How to use two forms and submit once? - php

Is it possible to have two forms with two submit buttons such that when I click on the button it saves the input fields in both forms?
I'm concerning to solve this in PHP / MySQL.
I tried my own way:
if ((isset($_POST["form-1"])) && (isset($_POST["form-2"])) {
//SQL Insertion
}

Nope, you can only submit one form at a time.
If you have to use two forms, the only way to do this would be to clone the second form's fields into the first one using jQuery. Won't work when JS is turned off, though.
See Copying from form to form in jQuery
Why do you need two forms?

If you have a problem like this, the design is flawed.
You can submit only one form at a time for a reason.
Change the design to use only one form; doing workarounds to submit two anyway is a horrible practice that would be better to avoid.

One way of achieving similar result would be to club the two forms into a single one and have 2 submit buttons with different values and same name="submit" field.
toFoo.html :
<form action="doFoo.php">
User <input type="text" name="username" />
Pass <input type="password name="password" />
<!-- Submit one -->
<input type="submit" name="submit" value="Create user" />
<!-- some more of your fields or whatever -->
<input type="text" name="blah" value="bleh" />
<!-- Submit two -->
<input type="submit" name="submit" value="Login user" />
</form>
doFoo.php :
<?php
if( $_POST["submit"] == "Login user" ) {
//do login foo
}
if( $_POST["submit"] == "Create user" ) {
//do signup foo
}
?>

You could submit both forms at the same time via Ajax but your php script would only receive one form at a time. Better to just convert your 2 forms into one big form if you need all inputs going to one script

A submit button submits only fields of the form it lives in. If you need content of both forms, you'll have to copy the fields from other form to some hidden field in the form where submit button was clicked. This can quite easily be done in JavaScript.

As far as i know only one form can be submitted at a time. You could try wrapping them in one form.

Related

php get data from another form using a different form

Is it possible to get data from another using a different form?
I don't want to use one form
<?php
echo $_POST['2'];
?>
<html>
<head>
</head>
<body>
<form method="post">
<input type="submit" name="submit" />
</form>
<form method="post">
<input type="text" name="2" />
</form>
</body>
</html>
No, that's not possible because browsers will only ever submit one form at a time (the one containing the clicked submit button, typically).
They can't possibly submit multiple forms at once because each form has its own action and method attribute which determines the request to send.
As #peter said, you can submit only one form at a time. But there are some workarounds for your needs.
Method 1
Post your form to a php script(say form_1_action.php) and then store the form input in a Session variable.
$_SESSION['form_data_1'] = $_POST;
Then you will be able to access it in different pages. Like,
$_SESSION['form_data_1']['field_name']
Method 2
Post your form to a php script(say form_1_action.php) and then store the form input in a PHP variable.
$formData1 = $_POST;
Then you can use the data from the first form in the second form (the second form should be on the same file form_1_action.php) like
<input name="name" value="{$formData1['field_name']"}>
You should pass the data from the first form in a hidden field on the second form if you need it on the form_2_action.php.
Method 3
Use Javascript to accomplish your requirements in a more userfriendly way.
try using jquery to Prevent the other form from submiting and try updating the value using event listening of the first form and update that input.
$( '#Submit' ).click( function ( event ) {
event.preventDefault();
var value = <?= $postedValue ?>;
$('input[name="input_name/2"]').val(value);
}

