Need help to passing label text to another page via session array and also i have tried to post radio button values but problem is that radio button values didn't display into another page?
Index Page Php Function
<?php
session_start();
if (isset($_REQUEST['demo'])) {
$_SESSION['size'] = array('Small', 'Medium');
}
?>
$(document).ready(function () {
$('.form-radios label').attr('checked', 'unchecked');
$(".form-radios label").click(function(){
$(this).attr('checked', 'checked').addClass('checked');
$('label').not(this).removeClass('checked');
});
});
</script>
Index Page Form
<form action="page2.php" method="post">
<div class="form-radios">
<input type="radio" name="size" id="size" value="Small" />
<label for="size">Small</label>
<input type="radio" name="size" id="size" value="medium"/>
<label for="size">Medium</label>
<input type="radio" name="size" id="size" value="large"/>
<label for="size">Large</label>
<input type="radio" name="size" id="size" value="xl"/>
<label for="size">Xl</label>
</div><br />
<input type="submit" name="demo" value="submit" />
</form>
Page2 Function
<?php
session_start();
echo $sizes=$_SESSION['size'];
?>
<input type="text" value="<?php echo $sizes ?>" />
Your session is an array you have to loop it.
foreach ($_SESSION['sizes'] as $size)
{
echo $size;
}
But I want to post radio values how could I send?
Radio values from your form are sent automatically to the action specified when a user press the submit button. In the action page they can retrieve them via the $_POST or $_GET global array. In your case, since you are using the POST method, you can retrieve your radio box via:
$_POST['size'];
Actually I want to send when someone click on label then label values to another page.
Well, you can do that by make Javascript trigger an event when a radio box is selected and send a request via AJAX.
Related
I am still working on my school project, which is almost finished.
With your help, I successfully created a working system that allows users to write, edit and delete data to/from the database.
The only problem I have right now us "user-friendly form." I managed to create auto-focus, insert correct values on edit so the user can see what was previously written in that field, etc.
I have my forms hidden with jquery. When a user clicks add, the form slides in. What I need to achieve is: "when a user clicks submit and the page refreshes and adds the element to the database, the form should appear again so users can add data faster."
Here is my code.
$(document).ready(function() {
$('#x').click(function() {
$('.y').toggle("slide");
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="x">Click</div>
<div class="y" style="display: none;">
<div class="container">
<form action="insertzunanja.php" method="POST" id="x">
<input type="hidden" name="narocilo" value="0.1412312">
<input type="hidden" name="id" value="id-1">
<input type="text" name="dolzina" style="width:49%;" placeholder="Dolzina (v cm)">
<input type="text" name="sirina" style="width:49%;" placeholder="Sirina (v cm)">
<input type="text" name="kolicina" value="1" style="width:49%;" placeholder="Kolicina">
<input type="text" name="opombe" style="width:49%;" placeholder="Opombe">
<input type="submit" value="Send">
</form>
</div>
</div>
Thanks and best regards.
You can use AJAX call to send the data to php file instead of form action. According to your code you will have something like this:
<div id="x">Dodaj</div>
<div class="y" style="display: none;">
<div class="container">
<input type="hidden" name="narocilo" value="<?php echo $randomNum; ?>">
<input type="hidden" name="id" value="<?php echo $id; ?>">
<input type="text" name="dolzina" style="width:49%;" placeholder="Dolzina (v cm)">
<input type="text" name="sirina" style="width:49%;" placeholder="sirina (v cm)">
<input type="text" name="kolicina" value="1" style="width:49%;" placeholder="Kolicina">
<input type="text" name="opombe" style="width:49%;" placeholder="Opombe">
<input id="sub" type="submit" value="Send">
</div>
</div>
$(document).ready(function(){
$('#x').click(function() {
$('.y').toggle("slide");
});
});
$("sub").click(function(){
$.post("insertzunanja.php",
{
dolzina: $("input[name=dolzina]"),
sirin: $("input[name=sirina]")
},
function(){
$("input[name=dolzina]").val("");
$("input[name=sirina]").val("");
if($('.y').is( ":hidden" )) {
$('.y').toggle("slide");
}
});
});
Basically, when you click on button you call php with AJAX POST request passing two values dolzina and sirin retrieved by the html code(note: you have more values to pass so change it accordingly) to php file. Jquery deletes the values of the input fields and check if input fields are shown. If not the inputs fields are shown.
If you are using PHP to process the form submission and generate the code in your question, and you always want the form to be displayed after submission, you can do this:
At the PHP code identify submission (e.g. isset($_REQUEST['id']) ).
[if #1 is true] On generating jQuery code, add $('.y').show(); within the ready function (but separated from the existing click function).
Example:
<?php
// your existing code...
?>
$(document).ready(function() {
<?= ( isset($_REQUEST['id']) ? '$(".y").show();' : '' ); ?>
$('#x').click(function() {
$('.y').toggle("slide");
});
});
<?php
// your existing code...
?>
Say I have a HTML form that lets you input a name and age and returns with a list of people with that name and age.
