First and foremost, this is my 1st time writing PHP code, so please forgive my newbie'isms.
I have a login form with 3 submit buttons (post method) named login, register and forgot. If I select login button, the login function in the PHP code gets called, however, the same is not true for the register and forgot buttons. Its almost like the only submit button that is working is the login buttons. My best guess at this point is there is id/name that not correct. For the sake of brevity I've removed the CSS portion. Any points in the right direction will be most appreciated.
<?php
function login(){
//do stuff
echo "Login";
}
function register(){
// do stuff
echo "Register";
}
function forgot(){
//do stuff
echo "Forgot";
}
if($_SERVER["REQUEST_METHOD"] == "POST") {
if(isset($_POST['login'])) {
login();
}
if(isset($_POST['register'])) {
register();
}
if(isset($_POST['forgot'])) {
forgot();
}
}
?>
<html>
<head>
<meta charset="utf-8">
<title>Login Page</title>
</head>
<body>
<div id="login">
<h1><strong>Welcome!</strong> Please login.</h1>
<form action="" method="post">
<fieldset>
<p><input type="text" name="username" required value="Username" onBlur="if(this.value=='')this.value='Username'" onFocus="if(this.value=='Username')this.value=''"></p>
<p><input type="password" name="password"></p>
<p><input type="submit" name= "login" value="Login"></p>
<p><input type="submit" name= "register" value="Register"></p>
<p><input type="submit" name= "forgot" value="Forgot Password"></p>
</fieldset>
</form>
</div> <!-- end login -->
</body>
</html>
HTML forms can not contain more than one input of type "submit" (which means others would not be functioning)
register and forgot are links rather than buttons.
Instead you could use:
Register
Another alternative is to use javascript to make these buttons act as links.
After copying and pasting my code into another file. It worked without any issues. The multiple submit buttons work without any issues.
Related
I have some numbered pages:
1.php
2.php
3.php
etc.
I want to create a textbox that the user enter any number: 2 for example, and hit enter or Go button, and they will go to the page 2.php depending on the number entered.
I know how to link to a specific page as in form action="....", but I am not sure how to echo the user input and translate it as link (whether using html or php).
Ex:
<form method="POST">
<input type="number" value="" />
<input type="submit" value="Go" />
</form>
You need to add an action attribute to your form and a name attribute to your number input. The file from your action attribute will "catch" the POST variables and do the logic needed to redirect your user. Change your form tag to:
<form method="POST" action="redirect.php">
<input type="number" value="" name="redirect" />
<input type="submit" value="Go" />
</form>
Then create the redirect.php file that gets the POST variables and does the redirection:
<?php
$redirectPage = (int) $_POST['redirect'];
$redirectUrl = "http://www.example.com/{$redirectPage}.php";
header("Location: $redirectUrl");
printf('moved.', $redirectUrl);
Beware that there's no input validation nor error handling included.
I think, the best available option in your case would be the one using client-side javascript to dynamically change the form's action attribute base on the number entered in the input box.
A quick and dirty solution to fulfil such a task might look like this
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">
function submitAction(formElement) {
var // get the input box element
el = document.getElementById('form-action-number'),
// get a number specified by user in the input box
num = parseInt(el.value),
// validate that it's really a number and is greater than zero
// (you don't want someone to request -666.php right? :)
// build a page url using the correct number
page = !isNaN(num) && num > 0 ? num.toFixed(0).toString() + '.php' : undefined;
if (page) { // the page url is valid
// set form's action attribute to an url specified by page variable
formElement.setAttribute('action', page);
// returning true will allow the form to be submitted
return true;
}
// you might think of a better way to notify user that the input number is invalid :)
console.error('INVALID NUMBER SPECIFIED!');
// returning false will prevent form submission
return false;
}
</script>
</head>
<body>
<!-- When user clicks Go, the return value of submitAction function will be used to decide if the form should be submitted or not -->
<form method="POST" onsubmit="return submitAction(this)">
<input id="form-action-number" type="number" value="" />
<input type="submit" value="Go" />
</form>
</body>
</html>
With PHP you can do something like this:
<?php
// Check if the POST value has been set
if(isset($_POST['my_number'])) {
// Redirect to the corresponding page
header('Location: ' . $_POST['my_number'] . '.php');
}
?>
<form method="POST">
<input name="my_number" type="number" value="" />
<input type="submit" value="Go" />
</form>
This is like DaMeGeX's answer but uses javascript to go to the new page.
<?php
// Check if the POST value has been set
if(isset($_POST['my_number'])) {
// Redirect to the corresponding page
echo "<script> window.location.href = '".$_POST['number'].".php' </script>";
}
?>
<form method="POST">
<input name="my_number" type="number" value="" />
<input type="submit" value="Go" />
</form>
I'm working on a database-driven quiz that lets users select a series of answers, then submit the results via a form. It was working great, when it suddenly blew up, and I haven't been able to find the problem.
So before I get into the more complex stuff, I'd like to go back to square one and just make something simple work - like passing a hidden value to another page that echoes that value.
Here's the code for my first page # mysite/form.php:
<html>
<head>
</head>
<body>
<!-- g1/form.php -->
<div id="quiz" rel="key">
<form action="form2.php" method="post" id="quiz">
<input type="hidden" name="PreviousURL" id="url" />
<input type="submit" value="Submit Quiz" />
</form>
</div><!-- quiz-container -->
</body>
</html>
And here's the code for the second page:
<html>
<head>
</head>
<body>
<?php ini_set('display_errors', 1);
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
echo $_POST['PreviousURL'];
}
echo 'XXX';
?>
</body>
</html>
I also tried moving the closing bracket, like this:
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
}
echo $_POST['PreviousURL'];
echo 'XXX';
In both cases, when I click the submit button and am forwarded to form2.php, I see "XXX," but there's no value for $_POST['PreviousURL'].
