On page reload display diffrent SWF file - html

I have a html page "first" playing 1.swf and on the next page
"reload" it plays 2.swf. How do I do that?
Code i now use for the first one:
<center><object width="500" height="500" data="1.swf"></object></center>

One really simple solution would be to create a cookie incrementer:
var number = parseInt(getCookie("number"));
if (number == "") {
document.cookie="number=1";
} else {
document.cookie="number=" + (number + 1);
}
function getCookie(cname) {
var name = cname + "=";
var ca = document.cookie.split(';');
for(var i=0; i<ca.length; i++) {
var c = ca[i];
while (c.charAt(0)==' ') c = c.substring(1);
if (c.indexOf(name) == 0) return c.substring(name.length,c.length);
}
return "";
}
Now, every time the page reloads, number gets incremented by 1.
You can now remove the previous swf and create a new swf with the new source.
JSFIDDLE

Related

Can Google apps script be used to randomize page order on Google forms?

Update #2: Okay, I'm pretty sure my error in update #1 was because of indexing out of bounds over the array (I'm still not used to JS indexing at 0). But here is the new problem... if I write out the different combinations of the loop manually, setting the page index to 1 in moveItem() like so:
newForm.moveItem(itemsArray[0][0], 1);
newForm.moveItem(itemsArray[0][1], 1);
newForm.moveItem(itemsArray[0][2], 1);
newForm.moveItem(itemsArray[1][0], 1);
newForm.moveItem(itemsArray[1][1], 1);
newForm.moveItem(itemsArray[1][2], 1);
newForm.moveItem(itemsArray[2][0], 1);
...
...I don't get any errors but the items end up on different pages! What is going on?
Update #1:: Using Sandy Good's answer as well as a script I found at this WordPress blog, I have managed to get closer to what I needed. I believe Sandy Good misinterpreted what I wanted to do because I wasn't specific enough in my question.
I would like to:
Get all items from a page (section header, images, question etc)
Put them into an array
Do this for all pages, adding these arrays to an array (i.e: [[all items from page 1][all items from page 2][all items from page 3]...])
Shuffle the elements of this array
Repopulate a new form with each element of this array. In this way, page order will be randomized.
My JavaScript skills are poor (this is the first time I've used it). There is a step that produces null entries and I don't know why... I had to remove them manually. I am not able to complete step 5 as I get the following error:
Cannot convert Item,Item,Item to (class).
"Item,Item,Item" is the array element containing all the items from a particular page. So it seems that I can't add three items to a page at a time? Or is something else going on here?
Here is my code:
function shuffleForms() {
var itemsArray,shuffleQuestionsInNewForm,fncGetQuestionID,
newFormFile,newForm,newID,shuffle, sections;
// Copy template form by ID, set a new name
newFormFile = DriveApp.getFileById('1prfcl-RhaD4gn0b2oP4sbcKaRcZT5XoCAQCbLm1PR7I')
.makeCopy();
newFormFile.setName('AAAAA_Shuffled_Form');
// Get ID of new form and open it
newID = newFormFile.getId();
newForm = FormApp.openById(newID);
// Initialize array to put IDs in
itemsArray = [];
function getPageItems(thisPageNum) {
Logger.log("Getting items for page number: " + thisPageNum );
var thisPageItems = []; // Used for result
var thisPageBreakIndex = getPageItem(thisPageNum).getIndex();
Logger.log( "This is index num : " + thisPageBreakIndex );
// Get all items from page
var allItems = newForm.getItems();
thisPageItems.push(allItems[thisPageBreakIndex]);
Logger.log( "Added pagebreak item: " + allItems[thisPageBreakIndex].getIndex() );
for( var i = thisPageBreakIndex+1; ( i < allItems.length ) && ( allItems[i].getType() != FormApp.ItemType.PAGE_BREAK ); ++i ) {
thisPageItems.push(allItems[i]);
Logger.log( "Added non-pagebreak item: " + allItems[i].getIndex() );
