Have this form:
<form action="<?= $_SERVER['PHP_SELF'] ?>" method="POST" id="form_pesquisa" data-toggle="validator" role="form">
<div class="radio"> <label> <input type="radio" name="cliente" id="73" value="ss"> "ss" </label> </div>
<div class="radio"> <label> <input type="radio" name="cliente" id="74" value="aa"> "aa" </label> </div>
<button type="submit" class="btn btn-warning" name="hchoose">Choose client</button>
How can I get the id of the radio when the button is pressed? $_POST['id'] does not work. Tks all.
The id doesn't get passes to the server for processing, only the input name and value get passed. PHP uses name to process the user input & $_POST['anything'] is used to get the user input of <input type="radio" name="anything" id="73"> & not the id="73". JavaScript can be used to get the id but not PHP actually.
If you really want to get whatever is in the id you can put it in value like value="ssId-73"
You have to set name for the button eg. <input type="radio" name="myRadioButton"/> then in PHP $_POST['myRadioButton']
There's no way to receive id in PHP, you can only set same name as id.
If you don't want to edit your form (which will be the easiest way), you'll have to use jQuery, because using native Javascript will make it even more difficult.
Below your form:
<script src="jQuery_url">
</script>
<script>
jQuery("#form_pesquisa").on("submit", function() {
var id = jQuery("input[name='cliente']:checked").attr("id");
jQuery(this).append("<input type='hidden' name='id' value='"+id+"'>");
});
</script>
Having done this, in your form handler file, $_POST['id'] will contain the value you want to catch.
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() {
$('.y').toggle("slide");
});
});
<?php
// your existing code...
?>
Here's the thing:
I put a <input type='checkbox' name='a'> in a <form action="b.php" method="GET"> but I can not get '?a=something' actually from url.
I don't know what's wrong because I use several <div>s by nesting them and also some css style and bootstrap classes.
My complete code is as below.
<form action="addfile.php" name = "fileinfo_form" method="GET">
<div class="input-group">
<span class="input-group-addon" id="basic-addon1">SHA1</span>
<span class="input-group-addon">
<input type="checkbox" id ="sha1checkbox" name="sha1checkbox"><!--I cannot get sha1checkbox's value from url -->
</span>
<input type="text" name="sha1code" class="form-control" placeholder="Sha1 Hash Code" aria-describedby="basic-addon1">
</div>
<input type="submit" class="btn btn-default" value="Submit"></input>
</form>
Your code seems to be working fine.
When the checkbox is checked, the result in the url will be addfile.php?sha1checkbox=on&sha1code=test.
If it is not checked, it will not return any value in the GET method via URL for that checkbox. Something like addfile.php?sha1code=test is returned only but if you use POST you can tell if it is checked or not in php code.
This is the normal behaviour of the checkbox in html forms. Don't be worried.
You can insert the following code in your destination file to test if the checkbox is checked or not.
<?php
if(isset($_GET['sha1checkbox'])){
echo "It is set";
}else{
echo "it is not";
}
So you do not need to worry about it appearing in the url unless you plan on using the url in some way.
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'm very interested in if it is possible to connect javascript variables to php. I know that in php we can write javascript code, but on the contrary we could not. To express my aim better , lets bring example like that:
<form name="some" action="<?php $_SERVER['php_self']; ?>" method="post">
<input type="submit" name="but" value="Action">
</form>
My question is how to make after pressing submit button to confirm (alert) with javascript and if it is confirmed do something (with php) and if isn't cancel (php operation).
You can do two things to pass javascript variable to php:
You can pass it as a hidden input field and submit it using POST
<input id="myHidden" name="myHidden" type="hidden"/>
Assign javascript variable to hidden input something like
var myVariable;
document.getElementById("myHidden").value = myVariable;
You can pass it as a query string with in your URL
As for confirming you can use javascript confirm, which will post on OK and not post/cancel on CANCEL
Something like:
<input type="submit" onclick="return confirm('Are you sure you want to submit?')"/>
<form name="some" action="<?php $_SERVER['php_self']; ?>" method="POST"
onsubmit="document.getElementById("response")
= confirm('some question') ? 'yes' : 'no';
return true;">
<input type="hidden" name="response" id="response" value="">
<input type="submit" name="but" value="Action">
</form>
BTW action="<?php $_SERVER['php_self']; ?>" opens up your page to XSS attacks.
