使用Ajax请求检查会话

Question

Im checking if a session is valid using Ajax request.

Im using the below code to accomplish it...

function checksession() {
        $.ajax({
            type:'POST'
            ,url:'CheckSession'
            ,success: validateresult
            ,data: { antiCSRF : '{{acsrf}}',
                   session_id: '{{session_id}}'}
            ,error: function(){ alert('Session check failed') }
            })

function validateresult(session_ind){
            alert('called_form validateresult = ' + called_for)
            if (session_ind == "Y"){
                    alert_message = "Y"
                    if (called_for == "SEARCH") {formSubmit()}
                    else if (called_for == "ENTER" ) {formSubmit();}
                    else if (called_for == 'DATA') { $('#datasetform').submit();}
                    else if (called_for == 'FEEDBACK') { createfeedback()}
                    else if (called_for == 'HELP') {createhelp()}
            }else{
                    if (alert_message == "Y") {
            alert("Your session has been timed-out ");
            }
               alert_message = "Y";
               window.location = './';

            }
        }

It all works fine...The only problem is when I have a FORM submission...

 <form id="bigsearchform_new" method="post" action="Paid">
                        <label style="display:none" for="search_string1">SEARCH</label>
                        <input id="search_string" name="search_string1" type="text" class="startnewsearch rounded" placeholder="Search..." maxlength="500" >
                        <input id="searchButton1"  type="button" class="searchButton" title="Click here to search the database">
                        <input type="hidden" name="antiCSRF" value="{{acsrf}}" />
                        <input type="hidden" name="session_id" value="{{session_id}}" />
                        <input type="hidden" name="commodity_id" id="commodity_id" />
                </form>

When I submit the above form I check if the session is valid using the below code -

 $('input[id=search_string]').on('keyup', function(e) {
    if (e.which == 13) {
        alert('called_for')
        called_for = "ENTER"
        checksession()
   }
   });

When I do this the POST request to run the CheckSession is cancelled.

Error from Chrome internals -

t=254662 [st=60]        HTTP_TRANSACTION_READ_RESPONSE_HEADERS
                        --> HTTP/1.1 200 OK
                            Content-Type: text/html; charset=UTF-8
                            Date: Mon, 11 Aug 2014 19:31:36 GMT
                            Server: Apache
                            X-Frame-Options: DENY
                            Content-Length: 1
                            Connection: keep-alive

Form submission:

 function formSubmit()
    {
    alert('calling main page = ' + $("#search_string").val().length )
    if ($("#search_string").val().length != 0) {
        $("#bigsearchform_new").submit();
   } else {
        alert("Please enter a search term!");
   }

   }

Answer 1

It gets cancelled because the form is submitted without waiting for the session check. If you want to test the form before submit it, you should use the on submit event and return false.

But why do you check the session before submit ? The search script should test the session itself.

You set a global variable called_from. This is not good practice. Rather give your session check function the callback function as parameter.

function checksession(callback) {
    ...
            if (session_ind == "Y"){
                alert_message = "Y"
                callback()
   ...
}

Then call it like this :

if (e.which == 13) {
    alert('called_for')
    called_for = "ENTER"
    checksession(formSubmit1())
}