this is my first time to use codeigniter to make a web site.
My problem is as follow:
My view file contains:
<label class="control-label ">Insert Number of photos : </label>
<?php echo form_open('project_controller/no_of_pics'); ?>
<input name="no_pics" placeholder="Insert a number" type="text" value="<?php echo set_value('no_pics'); ?>" >
<input type='submit' name='submit' value='Insert' />
<?php echo form_close() ?>
<?php $no_pic=''; ?>
<?php echo $no_pic; ?>
<br><br>`
My controller file contains:
public function no_of_pics(){
$this->load->view('control_panel');
$number=$this->input->post('no_pics');
$hoho = array('no_pic' => $number );
$this->load->view('control_panel',$hoho);
}
and I wrote that line the .htaccess file:
$route['project_controller/no_of_pics'] = "/project_controller/no_of_pics";
my problem is that I can't get the no_pics back to the view, it is supposed to check the number through validation rules in the controller but I want firstly to get the number back to the view file
when I wrote the code above I get the number back but in a plain page of the view file without any styling and in the top the number I inserted in the input form not even where I want to use.
3-01-2015
I think I knew what the problem is, but I still don't know how to fix it . I have an other method in that controller
"project_controller" is:
public function project($page = 'home'){
if ( ! file_exists ('application/views/'.$page.'.php')){
show_404();
}
$this->load->view($page);
}
and I wrote in the route.php file:
$route['(:any)'] = 'project_controller/project/$1';
so I think the problem is "function no_of_pics()" can not view the page properly when sending a variable to it as function project() is responsible for that , so what is the solution now?
You must supply the field name via the first parameter of the function. The second (optional) parameter allows you to set a value for the form.check here
set_value('no_pics',$no_pic);
You can try this technique
$hoho['no_pic'] =$number;
$this->load->view('control_panel',$hoho);
I think then you find it on view page.
You need to remove this line from view file
<?php $no_pic=''; ?>
It reset the variable that passes from the controller.
Related
I am making a website in CodeIgniter and for one of these pages I need to insert information into a database, however every time I enter information into my form and submit it, the page refreshes like it had been submitted but nothing enters the database.
Controller:
public function insertjob()
{
$this->load->helper('form');
$data['title']="Add a new job";
$this->load->view("insertjob", $data);
}
public function addingjob()
{
$jobtype=$this->input->post('jobtype');
$jobinfo=$this->input->post('jobinfo');
$this->load->model("cmodel");
if($this->cmodel->addjob($jobtype, $jobinfo)){
$data['msg']="New job addition successful";
}else{
$data['msg']="There was an error please try again";
}
$this->load->view("confirmation",$data);
Model:
function addjob($jobtype,$jobinfo)
{
$newjob=array("jobtype"=>$jobtype,"jobinfo"=>$jobinfo);
return $this->db->insert('clientjobs', $newjob); exit;
View:
</p>
<?php
echo form_open('client/insertjob');
echo form_label('Job:', 'Job');
echo form_input('jobtype');
echo form_label('Job information:', 'Job information');
echo form_input('jobinfo');
echo form_submit('Add job', 'Submit Post!');
echo form_close();
?>
Try removing the exit from your model:
function addjob($jobtype,$jobinfo)
{
$newjob=array("jobtype"=>$jobtype,"jobinfo"=>$jobinfo);
return $this->db->insert('clientjobs', $newjob);
}
It's not neccessary and could be breaking the database class, as well as halting any execution for the application.
Here's your problem:
echo form_open('client/insertjob');
If you look at your HTML code in your browser, you'll see something like this:
<form action="client/insertjob">
There will probably be a whole bunch of other attributes in your form tag - they're not important for this answer.
That action attribute is telling the browser where to go after you click submit. Where is it going? Back to the insertjob method. But it needs to go to your addingjob method - that's where the database update is actually being done. So change the form_open call to:
echo form_open('client/addingjob');
As I see your are using 2 controller functions for posting, page 1 to page 2. You have error on form open you should post your data to addingjob not insertjob.
echo form_open('client/addingjob');
will fix your issue but I highly recommend you to use, one controller for form submit. Below code will send post to same url. And you could add some attributes on it.