load form using AJAX will loose form and values after a failed submission

I have a PHP page that loads several parts of a form using AJAX. For instance, first check if the user is already registered, if so the script loads (with AJAX) the rest of the form. The form will not be submited using AJAX what can be a problem when the user submits the form (without AJAX) - imagine there are some errors - the form will loose all values.
I'm wondering if CSS hiding part of the form and after the successful login use JS to display the rest of the form, would be better.
Here some code:
<form action="some_action.php">
Email: <input type="text" name="email" id="email"> <br />
Password: <input type="password" name="password" id="password"> <br />
<button id="vrf_login">Verificar</button>
<div id="rest_form">
</div>
</form>
AJAX:
- CHECK login: if email and password matches then
- LOAD the form for div with id "rest_form"
(it is in another file, for instance:
<input type="text" name="place" id="place">
<input type="text" name="age" id="age">
<input type="submit" name="submit" value="submit">
)
The problem is if I submit the form (without AJAX) and there are errors I will loose the form loaded with AJAX
EDIT (again)
Thank you all for your constructive suggestions:
The solution I adopted is close to the first Alkis's suggestion:
almost all the form is hidden (CSS)
after some logic choices the (part of the) form is turned visible (jQuery) - to "remember" what parts should be visible in case of submission failed (server side validation) some session variables hold the information (AJAX) - and then, after the submission (failed) use jQuery to restore the prior form structure (get the session variables with JS this way: var xpto = "<?php echo $_SESSION['prior_xpto']; ?>" ; )
the fields of the form will remember theirs values (with PHP)
You have 3 options.
Stop loading the whole form by ajax. Hide it with css and show it if the the conditions are met. If the page is shown after some validation error, just show it (change the css inline or give it a different class)
Have a condition and every time the page loads check if it is a first load or if the page is shown after some validation error occured. If the latter is true then load again the form with ajax. This condition can be a hidden field that takes its value from the server and you check it on the client every time you serve the page.
The second solution can be done on the server too. Have the condition be checked on the server. If it's a first load, then don't populate the form and let it be populated from ajax as you do now. If it's after a validation error then pre-populate the form. It's just an if/else clause.
Please provide some codes for your question, but i guess your problem is sending result using a button with "submit" type !
if you have a form like this:
<form>
<inputs ...>
<input type="submit" value="Send data" onclick="SendDataUsingAjax()" >
</form>
after clicking on submit all values on input will reset regardless of what your ajax function is doing. to fix this problem you only need to change type="submit" to type="button".

One form two submit buttons

I have two similar forms on a site that I'd like to merge into one form with two submit buttons. They use most of the same data.
The first form uses GET to send the data to another server. The second form sends it in an email. I'd like to strongly encourage site users to use option one before trying option two.
I know how to do this with javascript, but not in a way that degrades well. Any other ways to have two submit options? Or other ideas for how to accomplish this? Thanks!
Snow Blind provided the good solution, but you can't determine which button was clicked.
Buttons must have different names, not the same.
Example:
<input type="submit" name="server" value="Server" />
<input type="submit" name="email" value="Email" />
<?php
if(isset($_GET['server']))
{
// Send to another server
}
else if(isset($_GET['email']))
{
// Send to email
}
else die("None of buttons was clicked.");
?>
Additionally, if you have a same code in both parts (server and email), you can do the following:
if(isset($_GET['server']) || isset($_GET['email']))
{
// Do something common to both methods
if(isset($_GET['server']))
{
// Send to server
}
else
{
// Send to email
}
}
Better solution, in my opinion, is to put only 1 submit button + a dropdown menu with method to choose.
<select name="sendMethod">
<option value="" disabled>Choose sending method...</option>
<option value="server">Send to another server</option>
<option value="email">Send to e-mail</option>
</select>
Like everybody said use either checkbox/radio button.
And set the one that use GET as the default option
If you don't want to use javascript you can always get the value and of the checkbox/radio and check user choice
You can create two submit buttons and give them the same name but different value.
<input type="submit" name="submit" value="Server">
<input type="submit" name="submit" value="Email">
Then you can check $_GET['submit'] or $_POST['submit'] (depending on form method) to see which submit was used.

How to check if a form is submitted via javascript?

I have this conventional submit button which submit a form like this:
<form method="post" id="form_submit">
...
<input class="button" type="submit" name="Submit" value="Submit">
</form>
And I check if the submit button is clicked using this:
if(isset($_POST['Submit'])){
//update DB
}
Now I have a submit link using jquery:
Submit
JS code:
$("#form_submit").submit();
What is the alternative way here to be used here for if(isset($_POST['Submit'])) since I'm submitting the form using javascript?
If I understand you correctly, try this:
if ($_SERVER['REQUEST_METHOD'] === 'POST')
{
// your code.........
}
You should add a hidden input <input type="hidden" name="formsubmit" value="yes" /> to the form which will always get submitted, and check for that instead of the button (which only gets submitted if it is clicked on ..)
If I understood your problem correctly that you can simply change input type to hidden.
<form method="post" id="form_submit">
...
<input type="hidden" name="Submit">
</form>
$_POST['Submit'] variable will be defined.
The best solution is "Don't do that". If you want to submit a form then use a submit button (don't do it as a side effect of clicking on a hyperlink to the top of the page). Any JavaScript you want to run can then be handled in the form's submit event.
If you really want to do it as a side effect, then check for the existence of any other field that you know will be set. You could add a hidden field to ensure there will be one of a given name/value combination if you like.

