Get submit button value in PHP - php

I would like to test my button can get the current value from the date's textbox? I got the value of submit button but I couldn't find any code get name of the button. So what should I call I'm able to view the value?
$form_button_layout = '<button class="btn" name="oprf" value="view-log" type="submit">View</button>';
<form class="well form-inline" method="post"<?= $form_action_override; ?>>

If you'd like to get the value of 'oprf' above, then:
$value = $_POST['obrf'];
Should do it in php.

I don't think this works consitently in all browsers. Better to work with a customized name attribute.

You can try this one
if(isset($_POST['oprf'])) {
echo $_POST['oprf'];
}
Hope this helps

Add type property with submit type="submit" to button tag this will submit the form then receive the form inputs value in php script with super global variable $_POST['input name'] or $_GET[] wdepends on from action property value by default GET.

You can do:
if ($_POST['oprf'] == '' || $_POST['oprf'] == NULL) {
//this checks to see if there has been a value that was sent with the
//form
}
I usually use isset for cookies.

Related

Doing an inline if in the action of a form

I need to redirect the user to a site that gets the "Short_proj_name" information. So i did this:
<form action="Main.php?short_proj_name=<?=$_REQUEST['short_proj_name']?>" method="post" name="formProjName" target="_blank" id='frmProjName'>
However, upon searching, i found out that there are several reasons NOT to use $_REQUEST, one of them being security and all that. However, simply doing $_POST['short_proj_name'] or $_GET['short_proj_name'] never returns the information i need.
Basically, how would i go about doing an if statement that checks if the $_GET is empty, and does a $_POST instead? Can i do that in the action method of my form?
EDIT:
Adittionally, is it possible that maybe using both $_POST and $_GET return null, yet using $_REQUEST doesnt? As far as i know, $_REQUEST is both get and post together, but none of them returns any information
It works if i do it as so:
if(!empty($_POST['short_proj_name']))
{
$projName = $_POST['short_proj_name'];
}
elseif (!empty($_GET['short_proj_name']))
{
$projName = $_GET['short_proj_name'];
}
else
{
$projName = $_REQUEST['short_proj_name'];
}
But i'm not sure if that solves the security problem at all
I think the answer here is to always use _GET.
A form can actually send both _GET and _POST data based on what you use in the "action" attribute of the form. The action part doesn't care what you set the "method" attribute as.
From what you are showing above, the params are all in the "action" part of the form so these are always passed into _GET anyway. If the inputs were inside the form then those would be received via _POST
Here's an example.
In PHP I would receive $_GET['monkey'] = '1' and $_POST['lion'] = 1
<form method='post' action='receive.php?monkey=1'>
<input type='text' name='lion' value='1' />
<input type='submit' />
</form>
There shouldn't really ever be an instance where you need to check if the answer is in _GET or _POST and as mentioned in a comment, it's quite a security risk to use $_REQUEST or check if it's in _GET or _POST.
Most times, you can just push the page request URL back into the form "action" to ensure all the same _GET params are included on the form _POST.
The big mistake most people do is try to move them from _GET into hidden input fields inside a form thinking they need to do that to carry all that data through.
However, this type of function call might help you but I wouldn't approve of it.
function getRequestParam($param){
if(isset($_GET[$param])){return $_GET[$param];}
if(isset($_POST[$param])){return $_POST[$param];}
return "";
}
you can like
<?php
if(!empty($_POST))
{
$projName = $_POST['short_proj_name'];
}
else
{
$projName = $_GET['short_proj_name'];
}
?>
<form action="Main.php?short_proj_name=<?=$projName ?>" method="post" name="formProjName" target="_blank" id='frmProjName'>
but i think it's ugly
Here is a simple code :
<?php
if (isset($_GET) && $_GET['short_proj_name'] != '')
echo $_GET['short_proj_name'];
else if (isset($_POST) && $_POST['short_proj_name'] != '')
echo $_GET['short_proj_name'];
else
echo $_REQUEST['short_proj_name'];
?>
But if you get the value from a post or get, it can be anything so be careful...
If the "short_proj_name" is a file name, a nasty person can get access to other files just by guessing their names...

Re-populate checkbox if it fails in validation in an edit form in Codeigniter

I works on the an data-edit form in codeigintier. And the problem is about re-populate checkbox
It works if it is an add form (that means I need not concern about the value in database):
<?= set_checkbox('is_default', '1'); ?> for checkbox
The problem is, in the edit form:
I can't repopulate the checkbox
<?php if ($customer_group[0]['is_default'] == "1") echo "checked"; set_checkbox('is_default', '1'); ?>
The checkbox will check even I have not check it in the edit => fail to validate in the form, thanks for helping
I have already set the validation rule in controller, the code in the add form is working , but how to handle the case for edit form?
In order to re-populate checkbox following code might be helpful:
set_checkbox('fieldName', 'fieldValue');
Where 'value' is the second parameter of the form_checkbox call. Like this:
form_checkbox('fieldName[]', 'value', set_checkbox('fieldName', 'value'));
Now if you are on edit form then below code might help you
$getVal=$valFromDb; //$valFromDb is actually value of the filed from db as you are on edit page
if($getVal!=0){
{
echo form_checkbox('fieldName[]', 'value', true);
}
else
{
echo form_checkbox('fieldName[]', 'value', false);
}
set_checkbox takes a third argument to set the default state, so basically you have to do something like this
echo set_checkbox('is_default', 1, $customer_group[0]['is_default'] == "1");
Can give one suggestion??
1. Hide all the checked value of checkbox in input box when you are directed towards edit page.
If checked box is checked in edit page, edit the value of hidden input field of textbox value.
Submit it, when validation failed, checked or repopulate the checkbox value according to hidden field value. send checkbox value of checked box field through array from controller to edit page view like this. e.g $data['repopulate_checks'] = $this->input->post('array name of checkboxs');
In view :
getit like this
$catch_checkbox = $repopulate_checks;
You can directly get through $repopulate_checks also.
Hope this help you.
You can use form_checkbox() function: Guide
$isChecked = False; // or True for default value
If have stored data then:
$isChecked = $customer_group[0]['is_default'];
echo form_checkbox('input_name', 'value', $isChecked);
or the hard way:
set_checkbox():
The first parameter must contain the name of the checkbox, the second
parameter must contain its value, and the third (optional) parameter
lets you set an item as the default (use boolean TRUE/FALSE)
<input type="checkbox" name="is_default" value="1" <?php echo ($customer_group[0]['is_default']) ? set_checkbox('is_default', '1') : '' ; ?>/>
set_checkbox takes a third argument to set the default state, so basically you have to do something like this
$checked = FALSE; if($customer_group[0]['is_default']){ $checked = TRUE; }
echo set_checkbox('is_default', 1, $checked);