}
return thisPageItems;
}
function shuffle(array) {
var currentIndex = array.length, temporaryValue, randomIndex;
Logger.log('shuffle ran')
// While there remain elements to shuffle...
while (0 !== currentIndex) {
// Pick a remaining element...
randomIndex = Math.floor(Math.random() * currentIndex);
currentIndex -= 1;
// And swap it with the current element.
temporaryValue = array[currentIndex];
array[currentIndex] = array[randomIndex];
array[randomIndex] = temporaryValue;
}
return array;
}
function shuffleAndMove() {
// Get page items for all pages into an array
for(i = 2; i <= 5; i++) {
itemsArray[i] = getPageItems(i);
}
// Removes null values from array
itemsArray = itemsArray.filter(function(x){return x});
// Shuffle page items
itemsArray = shuffle(itemsArray);
// Move page items to the new form
for(i = 2; i <= 5; ++i) {
newForm.moveItem(itemsArray[i], i);
}
}
shuffleAndMove();
}
Original post: I have used Google forms to create a questionnaire. For my purposes, each question needs to be on a separate page but I need the pages to be randomized. A quick Google search shows this feature has not been added yet.
I see that the Form class in the Google apps script has a number of methods that alter/give access to various properties of Google Forms. Since I do not know Javascript and am not too familiar with Google apps/API I would like to know if what I am trying to do is even possible before diving in and figuring it all out.
If it is possible, I would appreciate any insight on what methods would be relevant for this task just to give me some direction to get started.
Based on comments from Sandy Good and two SE questions found here and here, this is the code I have so far:
// Script to shuffle question in a Google Form when the questions are in separate sections
function shuffleFormSections() {
getQuestionID();
createNewShuffledForm();
}
// Get question IDs
function getQuestionID() {
var form = FormApp.getActiveForm();
var items = form.getItems();
arrayID = [];
for (var i in items) {
arrayID[i] = items[i].getId();
}
// Logger.log(arrayID);
return(arrayID);
}
// Shuffle function
function shuffle(a) {
var j, x, i;
for (i = a.length; i; i--) {
j = Math.floor(Math.random() * i);
x = a[i - 1];
a[i - 1] = a[j];
a[j] = x;
}
}
// Shuffle IDs and create new form with new question order
function createNewShuffledForm() {
shuffle(arrayID);
// Logger.log(arrayID);
var newForm = FormApp.create('Shuffled Form');
for (var i in arrayID) {
arrayID[i].getItemsbyId();
}
}
Try this. There's a few "constants" to be set at the top of the function, check the comments. Form file copying and opening borrowed from Sandy Good's answer, thanks!
// This is the function to run, all the others here are helper functions
// You'll need to set your source file id and your destination file name in the
// constants at the top of this function here.
// It appears that the "Title" page does not count as a page, so you don't need
// to include it in the PAGES_AT_BEGINNING_TO_NOT_SHUFFLE count.
function shuffleFormPages() {
// UPDATE THESE CONSTANTS AS NEEDED
var PAGES_AT_BEGINNING_TO_NOT_SHUFFLE = 2; // preserve X intro pages; shuffle everything after page X
var SOURCE_FILE_ID = 'YOUR_SOURCE_FILE_ID_HERE';
var DESTINATION_FILE_NAME = 'YOUR_DESTINATION_FILE_NAME_HERE';
// Copy template form by ID, set a new name
var newFormFile = DriveApp.getFileById(SOURCE_FILE_ID).makeCopy();
newFormFile.setName(DESTINATION_FILE_NAME);
// Open the duplicated form file as a form
var newForm = FormApp.openById(newFormFile.getId());
var pages = extractPages(newForm);
shuffleEndOfPages(pages, PAGES_AT_BEGINNING_TO_NOT_SHUFFLE);
var shuffledFormItems = flatten(pages);
setFormItems(newForm, shuffledFormItems);
}
// Builds an array of "page" arrays. Each page array starts with a page break
// and continues until the next page break.
function extractPages(form) {
