I've form tag like this
sample name:register.php page
<form id="formElem" name="formElem" action="form10.php" method="post">
<input id="pd" name="pd" type="text" AUTOCOMPLETE=OFF />
<input id="pd1" name="fname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="mname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="lname" type="text" AUTOCOMPLETE=OFF />
6 more input boxes
<button name="submit" type="submit">Register</button>
<button name="preview" type="submit">Preview</button>
</form>
I'm sending this info to next form10.php page and displaying all the 10 input values on that page
I'm using $pd= htmlentities($_POST['pd']); $fname= htmlentities($_POST['fname']); to fetch values from form tag and such 10 variables and I'm echoing those entered value
on form10.php file after successful submit button.
like i entered fname, mname, lname came from form tag and displayed on form10.php page.
first name <?echo $fname?>
but now problem is user can see the next page (form10.php) after entering only 10 textboxes values inside form tag.
but I want to give preview option to user so that user can preview that next page either filling any of 1 to 10 textbox values. means he has filled fname and lname but not rest of 8 fields and he clicks on preview button I want to open form10_preview.php which same as form10.php but as user has entered only fname and lname so echo only those values which he as supplied.
Now problem is how can i can have two submit button and two actions in one form?
I think it is better to control form submit rules clientside. Remove the action from your form, and change the button type to be button :
<form id="formElem" name="formElem" action="" method="post">
<input id="pd" name="pd" type="text" AUTOCOMPLETE=OFF />
<input id="pd1" name="fname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="mname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="lname" type="text" AUTOCOMPLETE=OFF />
6 more input boxes
<button id="register" type="button">Register</button>
<button id="preview" type="button">Preview</button>
</form>
Then let javascript control the flow of the submitting :
var formElem = document.getElementById('formElem'),
btnSubmit = document.getElementById('register'),
btnPreview = document.getElementById('preview');
function formSubmit() {
switch (this.id) {
case 'register' :
formElem.action='post10.php';
break;
case 'preview' :
formElem.action='preview10.php';
break;
}
formElem.submit();
}
btnSubmit.onclick = formSubmit;
btnPreview.onclick = formSubmit;
You could have the form point to its own page and handle each submit value separately. At the top of the file with the form, you'll need to start the output buffer and a session. This allows the use of header() to redirect, and storage of session variables.
<?php
ob_start();
session_start();
?>
The form will point to itself by removing the action attribute:
<form id="formElem" name="formElem" method="post">
<input id="pd" name="pd" type="text" AUTOCOMPLETE=OFF />
<input id="pd1" name="fname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="mname" type="text" AUTOCOMPLETE=OFF />
<input id="pd2" name="lname" type="text" AUTOCOMPLETE=OFF />
6 more input boxes
<button name="submit" type="submit">Register</button>
<button name="preview" type="submit">Preview</button>
</form>
We process each of the buttons via their name in the POST array:
<?php
if(isset($_POST['submit'])){
foreach ($_POST as $key => $value) {
$_SESSION[$key] = $value;
}
header("Location: form10.php");
}
if(isset($_POST['preview'])){
foreach ($_POST as $key => $value) {
$_SESSION[$key] = $value;
}
header("Location: form10_preview.php");
}
?>
And at the very end of the file, we flush the output buffer:
<?php ob_end_flush(); ?>
So, essentially the form has one action, which is to submit the values to itself. Finally, both form10.php and form10_preview.php will need session_start(); at the top of the file to access the Session variables we've created, like so:
<?php
session_start();
$inputs = array("pd", "fname", "mname", "lname", etc...);
foreach ($inputs as $input) {
echo $_SESSION[$input];
}
?>
Related
I have made a form in my site, which will allow me to get suggestions about Rubik's cube algorithms, but how to know what input the user has? For better understanding, I've given the code below:
<form method="POST">
<label>Your name: </label><br>
<input type="text" name="name" placeholder="Your Name" required><br><br>
<label>Your E-mail: </label><br>
<input type="email" name="email" placeholder="email#domain.com" required><br><br>
<label>Select puzzle: </label><br>
<input type="radio" name="2x2" value="2x2">2x2<br>
<input type="radio" name="3x3" value="3x3">3x3<br><br>
<label>Select set/subset: </label><br>
<input list="set"><br><br>
<datalist id="set">
<option>Ortega OLL</option>
<option>Ortega PBLL</option>
<option>CLL</option>
<option>EG-1</option>
<option>EG-2</option>
<option>F2L</option>
<option>OLL</option>
<option>PLL</option>
<option>COLL</option>
<option>WV</option>
</datalist>
<label>Your Alg: </label><br>
<input type="text" name="alg"><br><br>
<input type="submit" name="submit" class="w3-black w3-button w3-hover-white w3-hover-text-blue w3-text-white">
</form>
Please add action attribute to your form tag and on submit here is the example
<form action="getvalue.php" method="post">
</form>
Note: Every form element must have unique name .