<?php
$attributes = array('class' => 'form-horizontal');
echo form_open($this->uri->uri_string(),$attributes); ?>
It seems my code is correct, however the posted variables in the form will not echo in the update user settings page in the form. I have echoed the posted ids from the input in the database but I cannot get the variables to show.
I have done this in Codeigniter fine but am trying to do it in pure php with my own framework.
$users = new Users($db); comes from my init.php file that is called at the beginning of the file below.
when I
<?php var_dump($user['first_name'])?>
I get Null
<input type="text" name="first_name" value="<?php if (isset($_POST['first_name']) )
{echo htmlentities(strip_tags($_POST['first_name']));} else { echo
$user['first_name']; }?>">
Hoi Stephen,
Try print_r($_POST["first_name"]); instead of var_dump();
or just for all:
print_r($_POST);
best regards ....
add this at the top of your html page
#extract($_REQUEST);
and put is just to check and after checking remove the below line
print_r($_REQUEST);
hope this help .
After the user hits the submit button, how do I reset the drop down menu to the "blank" option of the the menu? I am using a MVC set up with php and HTML, and the concrete5 library. THANKS IN ADVANCE! Here is what I have so far:
Controller code:
public function authorize() {
$selectHost = array('' => '');
foreach ($this->host->getHostInfo() as $row) {
if (isset($row['HARDWARE_id'])) {
$selectHost[$row['id']] = $row['host'];
}
}
$this->set('selectHost',$selectHost);
$postCheck=array(array('param' => 'host',
'check' => '^[0-9]{1,50}$',
'error_msg' => 'Invalid Host ID'),
);
$post = scrub($_POST,$postCheck);
if (isset($post['host'])) {
$this->host->authorize($post['host']);
$this->set('test', "<p> The host has successfully been authorized.</p>");
}
else{
$this->set('failed', "<p>Invalid Host ID</p>");
}
}
view code:
<form method="post" enctype="multipart/form-data" action="<?=$this->action('authorize')?>">
<?php
$form = Loader::helper('form');
print $form->label('host', 'Host: ');
print $form->select('host', $selectHost);
?>
<?php
print $form->submit('Submit','Submit');
echo $test;
echo $failed;
?>
</form>
I'm pretty positive that there's no way to override C5's desire to take the POSTed value and use that as the default. Even if, as TWR suggested, you specify a value. (This is typically a good thing, because if the page is POSTed to and there's an error, you don't want to show the value from the database; you want to show what was in the POST).
You can override the form helper pretty easily.
However, I'd suggest that you do a redirect after successful submission (don't redirect after an error -- then the POSTed value will be useful) to a page. You can redirect to another page, or the same one, ideally with a confirmation message. See https://github.com/concrete5/concrete5/blob/master/web/concrete/core/controllers/single_pages/dashboard/blocks/stacks.php#L23 for an example of using redirect.
This is the best practice for your problem but also because, with your current code, if somebody hits refresh, it'll rePOST the data and reauthorize the host.
i think you could extend the form tag with a (javascript) onsubmit action which does the reset.
Since it's a form submit, you just want to change the value of the "drop box"/select in your view. After a submit, you'll have a fresh page load; so, in every case you'll want to display the default value, and not the current value of $selectHost
In your view, change this
print $form->select('host', $selectHost);
to this
print $form->select('host', $selectHost, null);
According to http://www.concrete5.org/documentation/developers/forms/standard-widgets
If the problem is that the Concrete5 form helpers are not behaving as you want, then just don't use them -- instead just use regular HTML form inputs instead.
<form method="post" enctype="multipart/form-data" action="<?=$this->action('authorize')?>">
<label for="host">Host: </label>
<select id="host" name="host">
<?php foreach ($selectHost as $value => $text): ?>
<option value="<?php echo htmlentities($value); ?>"><?php echo htmlentities($text); ?></option>
<?php endforeach; ?>
</select>
<input type="submit" value="Submit" />
<?php
echo $test;
echo $failed;
?>
</form>
I have a file called admin.php in which I have a button with the name send. What I want to do is when I click it, to make visible a link on the user's page, user.php. How can I do this?