var formItems = form.getItems();
var currentPage = [];
var allPages = [];
formItems.forEach(function(item) {
if (item.getType() == FormApp.ItemType.PAGE_BREAK && currentPage.length > 0) {
// found a page break (and it isn't the first one)
allPages.push(currentPage); // push what we've built for this page onto the output array
currentPage = [item]; // reset the current page to just this most recent item
} else {
currentPage.push(item);
}
});
// We've got the last page dangling, so add it
allPages.push(currentPage);
return allPages;
};
// startIndex is the array index to start shuffling from. E.g. to start
// shuffling on page 5, startIndex should be 4. startIndex could also be thought
// of as the number of pages to keep unshuffled.
// This function has no return value, it just mutates pages
function shuffleEndOfPages(pages, startIndex) {
var currentIndex = pages.length;
// While there remain elements to shuffle...
while (currentIndex > startIndex) {
// Pick an element between startIndex and currentIndex (inclusive)
var randomIndex = Math.floor(Math.random() * (currentIndex - startIndex)) + startIndex;
currentIndex -= 1;
// And swap it with the current element.
var temporaryValue = pages[currentIndex];
pages[currentIndex] = pages[randomIndex];
pages[randomIndex] = temporaryValue;
}
};
// Sourced from elsewhere on SO:
// https://stackoverflow.com/a/15030117/4280232
function flatten(array) {
return array.reduce(
function (flattenedArray, toFlatten) {
return flattenedArray.concat(Array.isArray(toFlatten) ? flatten(toFlatten) : toFlatten);
},
[]
);
};
// No safety checks around items being the same as the form length or whatever.
// This mutates form.
function setFormItems(form, items) {
items.forEach(function(item, index) {
form.moveItem(item, index);
});
};
I tested this code. It created a new Form, and then shuffled the questions in the new Form. It excludes page breaks, images and section headers. You need to provide a source file ID for the original template Form. This function has 3 inner sub-functions. The inner functions are at the top, and they are called at the bottom of the outer function. The arrayOfIDs variable does not need to be returned or passed to another function because it is available in the outer scope.
function shuffleFormSections() {
var arrayOfIDs,shuffleQuestionsInNewForm,fncGetQuestionID,
newFormFile,newForm,newID,items,shuffle;
newFormFile = DriveApp.getFileById('Put the source file ID here')
.makeCopy();
newFormFile.setName('AAAAA_Shuffled_Form');
newID = newFormFile.getId();
newForm = FormApp.openById(newID);
arrayOfIDs = [];
fncGetQuestionID = function() {
var i,L,thisID,thisItem,thisType;
items = newForm.getItems();
L = items.length;
for (i=0;i<L;i++) {
thisItem = items[i];
thisType = thisItem.getType();
if (thisType === FormApp.ItemType.PAGE_BREAK ||
thisType === FormApp.ItemType.SECTION_HEADER ||
thisType === FormApp.ItemType.IMAGE) {
continue;
}
thisID = thisItem.getId();
arrayOfIDs.push(thisID);
}
Logger.log('arrayOfIDs: ' + arrayOfIDs);
//the array arrayOfIDs does not need to be returned since it is available
//in the outermost scope
}// End of fncGetQuestionID function
shuffle = function() {// Shuffle function
var j, x, i;
Logger.log('shuffle ran')
for (i = arrayOfIDs.length; i; i--) {
j = Math.floor(Math.random() * i);
Logger.log('j: ' + j)
x = arrayOfIDs[i - 1];
Logger.log('x: ' + x)
arrayOfIDs[i - 1] = arrayOfIDs[j];
arrayOfIDs[j] = x;
}
Logger.log('arrayOfIDs: ' + arrayOfIDs)
}
shuffleQuestionsInNewForm = function() {
var i,L,thisID,thisItem,thisQuestion,questionType;
L = arrayOfIDs.length;
for (i=0;i<L;i++) {
thisID = arrayOfIDs[i];
Logger.log('thisID: ' + thisID)
thisItem = newForm.getItemById(thisID);
newForm.moveItem(thisItem, i)
}
}
fncGetQuestionID();//Get all the question ID's and put them into an array
shuffle();
shuffleQuestionsInNewForm();
}