after adding action attribute then create getvalue.php file and add following code in to it
<?php
print_r($_POST);
?>
Above code will give all the form field values
do let me know if it was helpfull...
I'm not sure exactly what you want to do, but here is an example of a form that submits to itself. This will allow you to remain on the same page after the form has been submitted. You can change what the user sees to indicate that the form was done successfully/etc. I have tested this code and it works.
<main>
<?php
// When the form is submitted
if (isset($_POST["submitButton"])){
//Do something with the data from the form - you don't have to print it out.
//This is all done on the server.
//Connect to DATABASE, send an EMAIL, VALIDATE the form, etc.
print("<pre>");
print_r($_POST); // for all GET variables
print("</pre>")
?>
<!-- This HTML will be visible to the user after the form has been submitted. -->
<h1>The form has been submitted successfully!</h1>
<?php
}else{ //If the form has not been submitted
?>
<form method = "post" >
First name:<br>
<input type="text" name="firstname" value="Mickey">
<br>
Last name:<br>
<input type="text" name="lastname" value="Mouse">
<br><br>
<input type="submit" id = "submitButton" name = "submitButton" value="Submit">
</form>
<?php
} //End else
?>
</main>
As the user fills out the form, I have some buttons that dynamically add new questions. However, I can't get the value of these buttons to Post to the subsequent PHP page (all the other information posts fine).
In the example below, I'm not able to get the value of "add_email" in "process-form.php" via $_POST['add_email'];
Thanks in advance for your help. In reduced form, it looks like this:
HTML form
<form id="form" class="form" action="process-form.php" method="POST">
<input id="name" name="name" type="text" placeholder="Name">
<!-- Yes/No buttons asking if the user wants to enter their email-->
<div id="form_email">
<h4>Do you want to enter your email?</h4>
<input type="button" class="btn" id="add_email" name="add_email" value="Yes" onclick="showEmailQ(this.value)"></input>
<input type="button" class="btn" id="add_email" name="add_email" value="No" onclick="showEmailQ(this.value)"></input>
</div>
<input type="submit" class="btn" value="Submit »">
</form>
Javascript
<script>
//Function to process whether user wants to enter email and then display value
function showEmailQ(value){
var table_row = document.getElementById("form_email");
if(value == "Yes"){
table_row.innerHTML = '<input id="email" name="email" type="text" placeholder="Email">';
}
else{
table_row.innerHTML = '<p>You have chosen not to enter your email</p>';
}
}
</script>
PHP
//process-form.php
session_start();
$enteredEmail = $_POST['add_email'];
echo $enteredEmail; // Nothing prints to screen
You're destroying your form by overwriting the inputs with
table_row.innerHTML = '<input id="email" name="email" type="text" placeholder="Email">';
So maybe change the input name to match
table_row.innerHTML = '<input id="email" name="add_email" type="text" placeholder="Email">';
This is my code that I am using to submit form with post value
<form action="<?php echo DOMAIN; ?>contact/booking-form.php" method="post">
<input type="text" name="name" value="<?php echo $name; ?>" />
<input type="text" name="email" value="<?php echo $email; ?>" />
<input type="submit" name="submit" id="submit" />
<script>document.getElementById('submit').submit();</script>
</form>
Can anybody help me to pass name="submit" value of submit button to another page?
A submit button is only going to be a successful control if it is used to submit the form (and even then only if it has a name and a value … which yours does not).
If you want submit=submit in your form data when you submit the form with JavaScript, then don't use a submit button to put that data in the form in the first place. Use a hidden input.
