HTML Input Checkbox Value Checked - php

I am processing an html form using php. My question is specifically about input type checkbox:
<input type="checkbox" name="checkme" value="checked" <***php echo $data['checkme']; ***> >
This works for me because, on load, $data['checkme'] = ""; and on error, $data['checkme'] = "checked". I have searched quite a bit regarding this, and there are plenty of suggestions for setting the value of the checkbox input. But not in this way (that I found). This works, but I want to make sure I am not creating a problem that I don't foresee.
My question: Is this good practice?

Related

html checkbox onChange methode in php

Here is my problem, I know only html and php and I have no clue about how to use javascript... And all the solutions about my problems seems to be resolved in javascript and I wondered if there was a way to do it with php so that I could understand what I do.
I want to put a checkbox on the corner of my page (for instance "hide information") that would refresh the page automatically when checked and that would hide information on the page.
What I currently do is :
<?php
if(isset($_GET['condition']))
$_SESSION['condition'] = true;
else
$_SESSION['condition'] = false;
?>
...
...
<form>
<input type="checkbox" name="hide" value="1" onChange="this.form.submit()" <?php if($_SESSION['hide']) echo "checked";?> > hide information
</form>
I am facing two problems :
the first one is that I want the checkbox to stay checked/unchecked when the page is refreshed.. I solve that poorly with my php code, but there surely exists something better to do that.
When the box is checked, the page is refresh with only "hide=1" as an url argument, but I would love to keep all the other arguments that were there before the page was refreshed. Is there a way to refresh the page and keep all the arguments while knowing that the box is checked/unchecked ?
thanks for your help, and sorry for my poor knowledge.
Regarding the second point of your question you can move the POST (or GET) array to the SESSION one and back with the following code:
if(isset($_POST) & count($_POST)) { $_SESSION['post'] = $_POST; }
if(isset($_SESSION['post']) && count($_SESSION['post'])) { $_POST = $_SESSION['post']; }
I use this to do exactly the same. When I reload the page I keep the posted values.
Regarding the first point you are already on the right path.
I don't see anything wrong with how you've tried to solve problem 1.
Regarding the URL problem 2, either put session_start(); at the top of the page to get the session to work correctly.
Alternatively have hidden inputs in this pages' form and echo out the previous pages' POST values.
<form action="" method="post">
<input type="hidden" name="condition" value="<?php echo $_POST['condition']; ?>" />
<!-- have hidden inputs from previous page here, plus your checkbox to retain post values from the previous page -->
</form>
Although I'd recommend POST for this, you can do GET although it gets a bit messy like so:
<form action="thispage.php?condition=<?php echo $_GET['condition'];?>" />

echo back $array[] elements to html form on Submit

After alot of digging around some very informative posts and info to try and find out how to solve this issue I thought I would ask around to see if anyone has any pointers.
I have an html form with various inputs (checkboxes, text boxes etc...). Each input section has its own submit or 'Upload' button. On Upload a php script is called and various bits of processing is done before data is sent over a pipe to a Python script for further stuff.
I am currently echoing back input variables to the form on submission so that the html page does not refresh (or should I say the inputted data is not lost to the users view) on an Upload event, however, I now have to do the same for a bunch of checkboxes and text boxes the values of which are stored in an array. The code I have written so far is as follows (I am new to both php and html so please excuse the inefficiency that I'm sure is obvious)
html/php
<margin>CH1</margin><input type="checkbox"name="ANout[]"value="AN1_OUT"
<?php if(in_array('AN1_OUT',$_POST['ANout']))echo'checked';?>>
Voltage<input type="text"size="5"name="ANout[]"
value="<?php $ANout[$i]=$_POST['ANout'];
if(!empty($ANout[$i]))echo"$ANout[$i]";?>">
<br>
The code above works fine for the checkboxes which happily remain after an Upload button is pressed but not for the array. When the Upload event occurs I simply get 'Array' written in the text box. I have tried existing code I have written to echo back other text input in the form (see below) and which works but these are for sole entries, not arrays. I have tried various configurations of syntax but I always seem to get the same result.
Working Code:
<margin>Duty Cyle</margin><input type="text"name="PWM1DC"size="3"
value="<?php $PWM1DC = $_POST['PWM1DC'];
if(!empty($PWM1DC))echo "PWM1DC";?>">
<br>
I'm sure it is something straightforward but I have been fiddling and staring at it for ages and can't seem to find the problem.
You are getting "Array", because you are trying to print out variable of type Array.
You probably want to give your fields separate names or indexes and do something like this:
<form method="post">
<input type="checkbox" name="ANout[1]" value="AN1_OUT"
<?php if(isset($_POST['ANout']) && in_array('AN1_OUT',$_POST['ANout']))echo'checked';?>>
Voltage<input type="text"size="5"name="ANout[2]"
value="<?php if(isset($_POST['ANout']) && !empty($_POST['ANout'][2])) echo $_POST['ANout'][2]; ?>">
<input type="submit" value="ok">
</form>
(Just added form tags, submit button and isset checks to show working example.)