I have a file with all my functions called functions.php in which I have a function called onSubmit($var); I initialize the variable $var is admin.php with the value $_POST['send'] but when I call the function in the file user.php I have no way of telling him who the variable $var is so I get 'undefined index'.
Is there another way to do this?
EDIT Added code
This is admin.php
<input type="button" name="send" value="Submit" /><br/>
require 'functions.php';
$posted = $_POST['send'];
onSubmit($posted);
This is user.php
require 'functions.php';
onSubmit($var); //here it says undefined index var because it doesn't know who the variable is
if($isSent == 1) {
<a style="visibility:visible;" href="test3.html" id="test3">Test3</a> <br/>
}
And this is functions.php
global $isSent;
function onSubmit($var) {
if(isset($var)) {
$isSent = 1;
}
}
Basically you need to use sessions like below:
if(isset($_SESSION['makeVisible']) && $_SESSION['makeVisible'] == true){
echo '<button>Button init</button>'; //you could also use html like the comment below.
}
/*
if(condition){?> <!-- this is now html --> <button>Button init</button><?}
*/
Then to set this variable on your admin page use:
if(isset($_POST['submitButton'])){
$_SESSION['makeVisible'] == true;
}
You'll also need a form for this method to work but there are other methods but I prefer this one.
<form name="buttonMakerThing" method="POST">
<input name="submitButton" value="Make button init Visible" type="submit"/>
</form>
Without an action the form defaults to 'POSTING' the form information to the current page. Making the condition if(isset($_POST)) return true.
You will need to add a $_SESSION declaration at the top of every php page you have on your site for this to work. It MUST go on the very first line of every page! for example:
01: | <?php session_start();
02: |//rest of script;
Please look more into $_SESSIONS for unnsetting/destroying your sessions and more uses for them :) http://php.net/manual/en/reserved.variables.session.php
Right I've done a bit of research on Caching and this is what I've come up with. It might not be 100% correct but it's a start as like I've said I've never tried it myself lol
In your admin.php I'd put this function in:
if(isset($_POST['send'])){
if($enabled == true){
$enabled == false;
}
else{
$enabled == true;
}
apc_add('enabled',$enabled);
}
Now to 'get' our $enabled var:
$enabled = apc_fetch('enabled');
Then to check the the var within your client page:
if($enabled == true){
echo ' button';
}
Now the only things I haven't fully looked at is the security of the apc_ function and the client usage. I believe it works for all clients of the server but I'm not 100% certain. Here the php manual to give better examples.
This is the method I was thinking of. But again I'm not sure on the security of it but I'm sure you can find something to keep it secure. The video is actually is tutorial for a Youtube API. But he does cover saving a variable to a cache text file which should be of use to you :)
If you have functions.php which defines functions, simply include it in admin.php file and then you can call the function from there and also pass value.
In my cakephp form I have following code
<p> <?php echo $form->input('option[]',array('size'=>13)); ?> </p>
<p> <?php echo $form->input('option[]',array('size'=>13)); ?> </p>
<p> <?php echo $form->input('option[]',array('size'=>13)); ?> </p>
<p> <?php echo $form->input('option[]',array('size'=>13)); ?> </p>
I am trying to get values from a set of input text boxes, the number of text boxes can be set by the user, so cant give individual names of each text box, but How can I get values from my controller to insert data to db table
Thank you
You can leave the form as it is (and use suggestions from #Wizzard and #Lee), but the best practice is to use an incrementing variable to construct the list. i.e.:
for($i=0;$i<$option_number;$i++){
echo $form->input("MyModel.{$i}.option");
}
This way your variable after posting the form will look like:
data[MyModel][0][option] = 'the value'
dataMyModel[option] = 'the value'
data[MyModel][2][option] = 'the value'
... and so on...
In the controller you can access the posted data by:
print_r($this->data);
Take a look saveAll() (search for saveAll in your browser and look for suggested data structure)
your input fields are all named the same thing: option[]. This is good. It causes php to automatically turn them into an array when the request is loaded in. So you can get them in your CakePHP controller like this:
$this->params['form']['option'][0]
$this->params['form']['option'][1]
... and so on ...
Pretty sure they're in the array $this->params['form'] in the controller.. or $this->data
In the method of your controller, do a var_dump($this); and you'll see where they show up