I want to create a checkbox and check whether checkbox is checked or not in another PHP page so I created form like :
<form action="heartbeat.php" method="post">
<input type="checkbox" name="keepme">
<input type="submit" value="log in">
</form>
and then I tried to extract this input on heartbeat.php like:
<?php
/*
* A PHP file for laying down a heartbeat JavaScript call.
*/
if(!isset($_POST["keepme"])){
$auto_logout = 10;
}
else{
$auto_logout = 1000;
}
?>
but it never gets $_POST["keepme"] value. Any idea??
If you don't pass value to checkbox tag, the POST array will be empty. You need to do something like that
<input type="checkbox" name="keepme" value='1'>
Make Sure you have submit button
If(isset($_POST['submit-btn'])){
echo $_POST['keepme'];
}
Related
Having issues with using a form's value in a different php file:
my firstpage.php
<form method="post">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
</form>
my secondpage.php is here
<?php
include("firstpage.php");
$result = $_POST['rdbbtn'];
if ($result == "1") {
echo 'thirdpage.php';
}
else {
echo 'fourthpage.php';
}
?>
problem:
Notice: Undefined index: rdbbtn in
how come I can't use "rdbbtn"? Should I have something like
$rdbbtn = $_POST['rdbbtn'];
in secondpage.php? Tried this but didn't solve my problem.
firstpage.php and secondpage.php are in the same directory.
Probably it's some pretty obvious thing that I don't see...thanks!
EDIT: I have accepted pradeep's answer as that helped me the most to figure what the problem should be. would like to say thank you for everybody else showing up here and trying to help!
When you change current page it reset the value and $_POST is empty.
You can try with set form action to next page . It will work
<form method="post" action="secondpage.php">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
<input type="submit" name="" value="Next">
</form>
Other wise you can make a function in a class and set each page action
to this function.
And set your each form data to session.
Finally when you change the page you read data form session.
Class FormAction{
public function setFormDataToSession(){
if(isset($_POST['rdbbtn']){
$_SESSION['rdbbtn'] = $_POST['rdbbtn'];
}
}
}
In your page simply get the session value.
echo $_SESSION['rdbbtn'];
Should be like this :
Check with isset method in
<?php
include("firstpage.php");
$result = isset($_POST['rdbbtn']) ? $_POST['rdbbtn'] : NULL;
if ($result == 1) {
echo 'thirdpage.php';
}
else {
echo 'fourthpage.php';
}
?>
and your form should be like this :
<form method="post">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
<input type="submit" name="submit" value="submit">
</form>
Sorry for not being able to comment in this post(less reputations). But seems like you are asking about storing the variables of the session. This way you can use the variables for a whole session. Just start the session by putting session_start() in the very beginning of secondpage.php file and then you can access the variables at any time during the session by simply calling $_SESSION['rdbutton] in any page like fourthpage.php or anything. Just make sure u put the session_start() at the top of each page where you want to use the variables. Don't forget the semicolons at the end. 😜 Hope this helps.
I have an input type text box as follows
<input type="text" name="deleteprofileconfirmation" id="deleteprofileconfirmation" class="editprofileinput">
Delete Account
I need to pass the value entered in the input type text to deleteaccount.php
I can do with help of jquery, no problem, i need a pure php solution...
I tried using sessions, but problem is how to read the value in input type when link is clicked.. $_POST is also not working...
i cannot use form because this is a form in another form so html5 is not allowing nested forms, sorry should have mentioned that earlier
the following is not working on deleteaccount.php
if (isset($_POST['deleteprofilebutton']))
{
$delete_profile = strtolower($_POST['deleteprofileconfirmation']);
}
Make your link as
href="../controllers/deleteaccount.php?id=$ID_VALUE"
and update the POST to GET
if (isset($_GET['id']))
{
$delete_profile = strtolower($_GET['id']);
}
That would be GET now.
Make sure users with no privileges can hit this url and delete the profiles. Do check the user rights before processing the delete operation.
You can do this using form
<form action="../controllers/deleteaccount.php" method="post">
<input type="text" name="deleteprofileconfirmation" id="deleteprofileconfirmation" class="editprofileinput">
<input type="submit" class="deleteprofilebutton" name="deleteprofilebutton" id="deleteprofilebutton" value="Delete Account">
</form>
You could also give each one a unique name and just check the $_POST for the existence of that input.
<input type="submit" name="deleteprofileconfirmation" value="Delete" />
<input type="submit" name="submit" value="Submit" />
And in the code:
if (isset($_POST['deleteprofileconfirmation'])) {
//delete action
} else if (isset($_POST['submit'])) {
//submit action
} else {
//no button pressed
}
I need to check whether checkbox is checked or not. Normally I would do it like this:
<?php
$checked = isset($_POST['checkbox']);
?>
But I don't know what is the name. More at screenshot (I'm using Laravel 4).
Screenshot
You simply can't. The data for unchecked checkboxes are not send to the server.
You could do a workaround with javascript where the JS appends some hidden fields before submit with the nonchecked boxes
Supposedly you should know what the list of checkboxes is/was that you asked the user to check. Checked checkboxes are submitted to the server, unchecked ones aren't. You can calculate the difference between these two lists.
if you are using jquery, and you know the id of the checkbox,
then, you can detect with following code:
var isChecked = $("#cbId").is(":checked");
<?php
if ( ! isset($_POST['checkbox_name']))
{
"Not checked";
}
?>
If checkbox didn't check - you will not have this variable in $_REQUEST.