<form method="post" action="/search_results">
<input type="text" name="personName">
<input type="text" name="personAge">
<input type="submit" value="Submit!">
</form>
And on the search results page I have a list of the results, but I only display 50 at a time and allow users to go forwards/backwards between the page with buttons. So the results page would also look for a POSTed 'pageNumber' value and default to 0 if there is none.
When they click a button, how would I resubmit the age and name and also submit the corresponding pageNumber from the button?
I'm using PHP
Add a hidden field to the form:
<form name=search"" method="post" action="/search_results">
<input type="hidden" name="pageNumber"
value="<?php echo isset($_POST['pageNumber']) ? (int) $_POST['pageNumber'] : 0; ?>">
<input type="text" name="personName">
<input type="text" name="personAge">
<input type="submit" value="Submit!">
</form>
Add a JavaScript function to modify the hidden field value:
<script>
function search(pageNumber) {
var form = document.forms.search;
if (!form) return;
form.elements.pageNumber.value = pageNumber;
form.submit();
}
</script>
Apply the JavaScript function for the page buttons:
<span onclick="search(1)">1</span>
<span onclick="search(2)">2</span>
Obviously, the buttons should be generated with PHP in the following manner:
<?php
for ($p = 0; $p < $pagesNum; ++$p) {
echo "<span onclick='search($p)'>$p</span>";
}
?>
<form method="post" action="/search_results">
<input type="text" name="personName" value="<?php echo (isset($formdata['personName'])?$formdata['psersonName']:"") ?>">
<input type="text" name="personAge" value="<?php echo (isset($formdata['personAge'])?$formdata['personAge']:"") ?>">
<input type="hidden" name="currentPage" value="<?php echo (isset($formdata['currentPage'])?$formdata['currentPage']:"0") ?>">
<input type="submit" value="Submit!">
</form>
formdata are the data which are the inputs of the previous submit. These data should be returned by the search_results page along with the view.
Below is the js
<script>
$(document).ready(function(){
$(".paginationBtns").click(function(){
var page = $(this).attr('pageValue');
$("input[name='current']").val(page);
$("form").submit();
});
});
</script>
Assumptions of the forward and backward button
<a href="#" pageValue=0>Back</a><a href="#" pageValue=50>Next</a>
You have to append the back and next pageValue while loading each page.
The best practice in pagination to use a simple link that contains the parameres (GET) and not (POST). That way, when you have the parameters in the url you can cache it, add to favorites, get indexed by google, share your page in email/facebook etc.
<a href='http://example.com/?personName=<?=$_POST['personName']?>&personAge=<?=$_POST['personAge']?>&page=<?=$next_page_number?>'>Next</a>
If for some reason you must or really want to use POST you can save the values in hidden inputs within a form in the page and then sumbit it when clicking on "next page" button.
Form example:
notice its just an example you should sanitize the variables and not put them directly from the $_POST
<form name="pagination" method="post" action="/search_results">
<input type="hidden" name="page" id="page" value="2">
<input type="hidden" name="personName" value='<?=$_POST['personName'];?>'>
<input type="hidden" name="personAge" value='<?=$_POST['personAge'];?>'>
</form>
notice that we set the next page numbers & do submit by using a command from the form of:
<button onclick='document.getElementById("page").value= "2";document.pagination.submit();'>Next page</button>
I have a table with radio buttons to get the row values and 2 buttons
1 button.)For printing data , which moves to "notice.php"
2 button.)For row details,which stays on the same page.
<form action="" method="POST">
<table border="1" >
<tr>
<th>sourceID</th>
....
<th>Status</th>
</tr>
<tr>
<td><input type="radio" name="ID[]" value="<?php echo $tot; ?>" /></td>
<td>1</td>
....
<td>open</td>
</tr>
<input type="button" name="issue" value="Issue Notice" onClick="location.href='notice.php'" />
<input type="submit" name="details" value="details" />
<?php
if(isset($_POST['details']))
{
$n=$_POST['ID'];
$a=implode("</br>",$n);
echo$a;
}
Notice.php:
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
$n=$_POST['ID'];
}?>
The problem here is: My code is working perfectly fine with the button details.
But it doesnt work with issue, i.e after selecting radio button and clicking on the issue notice button :it gives Undefined index: ID in D:\XAMPP\notice.php.
kindly help
Your details button is a submit button, so it submits the form. However your other button is just a regular button and you use javascript to send the browser to notice.php. As such, it does not post any data to notice.php.
You could include the data on the query string and send it that way, e.g.:
location.href="notice.php?id=<?=$tot?>"
Or you could also have the issue button post the page, and then have your receiving page check which submit button was used. If the issue button was used you could then have the php code post to notice.php.
Using the following code is the exact same as having a link:
<input type="button" name="issue" value="Issue Notice" onClick="location.href='notice.php'" />
As in, this will not change the form action and submit the POST data to your new page.
