Using isset from a name of an input

Question

I am not sure if I am going about this correctly. I have a set of checkbox inputs. If someone selects the last check box all_users_check, I want a new form to appear where I will be listing all of the users in a drop down (haven't added the drop down yet). I thought I could do this by using the name of the input, but I am mistaken apparently as I am getting this error..

How else could I structure what I am doing so that if someone checks that option the new form displays?

<div class="user_dropdown">
<form action="">
  <input type="checkbox" name="spectator_check" value=""> Spectators<br>
  <input type="checkbox" name="member_check" value="" checked> Team Members<br>
  <input type="checkbox" name="commissioner_check" value="" checked> Commissioner(s)<br>
   <label for="all_users_check">
  <input type="checkbox" name="all_users_check" value="" checked> Individual User<br>
  </label>
</form>
</div>

<script>
$(".user_dropdown").hide();
$(".all_users_check").click(function() {
    if($(this).is(":checked")) {
        $(".user_dropdown").show();
    } else {
        $(".user_dropdown").hide();
    }
});
</script>

This is how the page looks on load. Those fields are already checked for some reason.

Shouldn't it be `isset($_GET['all_users_check']`? You also have no submit button, are you going to process this with AJAX? — chris85, Sep 11 '15 at 16:10
I wasn't trying to submit anything with this yet. I just want what is inside of the isset to show up if that checkbox is checked. I do not want the user to have to hit a submit button for the future drop down box to show up. Is AJAX the only way I can make this show with just clicking the checkbox input? — Ralph, Sep 11 '15 at 16:11
PHP is server side. Once the page is loaded it is not accessible. An AJAX request will be required, or you could do this with JS/jquery. — chris85, Sep 11 '15 at 16:14
The data will already be there though. I am just wanting to basically hide it until the user clicks that check mark. Unless this cannot be done. I don't want to have the user click a submit button just to move to the next step. — Ralph, Sep 11 '15 at 16:15
Yes, so JS will be needed, not PHP. JS runs in the browser, PHP on the server. Start here, https://api.jquery.com/category/events/form-events/. — chris85, Sep 11 '15 at 16:17
Yes, I know that, I just though with the isset statement I could check this without submission. I was mistaken. Thanks for clarifying.....Does anyone know how I could do this with JS? — Ralph, Sep 11 '15 at 16:18
@chris85 I updated my question. I am trying this with jquery and it hides the div, but doesn't display it when clicked. Also side note, for some reason whenever I load my page, half of the inputs are already checked? This was happening before as well. — Ralph, Sep 11 '15 at 16:27
Sorry, didn't mean to do the chat. Thanks that helped. I can't believe I didn't catch that. Would you know why half of my checkboxex are loading checked? — Ralph, Sep 11 '15 at 16:32

score 2 · Answer 1 · edited May 23 '17 at 12:32

isset is a language construct and can't accept anything other than a variable as indicated by this warning on the linked to manual page:

Warning isset() only works with variables as passing anything else will result in a parse error.

You are not passing in a variable to the isset function, you are passing in a constant value, basically an array with a single string all_users_check. This is not a variable because you are not assigning it to a variable name. Try this instead:

if(isset($_POST['all_users_check']))

Here the variable being passed in is the superglobal $_POST, and you are checking to see if the index all_users_check is set inside of that array.

Update

To check if an input is empty or not via javascript, take a look at this question.

score 2 · Accepted Answer · answered Sep 11 '15 at 16:38

2

Issues in your code.

.all_users_check that is looking for a class. Your element doesn't have a class so this isn't found. You can use a different selector to use the name attribute, https://api.jquery.com/attribute-equals-selector/.
This $(".user_dropdown").hide(); hides your whole form. You might want to move around your divs, or remove that altogether.
The checked attribute checks the field it is on. https://developer.mozilla.org/en-US/docs/Web/HTML/Element/input

Use the checked attribute to indicate whether this item is selected

<div class="user_dropdown">
<form>
  <input type="checkbox" name="spectator_check" value=""> Spectators<br>
  <input type="checkbox" name="member_check" value=""> Team Members<br>
  <input type="checkbox" name="commissioner_check" value=""> Commissioner(s)<br>
   <label for="all_users_check">
  <input type="checkbox" name="all_users_check" value=""> Individual User<br>
  </label>
</form>
</div>

<script>
//$(".user_dropdown").hide();
$("input[name='all_users_check']").click(function() {
    if($(this).is(":checked")) {
        $(".user_dropdown").show();
    } else {
        $(".user_dropdown").hide();
    }
});
</script>

answered Sep 11 '15 at 16:38

chris85

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I was literally just about to click submit with the exact same solution when the notification appeared! – Andrew Willis Sep 11 '15 at 16:41
@chris85 The show hide is doing what I want. My only concern now is that when I load the page, some of the check box fields are already checked. I want it to load where nothing is checked. I put a pic in my question to illustrate. – Ralph Sep 11 '15 at 17:58
Did you try the code I posted, or update your code to remove the `checked` attribute? – chris85 Sep 11 '15 at 17:59
The checkbox issue with the inputs already being checked was happening before I added the jquery. I want that div to stay hidden until I click the checkbox, so how I had it works better. – Ralph Sep 11 '15 at 18:14
Yea, my comment and answer aren't referring to jquery with the `checked` attribute. Look at my HTML and yours ` Commissioner(s)
` <-- see it `value="" **checked**`. – chris85 Sep 11 '15 at 18:22
Oh, duh. Yep, that did it. Not sure how those got in there to be honest. Thanks for the help!! – Ralph Sep 11 '15 at 18:27

jqheart · Answer 3 · 2015-09-11T16:33:28.770

0

Try using this script, you have set the state of check boxes as checked by default.

<div class="user_dropdown">
<form action="">
  <input type="checkbox" name="spectator_check" value=""> Spectators<br>
  <input type="checkbox" name="member_check" value=""> Team Members<br><!-- removed 'checked' from this line -->
  <input type="checkbox" name="commissioner_check" value=""> Commissioner(s)<br><!-- removed 'checked' from this line -->
   <label for="all_users_check">
  <input type="checkbox" name="all_users_check" value="" > Individual User<br> <!-- removed 'checked' from this line -->
  </label>
</form>
</div>

<script>
$(".user_dropdown").hide();
$(".all_users_check").click(function() {
    if($(this).is(":checked")) {
        $(".user_dropdown").show();
    } else {
        $(".user_dropdown").hide();
    }
});
</script>

For the other issue of showing the hidden section again, try whether class all_users_check is visible to click.

edited Sep 11 '15 at 16:33

answered Sep 11 '15 at 16:11

jqheart

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5
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This doesn't throw any errors and hides what is in the isset, but if I check the `all_users_check` it doesn't do anything. – Ralph Sep 11 '15 at 16:13
@Ralph, if you have clicked on all_user_check and did not do any form submission, then you cannot find any difference. I think you are expecting a client side solution based on the checked state of the checkbox. – jqheart Sep 11 '15 at 16:16
@Ralph, check the solution above – jqheart Sep 11 '15 at 16:33
There is no `all_users_check` class. – chris85 Sep 11 '15 at 16:40

Using isset from a name of an input

3 Answers3