I am trying to update records with a image button. here is my code:
<?php
if(isset($_POST['unstar'])) {
echo "good!";
}
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' />
</form>
When I click on the image, it does not echo "good". what am I doing wrong?
I just tested your code and apparently when you click on the image button, php generates the following post values unstar_x and unstar_y, both of which correspond to the x and y coordinates of where you clicked the button.
Try changing it to the following and you'll see what I mean:
<?php
print_r($_POST);
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' />
</form>
EDIT
So if you just need to see if the button was pressed to submit the form, you can just do the following, which just checks if unstar_x or unstar_y was set.
<?php
print_r($_POST);
if(isset($_POST['unstar_x']) || isset($_POST['unstar_y'])){
echo "Good";
}
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' value="testing" />
</form>
You should note that the output of $_POST will not include the value Testing. At least on my system, I don't get a value when I submit the image button.
The problem is that you're not giving unstar a value, so it won't be detected as Set by the function.
Related
I'm using CodeIgniter form_validation and
Everything works good except the original notification still appears in the top right.
Can anybody advise me how to delete the original notification?
I've inserted the PHP code into to view page.
<?php echo form_error('uname'); ?>
The form_error message appears the way I want but the original error message still appears in the top right, and I don't know how to delete that.
EDIT 01/08/19
Below is edited coding I've taken from the CI UserGuide.
Controller
<?php
class Form extends CI_Controller
{
public function index()
{
$this->load->helper(array('form', 'url'));
$this->load->library('form_validation');
$this->form_validation->set_rules('username', 'Username', 'required');
$this->form_validation->set_rules('password', 'Password', 'required', array('required' => 'You must provide a %s.'));
if ($this->form_validation->run() == FALSE)
{
$this->load->view('myform');
}
else
{
$this->load->view('formsuccess');
}
}
}
View
<html>
<head>
<title>My Form</title>
</head>
<body>
<?php echo validation_errors(); ?>
<?php echo form_open('form'); ?>
<h5>Username</h5>
<input type="text" name="username" value="" size="50" />
<h5>Password</h5>
<input type="text" name="password" value="" size="50" />
<div><input type="submit" value="Submit" /></div></form>
</body>
</html>
If the above is loaded the result will be the Username Text Box in the top left corner, and the Password Text Box below it.
If the Submit button is clicked without any content in the text boxes the following notifications appear in the top left corner and above the Text Boxes.
You must provide a Username.
You must provide a Password.
However, if I change the view page to the following, the same notifications STILL APPEAR IN THE TOP LEFT CORNER, as well as next to the Text Boxes, so in effect each notification appears twice.
<html>
<head>
<title>My Form</title>
</head>
<body>
<?php echo validation_errors(); ?>
<?php echo form_open('form'); ?>
<h5>Username</h5>
<input type="text" name="username" value="" size="50" />
<?php echo form_error('username'); ?>
<h5>Password</h5>
<input type="text" name="password" value="" size="50" />
<?php echo form_error('password'); ?>
<div><input type="submit" value="Submit" /></div></form>
</body>
</html>
It is the notifications that still appear in the top left corner, that I want to delete.
I can't find any coding that positions the notifications so I guess that comes from somewhere within the CI system.
I can position the form_error notifications by using css coding.
I found the problem.
<?php echo form_error('username'); ?> // When this is inserted.
<?php echo validation_errors(); ?> //This must be deleted.
Or put another way the coding above provides the notifications at the top right.
I have to create a form with a submit bottom following a link
<form action="http://domain/**(((MY TEXT INPUT VALUE)))**.htm">
<input type="text" name="verb">
<input type="submit" value="Conjugate">
</form>
something like this.
please note that every link should be different.
I also want that the new page be opened in a new tab/window
could you please help me, and also make changes to the form code if there is sth under newer standards. Thank you!
You need to use javascript to get the TEXTBOX value and then place it into the form action.
You can create the submit button with an onclickevent.
Or you can use jQuery
$('#btnSubmit').click(function(){
var sTextValue = $("#MyText").val();
$('#MyForm').attr('action', 'htttp://domain/' + sTextValue + '.htm');
$('#MyForm').submit();
});
And the HTML
<form id="MyForm" action="">
<input id="MyText" type="text" name="verb">
<input id="btnSubmit" type="button" value="Conjugate">
</form>
There are many ways to accomplish this. That's just one of them.
