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future-architect
future-architect
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creating a button

May 25, 2012 5:51 AM

Is there any way to convert this to a submit button?  here is my code:

<a class="button small red" href="#" id="submitUpload"><span>Submit</span></a>

 

I followed this tutorial: http://acrisdesign.com/2011/06/create-a-multi-color-and-size-css3-butt ons/#

1132 Views   12 Replies   Latest reply: future-architect, May 27, 2012 9:23 AM
 
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  • MurraySummers
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    1. MurraySummers,
    May 25, 2012 11:10 AM   in reply to future-architect
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    <a class="button small red" href="#" id="submitUpload" onclick="document.forms['formName'].submit();"><span>Submit</span></a >

     
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  • future-architect
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    2. future-architect,
    May 25, 2012 2:30 PM   in reply to MurraySummers
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    that would only work with javascript enabled, right?  so should i create another button for users that have js disabled?

     
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  • MurraySummers
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    3. MurraySummers,
    May 26, 2012 9:13 AM   in reply to future-architect
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    Yes, that would only work with javascript enabled.  If you want to cover both eventualities, you could put a normal submit button inside a pair of <noscript> tags - then you'd have to figure out how to hide the other button when there is no javascript enabled.

     
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  • future-architect
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    4. future-architect,
    May 26, 2012 10:53 AM   in reply to MurraySummers
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    So if I had this form: <form enctype="multipart/form-data" name="fileUpload" id="fileUpload" method="post"> <a class="button small red" href="#" id="fileUpload" onclick="document.forms['submitUpload'].submit();"><span>Submit</span ></a >...

    when I do the "document.forms...", will it execute this?: if ($_POST['fileUpload']) {

    I seem to have issues actually submitting it. 

    And yes, that <noscript> is a good idea.  Will probably use that, as I have before.  All I do is set the visibility to none for the button, then if javascript is enabled, toggle the visibility.

     
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  • MurraySummers
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    5. MurraySummers,
    May 26, 2012 12:20 PM   in reply to future-architect
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    Is 'fileUpload' the name of your 'file' field?  You would have to test for that this way -

     

    if($_FILES['fileUpload']) {

     
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  • future-architect
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    6. future-architect,
    May 26, 2012 3:42 PM   in reply to MurraySummers
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    no - i am just concerned with getting the form submitted.  post fileUpload checks if the form was submitted.  I tried running the code above, but it didn't seem to do anything.  with the submit button, you would set the name to fileUpload to run the php validation.  I am not sure here how to do this though with a link.  I understand the code, and it should work.  I must me doing something wrong. 

     
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  • MurraySummers
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    7. MurraySummers,
    May 27, 2012 5:15 AM   in reply to future-architect
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    Sorry for my previous goofy answer.  The data was staring me in the face and I missed it!

     

    Try making your code look like this -

     

    onclick="document.forms['fileUpload'].submit();"

     

    You need the name of the form being submitted in that location, not the name of a field.

     
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  • future-architect
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    8. future-architect,
    May 27, 2012 7:53 AM   in reply to MurraySummers
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    so I have if ($_POST['fileUpload']) {...

    and

    <form enctype="multipart/form-data" name="fileUpload" id="fileUpload" method="post">

    and

    <a class="button small red" href="#" onclick="document.forms['fileUpload'].submit();"><span>Submit</span>< /a >

     

    but it does not do anything.  Is there anything else that I could be missing?

     
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  • MurraySummers
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    9. MurraySummers,
    May 27, 2012 8:02 AM   in reply to future-architect
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    Now you've confused me.  The PHP would have to be conditional on some element that is present in the $_POST array.  The form name is not. So, if(isset($_POST['form_field_name']) && $_POST['form_field_name'] == 'some_value') { ?>

     

    The javascript would submit based on the form's name attribute value, i.e., document.forms['fileUpload'].submit();

     
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  • future-architect
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    10. future-architect,
    May 27, 2012 8:30 AM   in reply to MurraySummers
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    oh, i see.  maybe i'm confusing myself.  what would i use to start off the php code so that it does not run every time the page loads? 

     
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  • MurraySummers
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    11. MurraySummers,
    May 27, 2012 8:40 AM   in reply to future-architect
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    I would use this -

     

    <?php if (array_key_exists('some_field_name', $_POST) {

    //processing code follows

    ...

    }

    ?>

     

    So far, we don't know the name of any of your form's fields - I usually use the name of the submit button, but you could also have a hidden or visible field with a name that you use here.

     
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  • future-architect
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    12. future-architect,
    May 27, 2012 9:23 AM   in reply to MurraySummers
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    oh yea - if($_POST) - forgot!  thanks. 

     
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