<input type="submit">
<input type="hidden" name="submit" id="submit" value="submit">
Then you have two other problems.
First, submit is a method of form elements, not inputs. So you need to change your script to call the right element.
<script>document.getElementById('submit').form.submit();</script>
Second, if a form has a control called submit then that will clobber the submit method. So you need to get one from a different form (not supported in old versions of Internet Explorer):
<script>
var form = document.getElementById('submit').form;
var submit_method = document.createElement("form").submit;
submit_method.call(form);
</script>
<form id=submit action="<?php echo DOMAIN; ?>contact/booking-form.php" name="form1" method="post">
<input type="text" name="name" value="<?php echo $name; ?>" />
<input type="text" name="email" value="<?php echo $email; ?>" />
</form>
<script >
document.form1.submit()
</script>
there is no sumit button in your code. first add in html
<input type="submit" value="submit" id="submit"/>
<script>
document.getElementById("submit").value = "newSubmitButtonValue";
</script>
I have a form with ,
In the form after user clicks on Submit , he will be taken to another page which would show all the data entered, Which would allow the user to verify the content.(User will have a look at the content , to see what he entered is correct).
Once user verifies he would submit the data and Insert to DB should be done.
I want to know a method in which i could carry on the approach, to do this.
How can i implement this
EDIT MORE EXPLAIN
addemp.php
The Main Div With Form
<div class="panel-body">
<form>
Employee : <input Type="text" id="name">
<input type="submit" value="check">
</div>
The Second Div in the same form should show once submit is clicked
<div class="submit panel-body">
<form>
Employee : <Employee Name asin main div>
<input type="submit" > <--! this submit would post data
</form>
</div>
how to pass the value from 1st div to the second , and from the second INSERT to db.how can i do without page refresh ?
Use following script on location file
$action='';
if(isset($_POST['submit'])){
//create form here
//Change the action of form
$action = 'save.php';
}
echo '<form method="POST" action="'.$action.'">
<input type="text" name="nric" value="'.isset($_POST['nric'])?$_POST['nric'].'" />
<input type="text" name="empName" value="'.isset($_POST['empName'])?$_POST['empName'].'" />
<input type="text" name="location" value="'.isset($_POST['location'])?$_POST['location'].'" />
<input type="submit" value="Submit"/>
</form>';
EDIT: You don't need to use a form in this case. You can simply use JQuery to show the data from text boxes in a DIV and a button that will POST the data for you on the server.
<input type="text" name="nric" id="nric" />
<input type="text" name="empName" id="empName" />
<input type="text" name="location" id="location" />
<input id="sndData" type="button" value="Submit" />
<div id="showData"></div>
JQuery:
$('#sndData').click(function(){
var makeData = "<p>NRIC: "+$('#nric').val()+"</p><p>Employee Name: "+$('#empName').val()+"</p><p>Location: "+$('#location').val()+"</p>";
$('#showData').html(makeData);
});
When you've done that, just create/show a HTML button that will POST the data for you.
If this answers your question, please mark it as an answer.
I have been looking through StackOverflow a lot and found SOME solutions but they don't work for me :( Probably because I can't place the strings of code correctly.
I have a form which submits content to a database - and a script that loads the database inputs into a div below the form. It works realtime with a small delay.
<div id="addCommentContainer">
<form id="addCommentForm" method="post" action="">
<div>
<input type="hidden" name="name" value="<?php echo $loggedInUser->display_username; ?>" id="name" />
<input type="hidden" name="email" value="<?php echo $loggedInUser->email; ?>" id="email" />
<textarea name="body" id="body" cols="20" rows="5"></textarea>
<input type="submit" id="submit" value="Submit" />
</div>
</form>
It works. I also have this script:
http://sandbox.jinoh.dk/script.js
I tried a solution onclick - but it made the content disappear not being posted. I tried onsubmit - but it didn't work. What I want is the "comment" to be posted and then disappear from the form aka the form to be cleared.
Chatchatchat-div is the one that the comments are loaded into.
$('#addCommentForm').submit(function(e){
var form = this;
.....
....
$.post('submit.php',$(this).serialize(),function(msg){
if(msg.status){
form.reset(); // if you want to reset on msg status
}
....
});
this.reset(); // just reset the form what happen
});