You would need something like:
<form method="post" action="" name="unique-form-name">
<input type="radio" name="ID[]" value="<?php echo $tot; ?>">
<input type="button" id="unique-btn-name" value="Issue Notice">
</form>
<script type="text/javascript">
document.getElementById('unique-btn-name').onclick = function(){
document['unique-form-name'].action='notice.php';
document['unique-form-name'].submit();
}
</script>
Then, once you get the data to notice.php, you'll have to use the data as an array (you won't be able to echo the data):
$IDs = $_POST['ID'];
echo '<pre>',print_r($IDs),'</pre>';
<input type="radio" name="ID" value="<?php echo $tot; ?>" />
Your error is the name attribute.
Also the other button is not related to the form at all. You may want to use ajax here.
I have this:
<form method="post" id="kl" action="step2.php">
<input type="radio" name="rubrik" value="bussines"></input>
<input type="radio" name"rubrik" value="private"></input>
<input type="image" value="submit" src="/images/submit.png" alt="Submit" />
</form>
What i bassicaly want is: When the second radio button is checked, to submit the form to step2a.php, a different file. How can i do this? Jquery, Javascript, php?
You could do this with JavaScript (bind a submit listener that checks the value of the radio button and then sets the action property of the form), but it would be simpler and more reliable to do something (server side) along the lines of:
<form ... action="step-selector.php">
and
<?php
if (isset($_POST['rubrik']) && $_POST['rubrik'] == 'bussines') {
include('step2.php');
} elseif (isset($_POST['rubrik']) && $_POST['rubrik'] == 'private') {
include('step2a.php');
} else {
include('error-state.php');
}
?>
you can do this by modifying the Form into:
<form method="post" id="kl" action="step2.php">
<input type="radio" class="radio" rel="step2.php" name="rubrik" value="bussines"></input>
<input type="radio" class="radio" rel="step2a.php" name"rubrik" value="private"></input>
<input type="image" value="submit" src="/images/submit.png" alt="Submit" />
</form>
I added rel attribute to radio buttons. each has a value of the url. I also added a class to get the element with jQuery.
Now, you will need some Javascript, i will use jQuery code:
$('.radio').click(function (){
rad = $(this);
radRel = rad.attr('rel');
$('form#kl').attr('action', radRel);
});
There are multiple ways of doing it, depending on what you want exactly.
Check this one out, it might help you get there; Radio Button to open pages
You can use form.submit() as onclick-handler (not onchange) and change the action, too.
<input type="radio" name"rubrik" value="private" onclick="this.parentNode.action='yourOtherFile.php'; this.parentNode.submit()"></input>
I have a form with various fiedls (Name, email, password, etc..). It also has a set of 5 check-boxes. As i have to update a database i prefer doing server-side validation and i am using php.
If a validation error occurs, the error is displayed on the top of the page along with the form and the previously entered data in the text fields. I am not able to retain the state of the checkboxes and radio buttons. The all revert back to being not selected. What should i do to retain the state of checkboxes and radio buttons?
The form looks something like..:
Password:<input type="text" name="password" size="16" maxlength="9" value="<?php echo $_POST['password']?>"/>
Retype Password:<input type="text" name="repassword" value="<?php echo $_POST['repassword']?>"/>
Select:<br />
<input type="checkbox" name="option1" value="on1" id="opt1"/> <label for="opt1">Option1</label><br />
<input type="checkbox" name="option2" value="on2" id="opt2"/> <label for="opt2">Option2</label><br />
<input type="checkbox" name="option3" value="on3" id="opt3"/> <label for="opt3">Option3</label><br />
<input type="checkbox" name="option4" value="on4" id="opt4"/> <label for="opt4">Option4</label><br />
<input type="checkbox" name="option5" value="on5" id="opt5"/> <label for="opt5">Option5</label><br />
Mobile No:<input type="text" name="mobileno" maxlength="10" value="<?php echo $_POST['mobileno']?>"/>
What you're talking about is Sticky Forms. You can implement sticky forms in many ways, my choice is to use session. If there is an error in your validation, simply dump everything in your POST data into session. On your form page, check to see if the value is set in session and set the default value of the form control it if it is:
Validation Page:
<?php
session_start();
if(/** some error condition **/) {
foreach($_POST as $k => $v)
$_SESSION['sticky_'.$k] = $v;
header('Location: http://site.com/yourform.php');
exit();
}
?>
Form Page:
<?php session_start(); ?>
<input type = "checkbox" name = "option1" value = "on1" id = "opt1" <?php
if(isset($_SESSION['sticky_option1']))
echo('checked = "checked");
?>/>
Example TextBox: <input type = "text" name = "textBoxName" <?php
if(isset($_SESSION['sticky_textBoxName']))
echo('value = "' . $_SESSION['sticky_textBoxName'] . '"');
?>
...
<?php
// Erase the POST values from session after the HTML is constructed.
foreach($_SESSION as $k => $v)
if( strpos($k, 'sticky_' !== false )
unset($_SESSION[$k]);
?>
Add a checked attribute to any checkbox who's value is in the submitted data.