I must have accidentally deleted or modified something, because it seems so simple, and it worked fine before. Can anyone tell me what the problem is?
there isn't a value for the hidden input.
In your form script you have missed out the value="" from the hidden input. This is the reason why nothing is displaying on the second page.
I have been trying to create a form that reads a post from an HTML form and displays an element from that post IF it detects that the post exists.
However, each time the post is submitted, it simply reloads the form as though no post were provided.
<!DOCTYPE html>
<html>
<head>
<title>Upload from Manifest</title>
</head>
<body>
<?php
if (isset($_POST['manifest'])) {
echo 'we are in the IF';
echo($_POST['manifest']);
}
?>
<h1>Submission from manifest into main db</h1>
<div class="container offset-top120">
<form method="post" action="https://nhsggc.cogiva.com/prism/loadFromManifest.php" enctype="multipart/form-data">
<input id="manifest" type="text" />
<input id="submit" value="Submit" type = "submit" />
</form>
</div>
</body>
</html>
Your form is going to either a different page (https://nhsggc.cogiva.com/prism/loadFromManifest.php so check for that first) if you wanted it to go to same page, you can give the action as just '#', or put in the whole URL like you have.
You're missing the name attribute from your submit input and text input. Read up on the name attribute!
<input id="manifest" type="text" name="manifest">
<input id="submit" value="Submit" type="submit" name='submit' />
Then your PHP should look like this:
<?php
if (isset($_POST['submit'])) {
echo 'Inside an if';
echo $_POST['manifest'];
}
Then it should work.
I have a simple php - password protecton / login based on a form which works. PHP-code is in page "secure.php" and includes the html-file "accessed.html" when user + pass is correct.
But i want the form to hide when hidden page (accessed.html) is shown.
I have tried wrapping the form in a div and using javascript and display: none to hide, but it doesnt work - not locally or on server.
What am i doing wrong? It doesnt have to be js hiding the form after login..
PHP in top
<?php
$user = $_POST['user'];
$pass = $_POST['pass'];
if($user == "a"
&& $pass == "a")
{
include("accessed.html");
echo "<script>
document.getElementById('wrap').style.display = 'none';
</script>";
}
if(isset($_POST))
?>
And the form in the body:
<div id="wrap">
<form method="POST" action="secure.php">
User <input type="text" name="user"></input>
Pass <input type="text" name="pass"></input>
<input type="submit" name="submit" value="access page"></input>
</form>
</div>
Move the <script> to after wherever you display "#wrap"
If you want to do it on the server side (PHP) just use an if statement.
I think this is what you are looking for my friend
<html>
<body>
<div id="wrap">
<form method="POST" action="temp.php" target = "hidden">
<input type="text" name="user">User</input>
<input type="text" name="pass">Pass</input>
<input type="submit" name="submit" value="access page" onclick = "closediv()"></input>
</form>
</div>
<iframe name = "hidden" id = "hidden" style = "display: none"></iframe>
<script>
function closediv(){
document.getElementById("wrap").style.display = "none";
}
</script>
</body>
</html>
With the above code, you'll need to do your PHP processing on the current page. If this isn't doing what you want it to do and I misread your question, let me know and we can work on your predicament.
maybe very easy!
I'm php coder and I don't have experience in js but I must do this for one of my codes
suppose I have sub1 in page after clicking it must be that sub1 but value now is sub2
<html>
<head>
<title>pharmacy</title>
</head>
<body>
<form method="post" action="pharmacy.php">
<?php
//some code
if(array_key_exists('update',$_POST)){
//somecode
}
?>
<input type="submit" name="update" value="<?php echo if(isset($_GET['update'])) ? 'Show' : 'Update' ?> ">
</form>
</body>
</html>
show as function name does not really make sense here (imo), but you could do:
<input type="submit" name="sub" value="sub1" onclick="show(this)">
and
function show(element) {
element.value = 'sub2';
}
Important:
But that will actually not solve your problem. As soon as you click the button, the form is submitted, meaning the browser initiates a new request and will load a new page. So every change you made the current page is lost anyway.
The question is: What are you trying to do?
It seems to me that you should change the value of the button on the server side. You have to keep track which form was submitted (or how often, I don't know what you are trying to do) and set the value of the button accordingly.
Update:
I see several possibilities to solve this:
You could keep using JavaScript and send and get the data via Ajax. As you have no experience with JavaScript, I would say you have to learn more about JavaScript and Ajax first before you can use it.
You could add a GET parameter in your URL with which you can know which label to show for the button. Example:
<form method="post" action="?update=1">
and
<input type="submit" name="sub" value="<?php echo isset($_GET['update']) ? 'Show' : 'Update' ?> ">
Similar to 2, but use a session variable (and not a GET parameter) to keep track of the state.
Update2:
As you are already having $_POST['update'] you don't need the URL parameter. It could just be:
<html>
<head>
<title>pharmacy</title>
</head>
<body>
<form method="post" action="pharmacy.php">
<input type="submit" name="update" value="<?php echo isset($_POST['update']) ? 'Update' : 'Show'; ?> ">
</form>
</body>
</html>
This should do it
function show(){
document.getElementsByName('sub')[0].value = 'sub2';
return false;
}
Edit: if you don't want it to submit the form, just add a return false, but then you'd need to change your onclick from your submit button to your forms onsubmit;
<html>
<head>
<title>test</title>
<script>
function show()
{
document.getElementById("sub").value= "sub2";
return true;
}
</script>
</head>
<body>
<form method="post">
<input type='submit' id="sub" name='sub' value="sub1" onclick="return show()">
</form>
</body>
</html>