in order to edit my entries i want to:
<form id="pregunta" name="pregunta" class="form_pregunta" method="post" action="pregunta.php?id=26">
<h2>Titulo de la pregunta</h2><input name="q" id="q" class="q" value="este es mi tÃtiulo " type="text">
<h2>Describe tu pregunta</h2>
<textarea name="texto" id="texto" style="width: 98%;"><p>esta es mi descripcion</p></textarea>
<h2>Etiquetas</h2>
<input name="tags" id="tags" onmouseover="mostrar_tooltip('nube_e','','0','70','')" onmouseout="ocultar_tooltip('nube_e')" value="dos,tres,una,">
<input name="responde_a" style="display: none;" id="responde_a" value="0">
<button name="pregunta" id="pregunta" type="submit">form_edit_question_button</button>
</form>
And then in file.php
i'd like to $_get['id'] and $_post['inputs']
but when i go:
if(isset($_POST['edit_pregunta'])){
echo 'lalalalalalalalalalalalalalala';
post_edit_pregunta();
}
it won't ever enter :S. is that normal or i'm missing something... i wanted not to have a hidden input with the id of the post i want to edit..
I'm not 100% sure, but forms don't send their name when submitted, much less their id.
You could do the following instead:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
<input type="hidden" name="edit_pregunta" value="anything">
... //inputs here
</form>
and your if should now enter.
It looks like you're checking for your form's "id" attribute. This is not sent when the form is submitted, only values in <input>, <select>, <textarea> and <button> are sent.
You should check for one of those.
Edit: Your button name is "pregunta", so that is the POST variable you should be checking for, eg
if(isset($_POST['pregunta'])){
Just to comment in general on mixing params in the form's "action" and inputs, you can mix them as long as the form method is "post". You cannot set GET params in the form's action and use the "get" method
<!-- Good -->
<form action="proc.php?id=123" method="post">
<input name="foo" value="foo">
<input type="submit">
</form>
<!-- Bad -->
<form action="proc.php?id=123" method="get">
<input name="foo" value="foo">
<input type="submit">
</form>
There should be no problem at all with having get and post variables in the same request, but are you sure your syntax is correct? If this is normal php, shouldn't you write
<form id="edit_pregunta" method="post" action="file.php?id=<?php echo $this->id; ?>">
... //inputs here
</form>
[Edit]
The problem is (if I'm correct and this is standard php) that you generate a form that looks something like this:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
... //inputs here
</form>
This will make id look like this: '$this->id' (including the '-signs). When what you want is something like this:
<form id="edit_pregunta" method="post" action="file.php?id=51">
... //inputs here
</form>
Then $_GET['id'] would be 51.
[Edit2]
Also, I think you need to change
if(isset($_POST['edit_pregunta'])){
with
if(isset($_POST['pregunta'])){
If I'm not mistaken the name of a form doesn't get sent to the server, however, the name of the submit-button does, but I might be wrong about that part.
Yes you can, I've done it several times.
Probably something else is wrong with your code.
Is there any control with name="edit_pregunta" or is it just the id of the form? IDs are not sent to the server.
Simply adding the id to the form will not create the $_POST['edit_pregunta'] you verify.
Instead, inside the form tag, add an <input name="foo" />; in the php script verify $_POST['foo']
While the HTTP spec doesn't disallow query parameters in POST methods, it is somewhat unusual. You'd be better off using a hidden input field in the form to pass any non-user values up to the script.
That said, the syntax for your form is wrong. You need to use "echo" to insert the value of $this->id into the action.
Use input type="submit" in place of button tag.
You need name for form submission and activate php script!
HTML Code:
<form action="change.php" method="POST">
<input type="password" name="p1" class="change_text" placeholder="New Password"/></br>
<input type="password" name="p2" class="change_text" placeholder="Re-Password"/></br>
<input type="submit" name="change" value="Change Password" id="change" />
</form>
PHP Code:
<?php
if (isset($_POST['change']) {
$p1=$_POST['p1'];
}
?>