Passing checkbox state to PHP

<input type="hidden" name="check_box_1" value="0" />
<input type="checkbox" name="check_box_1" value="1" />
This works fine, however when you click on submit, and the checkbox is ticked, it passes BOTH the hidden value and the original checkbox value to the $_POST variable in php, can this be avoided?
I have the hidden value there, so that unticked checkboxes are passed to the $_POST variable as well as the ticked ones.
The better approach is to remove the hidden field, and simply have a check in PHP:
if ($_POST['check_box_1']=='1') { /*Do something for ticked*/ }
else { /*Do something for unticked*/ }
You shouldn't need the hidden field. You should in fact not trust any of the form fields sent in the first place. What this means is that you cannot make code which takes the sent fields and trust them to send the correct data (which I assume you do now).
What you should do is to handle all fields you expect to get. That way if you don't get the checkbox value you can still handle that as if it was unticked. Then you also get the added inherent feature of throwing away form data you don't expect in the first place.
No, it will pass all the form data, whatever it is. The right way to do this is not to set the checkbox via a hidden field but to set the checkbox with whatever its state actually is!
I mean... why are you adding the hidden field to begin with?
Your PHP is receiving two fields named check_box_1, and last time I checked there was no way to guarantee that the params would get read into the REQUEST hash in the exact same order as you sent them, so, there's no way to tell which one will arrive last (that's the one whose value will get set). So... this is not the right approach to whatever problem you're trying to solve here.
Welcome to Stack, btw! If you find answers useful or helpful, make sure to mark them as correct and vote them up.
That's normal.
They must be both type="checkbox" to pass only 1 value.
If you want to get only 1 in any cases you can do:
<input type="checkbox" style="display:none;" name="check_box_1" value="0">
Make sure the first input field is of type Checkbox, or else it won't behave like one.
<input type="checkbox" name="check_box_0" value="0" />
<input type="checkbox" name="check_box_1" value="1" />
Everything is working normal with your code so far.
I'm assuming you are creating the hidden field so that 0 is passed to the server when the checkbox is not checked. The problem is that they both get passed when the check box is checked.
As Death said, the way you should be doing it is with a single checkbox and then checking if the value has been sent to the server or not. That's just how checkboxes work.
If you want to have a default set then you will have to handle all that on the server side based on weather the checkbox has a value.
For example:
$myValue = "";
if(isset($_POST['check_box_1']))
{
$myValue=$_POST['check_box_1'];
}
else
{
$myValue="0";
}

Language selection using images in PHP

I have found an open source PHP script that uses a select menu to choose the language.
In the code session makes use of a loop to fill in the select menu, as above...
$la = 0;
foreach($trans['languages'] as $short=>$langname) {
$params['LANGSEL'][$la]['LANG_SHORT'] = $short;
$params['LANGSEL'][$la]['LANG_LONG'] = $langname;
$la++;
}
In the php template creates the select menu like that...
<td><select class="select" name="lang"><#FOR LANGSEL#>
<option value="<#LANG_SHORT#>"><#LANG_LONG#></option>
<#/FOR LANGSEL#></select></td>
So this code works fine but i find it kinda ugly so i am trying to make an image input instead
So i thought something like that would work..
<input type="image" name="lang" value="http://localhost/index.php?lang=en" src=" <#IMAGES_DIRECTORY#>/english.png"></td>
<input type="image" name="lang" value="http://localhost/index.php?lang=it" src=" <#IMAGES_DIRECTORY#>/italian.png"></td>
But nothing changes, the page always shows up in italian that is the default language..
Thanks in advance from a newbie that is struggling to learn some html and php.
The value of your name field should be <#LANG_SHORT#>. You don't say what it looks like after being processed but I'm pretty sure it's something like en or it. However, you provide a URL. You also prepend several white spaces to the image's URL.
This will probably work:
<input type="image" name="lang" value="<#LANG_SHORT#>" src="<#IMAGES_DIRECTORY#>/english.png"></td>
Remember to test it in Internet Explorer. It's traditionally had several problem with <input type="image"> elements.
The problem here is that you have two images with different names, whereas you need to only have one form element, called name, whose value is the correct <#LANG_SHORT#> string. In this regard, select and radio form elements are perfectly suited to the job, whereas inputs are not.
It also seems unlikely to me that the form element really has a value of http://localhost/index.php?lang=en. Isn't that merely the URL that results from changing the language? It seems more likely that the proper value for your form fields is just en/it.
Ultimately, I reckon you're going to need a hidden form field, and some Javascript on your images to set that field's value when required. And be aware that the usability/accessibility of your site just went from [potentially] high to [definitely] very low.

codeigniter set_checkbox problem on an optional input

i have an optional checkbox called checkall
<input name="checkall" type="checkbox" value="ON" <?php echo set_checkbox('checkall', 'ON'); ?> />
i'm sure that the form is submitting that
if i give it a validation rule $this->form_validation->set_rules('checkall', 'Checkall', 'required');it works, but without a rule nothing worked out !
did i miss something? i think form helper doesn't require that for this function to work right ?
I think you're talking about the value being persisted without validation rules. This is STILL a problem in CI 2.x if I recall correctly, and jbreitwiser's patch from January 2010 is still necessary:
http://codeigniter.com/forums/viewthread/96617/P15/#689642
If this is still a problem in CI 2.x it is completely absurd, and I totally agree. But that patch will solve your problem.
If I understand you correctly your checkbox is not submitting with the form?
If that is the case, I had a fun time with this question on this thread HERE
PHP wants you to check to see if a checkbox is set or not by verifying whether or not there is a corresponding element in the POST array. If the checkbox was checked, there will be an element of the same name in the POST array (that element will have a NULL value), if the checkbox was NOT checked, then there will be NO matching element in the POST array.
The code would look something like this:
Your input element remains the same --
<input name="checkall" type="checkbox" value="ON" <?php echo set_checkbox('checkall', 'ON'); ?> />
Postback Handler Page gets a new way to validate a checkbox --
if(isset($_POST["checkall"])
{
$checkall = TRUE;
}
else
{
$checkall = FALSE;
}
Hopefully I helped, its late and your question is sparse on details.
Regards

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