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I'm having issues posting JS object into a PHP page on submit and then store it into mySQL db.

Here's the script:

var idToDb = [];
var nameToDb = [];
var friendToDb ={};
$('.btn-add').click(function(e){
    e.preventDefault();
    $(this).toggleClass('btn-info btn-success');
    $(this).toggleClass('friend-selected');
    $(this).html($(this).html() == "Selected" ? "Add" : "Selected");         
    addOrRemove(idToDb, $(this).parent().data('id'));
    addOrRemove(nameToDb, $(this).parent().data('name'));
    friendToDb = _.object(idToDb, nameToDb)

I want to post this array "friendToDb" to a separate PHP page using post after clicking submit button and then store it in mySQL db.

Here's what I'm trying to post in PHP:

<form method="post" action="friend_input.php">
    <div id="submitBtnRow" class="row top-plus">
      <div class="col-md-12">
        <div class="form-group">
          <input type="submit" id="submitBtn" class="btn btn-info btn-lg btn-block" value="SUBMIT">Submit</a>
        </div>
      </div>
    </div><!-- /submitBtnRow -->
</form>

And then in friend_input.php, I'm trying to fetch friendToDb and store in mySQL db and this I'm not sure how to do.

Please advice me how this should be done.

Thank you.

share|improve this question
1  
Have you heard of $.ajax?.. –  Joke_Sense10 Oct 14 '13 at 16:38
 
What issues are you having? Where is the code that tries to submit it? –  Barmar Oct 14 '13 at 16:38
 
If i am not wrong friendToDb is an object not an array.Your code says so. –  Joke_Sense10 Oct 14 '13 at 16:44
 
@NabeelSheikh Yeah sorry that's an object. And I'm not aware of ajax at all. –  rrg Oct 14 '13 at 16:46
 
@Barmar Sorry I forgot to include that. I've updated my question. –  rrg Oct 14 '13 at 16:52
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2 Answers

up vote 0 down vote

Add this to your form

<input id="friendDb" name="friendDb" type="hidden" />

Then using use

$("#friendDb").val();

To set the value before posting to your script.

share|improve this answer
 
Thanks. But what is the connection between <input> tag and setting the variable $("#friendDb").val();? I followed your approach and changed the input tag and used another variable to store #friendToDb and tried to echo it in php. But nothing was displayed. –  rrg Oct 14 '13 at 20:10
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up vote 0 down vote

Make form action="" and Use ajax to pass your object to php: 

$.ajax({
url:"friend_input.php",
type:"post",
data:{value:JSON.stringify(friendToDb)},   //Converts a JavaScript value to a JavaScript Object Notation (JSON) string.
success:function(result){
// response from php page
}
});

In php page:

echo json_decode($_POST['value']);  //Takes a JSON encoded string and converts it into a PHP variable.
share|improve this answer
 
Thanks for this, but how do I make sure that this happens only after I click the submit button? And could you also explain what does "//response from php page" means? Thanks –  rrg Oct 14 '13 at 17:08
 
Use this function once friendToDb is declared. i.e after assigning value to friendToDb. "//response from php page" means the response that your getting from your php page.i.e in this case $_POST['value'] will be displayed if you alert variable result –  Joke_Sense10 Oct 14 '13 at 17:17
 
Ok. So I did what you told me and after I click the submit button, the page gets redirected to friend_input.php and I get an error: "Undefined index: value in C:\xampp\htdocs\project-1\friend_input.php on line 37" –  rrg Oct 14 '13 at 17:38
 
Is it getting redirected via ajax or form action?? –  Joke_Sense10 Oct 14 '13 at 17:44
 
It should get redirected via form action because the variable friendToDb is dynamic and is not final until user clicks submit button. –  rrg Oct 14 '13 at 17:46
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