How can PHP determine if the user pressed the Enter key or Submit button?

The problem I have is that I have multiple submit inputs in a single form. Each of these submit inputs has a different value and I would prefer to keep them as submit.
Whenever the user presses Enter, it is as though the topmost submit input is being pressed, and so it is causing problems for the code checking which input was clicked.
Is there a way for PHP to determine whether or not the input was clicked, or was just the input that was selected when the user pressed the Enter key?
You can identify which button was used provided you structure your HTML correctly
<input type="submit" name="action" value="Edit">
<input type="submit" name="action" value="Preview">
<input type="submit" name="action" value="Post">
The $_POST array (or $_GET/$_REQUEST) will contain the key "action" with the value of the enacted button (whether clicked or not).
Now, "clicking" is explicitly a client-side behavior - if you want to differentiate between a click and a keypress, you'll need to add some scripting to your form to aid in that determination.
Edit
Alternatively, you can be "sneaky" and use a hidden submit that should correctly identify a key-pressed for submission, but this probably has some significant impact on accessibility.
<?php
if ( 'POST' == $_SERVER['REQUEST_METHOD'] )
{
echo '<pre>';
print_r( $_POST );
echo '</pre>';
}
?>
<form method="post">
<input type="text" name="test" value="Hello World">
<input type="submit" name="action" value="None" style="display: none">
<input type="submit" name="action" value="Edit">
<input type="submit" name="action" value="Preview">
<input type="submit" name="action" value="Post">
</form>
Roberto,
PHP is server-side technology and therefore runs on the server - so there is no way for it to determine what keys where pressed at the client (aka the user). That is, of course, unless you specifically code the client-side to include such information with the server requests (posting form data is a form of request too).
One way to accomplish that is by using Javascript in your client code.
See this page as a starting point regarding handling form submit events using Javascript.
http://www.w3schools.com/jsref/jsref_onSubmit.asp
You may also have to add a listener for key press events on your page in order to capture the user pressing the Enter key and then recording this information.
http://www.howtocreate.co.uk/tutorials/javascript/domevents offers a discussion on the topic of adding/removing event listeners in Javascript but you have to be very careful when using events because improperly used they can be the source of memory leaks which are hard to debug and cause for unhappy users :)
PHP alone can't determine how the form submit event was triggered, because that happens on the client-side while PHP is a server-side language. You'd have to implement Javascript to listen for -- and log to the server-side -- key presses and mouse clicks, and then analyze that data to find what you're looking for.
Now, PHP can tell which submit input was triggered, as it will appear in the form data while the others will not. Most browsers make the first submit input the default (the one that is triggered on an Enter key press). You could re-order all your submits so as to control which submit is triggered.
PHP can't really know what happened on the client side.
I'd recommend using javascript. When the user do the action, catch it and store it in an hidden field that will be submited with the form. You can also keep track of what input is active and store it in an hidden field.
The code would go a bit like that (i didnt checked the syntax)
<input type="text" onfocus="setCurrent(this)" id="1" />
<input type="hidden" id="hid" />
function setCurrent(o){
$('hid').value = o.id;
}
I think that playing around with events catching and hidden fields should give you the result that you want.
Hope that helps
It's how you write the markup on the client side.
For example, here is one (non-XHTML) way you could do this:
In the HTML file:
<form method="post" action="myform.php" id="myform">
... form items here ...
<input type="submit" name="enter_key" value="true" style="display:none">
<input type="hidden" name="pressed_button" id="pressed_button" value="false">
<input type="button" value="Submit"
onclick="document.getElementById('pressed_button').value='true';document.getElementById('myform').submit();">
</form>
In myform.php:
if ($_POST['pressed_button']=='false') {
// Logic for enter key
} else {
// Logic for button press
}

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