how to html form post - file uploading and reading json response from php server

Question

i am trying file uploading to php my server.file and data uploading through multi part /form-data ,the file and data received on php server but in my php server return json

Answer 1

You should probably use an AJAX call. Here's a solution using jQuery:

<script type="text/javascript">
$(document).ready(function(){
    $("#browseButton").click(function(){
        var url = "";
        var formdata = $("#myForm").serialize();
        $.ajax({
            url: url,
            type: 'POST',
            data:  formdata,
            dataType: 'json',
            cache: false,
            contentType: false,
            processData: false,
            success: function(response){
                if(response.status == "success"){
                    // Success

                } else {
                    // Failure

                }
            },
            error: function(response){
                // Error

            }          
        });
    });
});
</script>

In order to redirect the user, you can use: window.location.href = " ... your_url ...";

Here's an explanation on how to use jQuery AJAX and multi-part data:

Sending multipart/formdata with jQuery.ajax

Answer 2

I think it should work for you. Using AJAX, as I do

     //Your php code
        $arrToJSON = array(
        "dataPHPtoJs"=>"yourData",
        "asYouWant"=>"<div class=\".class1\">soemting</div>"    
        );  
        return json_encode(array($arrToJSON));




    //Your javaScript code
    $(document).on("event", "#idElement", function(){
        //Data you want to send to php evaluate
         var dt={ 
                  ObjEvn:"btn_Login",
                  dataJsToPHP: $("#txt_EmailLogin").val()
                };

        //Ajax      
         var request =$.ajax({//http://api.jquery.com/jQuery.ajax/
                                url: "yourServer.php",
                                type: "POST",
                                data: dt,
                                dataType: "json"
                            });

        //Ajax Done catch JSON from PHP 
            request.done(function(dataset){
                for (var index in dataset){ 
                     dataPHPtoJsJS=dataset[index].dataPHPtoJs;
                     asManyasYouWantJS=dataset[index].asYouWant;
                 }

                 //JavaScript conditions. Here you can control the behaivior of your html object, based on your PHP response
                 if(dataPHPtoJsJS){
                    $( "#idYourHtmlElement" ).removeClass( "class1" )
                    $( "#idYourHtmlElement" ).addClass( "class2" )
                 }


         }); 

        //Ajax Fail 
            request.fail(function(jqXHR, textStatus) {
                alert("Request failed: " + textStatus);
            }); 
    }

Answer 3

Your JSON response would be a kind of associative array in php. Encode your array data into JSON using "json_encode" and return values as you want .

   $arr = array('status' => $status, 'status2' => $status2, );
   echo json_encode($arr);

NOTE: If you are using ajax to call php file then do not use any php echo/print in that file and not even HTML. ECHO only "json_encode();" Nothing else.

Answer 4

try json_decode.

    $data = ({"code":0, "message":"success"});
    $array = json_decode($data, true);

by passing 2nd parameter to true you will get response in array instead of object.

the array will be then populated as follow:

    array (size=2)
    'code' => int 0
    'message' => string 'success' (length=7)

Answer 5

To sum it up:

1. Upload your data to server using AJAX with native JS (>=IE10) or jQuery
2. Catch (xhr.responseText in native JS) and parse the response
3. Redirect with window.location.href="success.php"