I have a simple photo gallery using CodeIgniter that displays thumbnails and select buttons for each image.
On this page I also have a choose file and 'upload' button and a 'delete selected items' button.
<form action="http://localhost:8080/PhpProject1/gallery"
method="post" accept-charset="utf-8" enctype="multipart/form-data">
<input type="file" name="userfile" value="">
<input type="submit" name="upload" value="Upload">
<input type="submit" name="delete" value="Delete Selected">
</form>
My check boxes are grouped using the following style (i.e. 'photos[]' for group):
<input type="checkbox" name="photos[]" value="IMG_20120709_151023.jpg">
When debugging with Netbeans I am definitely calling the right method by getting the name value from the post data but with the 'delete' method the post data contains nothing else, just the input name and value (key= delete, value = Delete Selected)using.
Here is the php code:
$this->load->model('gallery_model');
if ($this->input->post('upload')) {
$this->gallery_model->do_upload($order_no);
redirect('gallery');
}
if ($this->input->post('delete')) {
$this->gallery_model->do_delete($order_no); // this is getting called ok, just no $_post data??
redirect('gallery');
}
Is there something else I need to do to ensure the post request is picking up the selected items?
I'd like to do this with php but if I have to go down the ajax route so be it, thanks.
Mick.
Related
In this code I have 3 button update,back,delete when I click anyone of it, $page should assign to action attribute in form element and post data should go based on which button is clicked.
<form action='<?php echo $page; ?>.php' method='post'>
<?php
if(isset($_POST['update'])) {
$page='book_update';
}
if(isset($_POST['delete'])) {
$page='book_delete';
}
if(isset($_POST['back']))
{
$page='patientpage';
}
?>
In PHP you can access only to the data that were already sent by the browser in the request.
If you want to detect on which button is clicked, you can name your buttons, like this:
<form action="/page.php" method="post" id="form_id">
<input type="submit" value="edit" name="action">
<input type="submit" value="delete" name="action">
<input type="submit" value="back" name="action">
</form>
After click on the button, the form is submitted and you will find the value in PHP in $_POST['action'].
If you want to modify the action attribute dynamically based on the button before the form is submitted, you have to do with javascript, e.g. document.getElementById('form_id').action = 'something-else.php';
Form 1:
<?php
echo $this->upload_message;
?>
<form enctype="multipart/form-data" method="post" name="mfuploaderwp-uploadform" action="?uploadfile">
Upload file: <input name="mfuploadwp-filename" type="file">
<input class="mfuploadwp-submit" type="submit" value="Upload" name="submit"></form>
Form 2:
<?php
echo $this->upload_message;
?>
<form enctype="multipart/form-data" method="post" name="mfuploaderwp-uploadform" action="?uploadfile">
Upload file: <input name="mfuploadwp-filename" type="file">
<input class="mfuploadwp-submit" type="submit" value="Upload" name="submit"></form>
Is it possible to fetch which form is submitted without giving any extra attributes or so to above forms? The forms are created dynamically based on what user enter for amount number of forms. (In this case user has entered 2 forms)
I want to do this so $this->upload_message would be accurate only for the form that is used for uploading.
Alter the name tags on your <input type="submit"> buttons. Have one as name="submit" and the other as name="submit_two" (for example, bad naming convention), then process code as
if (isset($_POST['submit'])) {
// do stuff
} elseif (isset($_POST['submit_two'])) {
// do other stuff
}
Yes, it's possible.
The cleanest way, in my opinion, is to put an hidden input tag in each form:
<form enctype="multipart/form-data" method="post" name="mfuploaderwp-uploadform" action="?uploadfile">
<input type="hidden" name="active_form" value="1">
(...)
and
<form enctype="multipart/form-data" method="post" name="mfuploaderwp-uploadform" action="?uploadfile">
<input type="hidden" name="active_form" value="2">
(...)
then, in the page that process the form, you can check it in this way:
if( $_POST['active_form'] == 1)
{
(...)
}
elseif( $_POST['active_form'] == 2)
{
(...)
}
If your form is generated dynamically based on the user input(The forms are created dynamically based on what user enter for amount number of forms), in this case you can use three type of solution as far as I know,
You can introduce a new hidden field for each form based on the form number.
Eg:
Upload file: <input name="mfuploadwp-filename" type="file">
<input class="mfuploadwp-submit" type="submit" value="Upload" name="submit">
<input type="hidden" value="1" name="form_id"/>
</form>
in php
switch($_POST['form_id']) {
//the form data to be processed..
}
or
You can update the input field submit button naming based on the form number.