<form action="">
<input type="checkbox" name="ch1"/>
<input type="checkbox" name="ch2"/>
<input type="checkbox" checked="checked" name="ch3"/>
<input type="submit" name="Post" value="Post">
</form>
When you click on "Post". In backend you'll see:
<?php
if(isset($_REQUEST['ch1']))
echo 'ch1 is checked!';
if(isset($_REQUEST['ch2']))
echo 'ch2 is checked!';
if(isset($_REQUEST['ch3']))
echo 'ch3 is checked!';
?>
In my case you'll see: "ch3 is checked!".
When I create i form - I do something like this:
<form name="form-name" method="post" action="?<?=$_SERVER['QUERY_STRING']?>">
[...some elements...]
<input type="submit" name="form-name" value="button">
</form>
Now I need to get the value of the name="" of the submit button, and not the actual value="".
In this case : "form-name".
And here's why:
When I submit a form; I write the action to database - and therefor need the name of the form submitted.
I know I can just have a hidden field with the form name. But I would like to make it simpler by just extracting the name from the submit button because I have a couple of other hidden form elements that I need to add on every single form I create to make my template system work.
And no javascript...
So, let's say your HTML form is this:
<form name="form-name" method="post" action="">
<input type="submit" name="form-name" value="button">
</form>
And you want to get what is inside name="form-name" in this case the form-name
Well, then in the PHP side you can, treat the $_POST global as associative array, and extract the key from it like this:
<?php
if(isset($_POST)){
foreach($_POST as $key=>$each){
echo $key; // this will output "form-name"
}
}
I might have come up with a solution to my question...
Here's a example form:
<form name="vehicle-vinNr" method="post" action="?<?=$_SERVER['QUERY_STRING']?>">
<input type="hidden" name="hello" value="world" readonly>
<input type="text" name="element">
<input type="submit" name="vehicle-vinNr" value="send">
</form>
First I need to extract and place the element-names into a new array:
<?php
if ($_POST){
foreach($_POST as $_FORM_ELEMENT_name => $_FORM_ELEMENT_value){
$_FORM_ELEMENT_names[] = $_FORM_ELEMENT_name;
}
}
?>
In this case the array now contains:
hello
element
vehicle-vinNr
If the submit-button is, and always is, the last element in the form - this would work:
$_FORM_name = end($_FORM_ELEMENT_names); // vehicle-vinNr
But sometimes the submit-button is not the last element, so I needed to make a change:
If I always start the name of the submit-button with submit_ - e.g. submit__vehicle-vinNr or with multiple submit buttons for different actions like submit_update__vehicle-vinNr/submit_delete_vehicle-vinNr I can just do this:
if ($_POST){
foreach($_POST as $_FORM_ELEMENT_name => $_FORM_ELEMENT_value){
if(strstr($_FORM_ELEMENT_name,'submit_')){
$_FORM_ELEMENT_submit_name = explode('__',$_FORM_ELEMENT_name);
$_FORM_name = $_FORM_ELEMENT_submit_name[1]; // vehicle-vinNr
}
}
}
This is the solution I came up with - any thoughts?
Does anyone know why this code will not work? I simply want to print which radio button is selected. It always prints 'Null' no matter what is selected. PHP code is below.
<?php
$conn = mysql_connect('localhost','student','student') or die(mysql_error());
mysql_select_db('vgs',$conn);
//Get Question 1
if (isset($_GET["q1option"]))
{
$q1option = $_GET["q1option"];
}
else
{
$q1option = "Null";
}
//Process Question 1
echo "".$q1option;
The HTML code is below.
<form action="" method="get" >
<div id="Q1">
<label><input type="radio" name="q1option" value="Less_than_16" />Less than 16</label><br />
<label><input type="radio" name="q1option" value="16_or_more" />16 or more</label>
</div>
Any help with this would be greatly appreciated. Note I have many tables in the 'vgs' database, if that makes a difference.
Thank you,
Daniel
Additional Code
HTML below
<input type="button" value="Submit" onclick="result();" />
<input type="reset" value="Reset" />
</form>
Embedded JavaScript below. It uses http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js
function result()
{
$('#Suggestion').load('process_answers.php');
}
Looks like your submit button is outside of the <form> tag, it should be within the form:
<form action="" method="get" >
<div id="Q1">
<label><input type="radio" name="q1option" value="Less_than_16" />Less than 16</label><br />
<label><input type="radio" name="q1option" value="16_or_more" />16 or more</label>
</div>
<input type="submit" value="Submit" />
</form>
You are always getting Null in return because your query string is not being attached anywhere. If you are using AJAX, you need to pass the query string data. Your cycle is not working properly because of a misconfiguration.
Take the following code:
//Get Question 1
if (isset($_GET["q1option"]))
{
$q1option = $_GET["q1option"];
}
else
{
$q1option = "Null";
}
Using $_GET will attempt to extract from your query string a variable labeled 'q1option'. If this variable is not set, then return 'Null'. So your code is working as expected as it is currently structured.
If you want to pass the variable between pages you will either need to submit the form and the variable will be automatically passed, or you need to modify your jQuery to allow for the passing and processing behind the scenes of your variable. Once processed, you would then handle the results accordingly and render them to the client as desired.