isset function doesn't work on my form and form is nod getting submitted

I am submitting form to page and checking if submit button value InsertMe isset but non of the code inside is executed
<?php
if (isset($_POST['InsertMe'])){
//code to execute
echo "Test";
}
?>
and insert buttons looks like that
<input style="float:right;" name="InsertMe" id="InsertMe" type="submit" class="rectangular-button" value="Add User" />
if (!isset($_POST['InsertMe']) || empty($_POST['InsertMe'])) {
// error message here
} else {
// What you want to do if not empty and is set.
}
This code will check if the variables is set and if if it's empty.
"||" is the PHP operator to check or. So in this case it's checking if it's set OR empty.
Make sure you have the form tag set to post.
<form method="post">
you are selecting Id I think we should be use name tag parameters in isset()

how NOT to store submit button value as post data php

ok i have dynamically created form elements with unique names and when validated i want to store all the form data into SESSION.
This is the code to do it:
if(isset($_POST["address_submit"])){
insertSessionVars();
header("Location: http://localhost/project%20gallery/notes.php");
exit;
}
function insertSessionVars(){
foreach($_POST as $fieldname => $fieldvalue){
$_SESSION['formAddresses'][$fieldname] = $fieldvalue;
}
}
This works almost fine but stores also submit button value. I output it like this:
foreach($_SESSION['formAddresses'] as $value){
echo $value;
}
Any help will be greatful :)
Don't assign a name attribute to your submit button. If you assign a name then it is passed as part of the $_POST array.
<input type="submit" value="My Button" />
Since you no longer have the submit button in post, instead of checking if the submit button is set check to see if the post array contains data using count().
if(count($_POST) > 0)
The $_POST values are just array so after the for each why don't you use unset to drop the submit key from the array
That way you can check the form has been submitted and the either choose to not set the submit key or drop it rather then not giving it a name at all
Either way you can in your foreach do something like this (see *) and pass "$except" as parameter.
Where $except has to be the name of your submit button in insertSessionVars();
if(!in_array($key, $except)){...}
And ofc above your foreach something like this :
if(!is_array($except)) $except=array($except);

How to get post values from first page to third page ,as post

i have 3 pages
Page1.php,page2.php,page3.php
In page1.php, i have some hidden values, for example 'name'
After the submission of page1.php, it will go to page2.
Then after some process in page2.php, it should need to automatically submit to page3.php(where page3.php is in another sever)
Finally,when i print the $_POST variables in page3.php, i need to get the variable 'name'
you could stick it in the session
<?php
session_start();
if (!isset($_SESSION['count'])) {
$_SESSION['count'] = 0;
} else {
$_SESSION['count']++;
}
?>
or you could pass them in hidden vars on page2.php if it has a form...
You will want to look into sessions.
If you need them in POST, try this:
$display = "";
$saveFields = array('one', 'two'); // whitelist of fields to add to the form hidden
foreach ($_POST as $key => $val) {
if (!empty($val) && in_array($key, $saveFields))
$display .= '<input type="hidden" name="'.$key.'" value="'.$val.'" />';
}
echo $display;
Should get you where you want to go. The whitelist just ensure's that random stuff is not injected that does not need to be.
(1) Option is to add hidden input on page2 too.
(2) Option is to set the value from page1's name into session and use it on page3
There are several solutions:
PHP sessions
cookies
passing arguments as GET/POST parameters
storing data in database
In simple cases passing arguments as GET parameter page2.php?name=... or using hidden form field is the best solution
This seems straightforward to me, page one has a hidden value called name. Page 2 should retrieve the post $_POST['name'] and print it on page 2 as a hidden field. Once you post it to Page 3 you can retrieve it the same way $_POST['name'].
Realistically if the data is exactly the same and is being carried all the way to page 3, why do you even need it? Can you not just declare it on page 3?
Okay, the way I read this is that on your first page you have a UI with a form. The form is then submitted for processing to page 2. After processing is done, you want to redirect, if you will, the user to another site (or server, doesn't necessarily have to make a difference).
If I got that right, here is what you should do; instead of using the header(); function (php), print an empty page with a hidden form with all of the details you want to send over and use javascript to emulate the user 'submitting' the form.
< div style="display: none;">
< form action="https://mywebpage.com/myscript.php" method=POST>
< input type=hidden name="key_1" value="value_1">
< input type=hidden name="key_2" value="value_2">
< input type=hidden name="key_3" value="value_3">
< input type=submit id="formButton" style="visibility: hidden; ">
< script language="javascript">
document.getElementById("formButton").click()
< /form>
< /div>

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