Eg:
Upload file: <input name="mfuploadwp-filename" type="file">
<input class="mfuploadwp-submit" type="submit" value="Upload" name="submit_{form_id}">
you can add a additional parameter in the form method.
...
Say I have a simple form on a page called photo/delete.php. This form deletes an image specified by the user. All it is, is this:
<form action="?" method="post">
<input type="checkbox" name="confirmDelete" value="confirm" />
<input type="submit" value="delete" />
</form>
So, this form contains a confirmation check box that must be ticked to ensure the image is deleted. How can I dynamically choose what page to POST this form to, based on its contents?
For example, if the checkbox is not checked, yet the submit button is clicked, I'd like to stay on the same photo/delete.php page and display an error, since its possible they really do want to delete the image and simply forgot to tick the box.
But otherwise, if everything is successful and the checkbox is ticked, I'd like to POST it to another page, say home.php since it makes no sense to stay on the same page of a just-deleted image.
How can I implement this?
You may try something like this
HTML:
<form action="delete.php" method="post">
<input type="checkbox" name="confirmDelete" value="confirm" />
<input id="btn_delete" type="submit" value="delete" />
</form>
JS:
window.onload = function(){
document.getElementById('btn_delete').onclick = function(e){
var checkBox = document.getElementsByName('confirmDelete')[0];
if(!checkBox.checked) {
if(e && e.preventDefault) {
e.preventDefault();
}
else if(window.event && window.event.returnValue) {
window.eventReturnValue = false;
}
alert('Please check the checkbox!');
}
};
};
DEMO.
I've multiple forms with a single action, a single php page that gets called by all the forms.
How can I differentiate which form was sent to the php page?
Using a different unique input type="hidden" for each form.
HTML:
<input type="hidden" name="form_id" value="1">
<input type="hidden" name="form_id" value="2">
PHP:
$myform = $_POST["form_id"];
You can also use the submit button but note that the "value" parameter is what gets displayed to the user so you won't be able to modify it (assuming you want the same text to be displayed on every button).
<input type="submit" name="action" value="the user saw this">
PHP:
$_POST["action"] // -> "the user saw this";
Add a hidden field (action or the like) to each field, then check for it.
<form id="num1">
<input type="hidden" name="action" value="first_action" />
</form>
...and the check:
<?php
if(!empty($_REQUEST['action']) {
switch($_REQUEST['action']) {
case 'first_action':
// first action code
break;
}
}
?>
Give each submitbutton an other name or put a with different values in each form.
You can detect this from the submit button itself too, if submit has different values like:
update name, update profile, delete users...
On the submit button for each form, use different names. Something like:
<input type="submit" name="submit_1" value="Submit" />
<input type="submit" name="submit_2" value="Submit" />
<input type="submit" name="submit_3" value="Submit" />
...
Then on your PHP, you'll have:
$_POST["submit_1"]
$_POST["submit_2"]
$_POST["submit_3"]
I have read the answer to this question, to execute PHP scripts with the click of a button. But what if I have a "nested button", like this :
<?php
if(!empty($_POST['act'])) {
echo "Ready to rock!";
$someVar = "Rock n Roll";
if(!empty($_POST['act2'])) {
echo $someVar;
} else {
?>
<form method="POST" action="">
<input type="hidden" name="act2" value="run">
<input type="submit" value="Rock It!">
</form>
<?php
}
} else {
?>
<form method="POST" action="">
<input type="hidden" name="act" value="run">
<input type="submit" value="Show It!">
</form>
<?php } ?>
I heard my problem can be solved with jQuery, but I no idea.
anyone please.
To execute a script on the server you use the action property of your form:
<form method="POST" action="myscript.php">
When clicking a input type="submit" the browser will go to to action of the form surrounding the input type="submit"
Nesting is not a issue, as the browser always will look for the 'surrounding' form.
Problem is in second form, so it will never calls in this code, because it fails in first $_POST variable IF statement, because in second form you do not POST variable "act". so you need to add it
<form method="POST" action="">
<input type="hidden" name="act" value="run">
<input type="hidden" name="act2" value="run">
<input type="submit" value="Rock It!">
</form>
with this form you should see echo $someVar;
p.s. if form action property is emtpy, by default it submits form to the same php script
Just like #DTukans said here, you need the hidden field. If you would post the second form, the value of act will be lost if you are not having a hidden field with the value of act from the first form.
In php you can also check which submit button you submitted by giving the input[type="submit"] a name, such as <input type="submit" name="form2">, then you could check if you submitted that form by:
if (isset($_POST['form2'])) {}, but this is not the case here.
Use the hidden input and you will be good to go.