<form action="http://domain/**(((MY TEXT INPUT VALUE)))**.htm" id="btnForm">
<input type="text" name="verb" onchange='javascript:document.getElementById("btnForm").action = "http://domain/"+ this.value +".htm"'>
<input type="submit" value="Conjugate" >
</form
This would update as soon you type the text. It wouldn't require jquery. it makes use of onchange event handler of input type text
<form action="http://domain/**(((MY TEXT INPUT VALUE)))**.htm" id="btnForm">
<input type="text" name="verb" onchange='updateFormAction(this.value)'>
<input type="submit" value="Conjugate" >
</form>
<script type="text/javascript">
function updateFormAction(value){
var btnForm = document.getElementById("btnForm");
btnForm.action = "http://domain/"+ value +".htm";
}
</script>
This is more explanatory form. Its based on onchange event handler for text types.
In this form I want to get the value of the text box Enter Techs name: in to the URL but my form method is post.So that my url would look like
http://localhost/cs_3/index.php?page=chat&tech=
Is there any way I could do it.Can't use get method because I don't want the message to be seen in the URL.
<form method="post" >
<label>Enter Username:<input type="text" name="sender"/></label>
<?php if($user['Level'] == 3){?>
<label>Enter Clients name:<input type="text" name="tech"/></label>
<?php }else{ ?>
<label>Enter Techs name:<input type="text" name="tech" /></label>
<?php } ?>
<label>Enter Message:<textarea name="message" rows="8" cols="70"></textarea></label>
<div id="submit"><input type="submit" name="send" value="Send Message"></div>
</form>
You may use action param at your form.
Change it on key press in tech input by javascript (easy with jquery).
I am used to get the echo message of the php script using the jquery ajax success function now I use a plugin but I dont know how to use it
my approach is this but definitely do not work how should the right way? It just go directly to my php script
$(function() {
$('form').ajaxForm(function() {
success: function(data){
$('.new-profile-pic').html(data);
}
});
});
HTML:
<p class="new-profile-pic">
<!--Image should be here-->
</p>
<form enctype="multipart/form-data" action="upload-image.php" method="post">
<input type="hidden" name="MAX_FILE_SIZE" value="100000">
<input class="profile-pic-name" name="uploadedfile" type="file">
<input type="submit">
</form>
PHP:
<?php
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo '<img alt="" src="'.$target_path.'">';
} else{
echo "There was an error uploading the file, please try again!";
}
?>
You need to prevent the default action from occuring.
I believe this post has the answer you are looking for:
How do I submit a form using jquery (ajaxForm plugin) to an external site but not reload to a different page?
You need to use .preventDefault() on the submit event, or get it to return false on the form submit, e.g.:
<form enctype="multipart/form-data" action="upload-image.php" method="post" onsubmit="return false;">
I am attempting to post form data to my app using Facebook. My form is integrated with a dialog and I am using form.submit() to submit the form. Please see my code. The post data does not appear on the backend (e.g. load-cargo-radio, city-id, train-id), but the fb_sig* data does. Does anyone know if there are any caveats regarding posting form data with Facebook? Thanks!
Note: I have tried both using my server domain and my fb callback url for the form action. Neither works.
<div id="action_prompt">
Loading cargo...
</div>
<fb:js_string var="fbjs_load_cargo_select">
<div id="load_cargo_select">
<form id="load_cargo_select_form" action="http://railsacrosseurope.com/turn/load_cargo_select" method="POST">
<p>Your train has stopped in the city of Arhus.</p>
<p>Arhus produces the following goods:</p>
<ul>
<li>Dairy</li>
</ul>
<p>Your train is hauling the following goods:</p>
<ul>
<li>Timber</li>
</ul>
<p>What would you like to do?</p>
<input type="radio" id="load_cargo_radio" value="1">Load new goods</input>
<input type="radio" id="load_cargo_radio" value="2">Discard existing goods</input>
<input type="hidden" id="city_id" value="3" />
<input type="hidden" id="train_id" value="15" />
<input type="submit" id="submit" value="Submit" />
</form>
</div>
</fb:js_string>
.
.
.
<script type="text/javascript">
var dialog = new Dialog().showChoice('Load Cargo', fbjs_load_cargo_select, 'Okay', 'Pass');
dialog.onconfirm = function() {
// Submit the form if it exists, then hide the dialog.
frm = document.getElementById('load_cargo_select_form');
if (frm) { frm.submit(); }
dialog.hide();
};
dialog.oncancel = function() {
form = document.getElementById('redirect_form');
form.setAction('http://apps.facebook.com/rails_across_europe/turn/move_trains_auto/');
form.submit();
}
</script>
[/code]
The problem was not with Facebook. The problem was that I left out the "name" attribute for my input elements.