u_email" placeholder="Email" id="user_email"> <br>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
HTML
<html>
<head>
<script type="text/javascript" src="jquery-1.11.2.min.js"></script>
<script>
$(document).ready(function(){
//Set form variable
var form = $("#user_form");
form.submit(function(event){
//Set data variables
var user_name = $("#user_name").val();
var user_lastname = $("#user_lastname").val();
var user_email = $("#user_email").val();
//Check if values are set
if( ($.trim(user_name) != "") && ($.trim(user_lastname) != "") && ($.trim(user_email) != "") ){
$.post("script.php", {u_name: user_name, u_lastname: user_lastname, u_email: user_email}, function(data){
$("#results").html(data);
});
}
event.preventDefault();
});
});
</script>
</head>
<body>
<?php echo $final_content ?>
<div id="results"></div>
</body>
</html>
Add an exit(); right after $conn->close();, just above the else.
This will make sure that once you have output the results, you do not continue with the rest of the code (not included in your question) where you output the full HTML page with script tags and <div id="results"> ...etc.
Apparently that display code is not all inside the else block, but also further down, after the else block.
I am trying to simply just pass (or POST) a variable indicated by the user in a textbox to another .php for processing.
This is my POST function in javascript.
function POST() {
var name = $('#name').val();
$.post('pages/searchResult.php',{name:name}, function (data) {
$('#name_feedback').html(data);
});
}
This is how the user indicates the variable through a textbox and submit.
<div id="name_feedback"></div>
<div style="text-align: center"><input type="text" placeholder="Enter first or last name" name="name" id="name" class="textbox"/></div>
<div style="text-align: center"><input type="Submit" value="Search" class="css_button" id="searchBtn" onclick="POST();"/></div>
I am not sure why but my .php wont receive the variable..
<?php
//include('connect.php');
$pointData = array();
$dateData = array();
$name = $_POST['name'];
$query = "SELECT * from `users` WHERE firstName = '$name' OR lastName = '$name' LIMIT 1";
$result = mysql_query($query); //<<<
$row = mysql_fetch_assoc($result);
$userid = $row['id'];
echo mysql_error();
?>
I am using this video as a reference:
http://www.youtube.com/watch?v=-EHbBHm5l4g
try this way
var name = $('#name').val();
$.post('pages/searchResult.php',{"name":name}, function (data) {
$('#name_feedback').html(data);
}) .fail(function(err) {
alert( "error" +err);
});
At your php script side, you can echo the data which you want to return back based on success of the query search.
remove the single around the table name users and around $name from the below
"SELECT * fromusersWHERE firstName = '$name' OR lastName = '$name' LIMIT 1";
Happy Coding :)
the header(location: agentverification.php) does not work when uploaded to godaddy server while it works on localhost server. I've try to edit my codes but the result is still the same. if any of you could help me, I'd appreciate it and thanks in advance.
agentlogin.php
<form method = "post" action = "agentverification.php" >
<table>
<input type="text" name="ID" size=20 ><br></td></tr>
<tr>
<td>Password <font color=red>*</font></td>
<td><input type="password" name="pass" size=20><br></td></tr>
</table>
<input type = "hidden" name = "login">
<input type = "submit" name = "login" value = "submit" id="pop">
agentverification.php
<?php
session_start();
$link = mysqli_connect('localhost', 'root', '','db5') or die(mysqli_error());
if(isset($_POST['login']))
{
extract($_REQUEST);
$id = $_POST['ID'];
$pass = $_POST['pass'];
$query= "SELECT * FROM agentReg WHERE AgentID = '$id'";
$record = mysqli_query ($link,$query);
$check=FALSE;
while($row=mysqli_fetch_array($record))
{
if($id === $row['AgentID']&& $pass === $row['password'] )
{
$check=TRUE;
}
}
if($check == TRUE)
{
$_SESSION['AgentID'] = $id;
$_SESSION['password'] = $pass;
header("Location: agentpage.php");
}
else
{ ?> <script>
alert ("Wrong combination of ID and Password. Please try again.");
</script> <?php
session_destroy();
header("location: agentlogin.php");
}
}?>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script>
agentReg database structure
CREATE TABLE agentReg
(AgentID varchar (100) primary key,
password varchar (100));
You cannot use header('Location: ...'); after you've printed an output.
In your last else you should only include session_destroy(); and for instance header('Location: agentlogin.php?error=idpassword');. The error message should be in agentlogin.php.
Use
window.location.href = 'http://www.google.com'; //Will take you to Google.