I need my dropbox not to change after refreshing or opening the page

I have a dropdown selection of two options. In stock or out of stock. I need the dropdown to stay the same as what was selected after someone leaves the page. Im sorry if there is a simple fix to this. I am new to web development and cant seem to get it to work. Everytime I refresh the page it goes to the default.
Thank you.
HTML alone is great if you want to give everyone the same exact page. But in order to do what you want to do, you will most likely need Javascript. If you haven't used Javascript before, Codecademy can teach you how to use it.
You will need 3 parts if you go with this method.
Part 1. Store a cookie with the drop-down selected value.
<script>
function storeCookie(name, value){
document.cookie = name + "=" + value + "; expires=Fri, 31 Dec 9999 23:59:59 GMT";
}
Part 2. Check if a cookie already exists, and if it does then get the value that should be selected in your drop-down.
function cookieExists(name){
return -1 != document.cookie.indexOf( name + "=");
}
function getCookie(cname) {
var name = cname + "=";
var decodedCookie = decodeURIComponent(document.cookie);
var ca = decodedCookie.split(';');
for(var i = 0; i <ca.length; i++) {
var c = ca[i];
while (c.charAt(0) == ' ') {
c = c.substring(1);
}
if (c.indexOf(name) == 0) {
return c.substring(name.length, c.length);
}
}
return "";
}
Part 3. Tie it all together.
function main(){
var cookieName = "yourcookiename";
if(cookieExists(cookieName){
var value = getCookie(cookiename);
//This is where you set the value in your drop-down with Javascript
}
}
</script>
Furthermore, you should call storeCookie("yourcookiename", "yourcookievalue") when the webpage knows what value should stay selected, and call main() when the page loads.

find the length of a string in google script

I'm trying to make a script for google sheet, who can count a letter in a text. But it seems that .length doesn't work. Anyone who can give directions on where to find the the solution.
function Tjekkode(tekst , bogstav){
var test = "";
// find the length of laengdeTekst
var laengdeTekst = tekst.length;
var t = 0;
// through the text character by character
for ( var i = 1; i<laengdeTekst ; i++) {
var test = tekst.substr(i,1);
if (test == bogstav) {
// if the letter is found, it is counted up
// REMEMBER == means compare
var t = t + 1;
}
}
// returns percent appearance of the letter
Return = t / længdeTekst * 100
}
Thanks in advance
length is ok in your code. To test it, run this script:
function test( ) {
var test = "123456";
// finder længden på teksten
var laengdeTekst = test.length;
Logger.log(laengdeTekst);
}
After you run it, check Log, press [Ctrl + Enter]
The correct code in your case:
function Tjekkode(tekst, bogstav) {
var test = "";
var laengdeTekst = tekst.length;
var t = 0;
// start looping from zero!
for ( var i = 0; i<laengdeTekst; i++) {
var test = tekst.substr(i,1);
if (test == bogstav) {
var t = t + 1;
}
}
// JavaScript is case sensitive: 'return != Return'
return t / laengdeTekst * 100;
}
Please, look at this tutorial for more info
thanks
I'll guess that I might get the one with the R instead of r at the end, but the script didn't run that line, because it kinda stopped at the .length line :/
the comments in danish is for my pupils (I'm a teacher in elementary school)
I'll see if google wants to cooperate today :|
This is the google script that worked for me. Note the 24 - that's the length of an empty message that has markup like <div>...</div>
function TrashEmptyDrafts() {
var thread = GmailApp.getDraftMessages();
for (var i = 0; i < thread.length; i++) {
b=thread[i].getBody();
if (b.length <= 24.0){
thread[i].moveToTrash();
}
}}

preload jQuery building of JSON results

So I am currently building an activity feed/news feed or sorts using JSON and jQuery (and of course PHP). Everything works really well, especially fetching new results. The only issue is the first load - and I'm wondering if there is some way to sort of preload the results to make it more slick?
jQuery code below:
for (var j = 0; j < jsonData.items.length; j++) {
var entryData = jsonData.items[j];
var entry = template.clone();
entry.removeClass("template");
entry.find(".message").text(entryData.statusid);
entry.find(".actName").text(entryData.name);
entry.find(".actContent").text(entryData.content);
//get the users ProfilePic
var profileImg = $("<img />");
profileImg.attr("src", "./img/" +entryData.profilePic);
profileImg.addClass("feed-user-img");
entry.find(".actProfilePic").append(profileImg);
//Get user-uploaded images.
entry.find(".actImage").text(entryData.imageKey);
if (entryData.imageKey != "")
{
var img = $("<img />"); // Create the image element
img.attr("src", "http://spartadev.s3.amazonaws.com/" + entryData.imageKey); // Set src to the s3 url plus the imageKey
entry.find(".actImage").append(img); // Append it to the element where it's supposed to be
}
spot.prepend(entry);
spot.find(".entry").first().hide().slideDown();
}

A* algorithm works OK, but not perfectly. What's wrong?

This is my grid of nodes:
I'm moving an object around on it using the A* pathfinding algorithm. It generally works OK, but it sometimes acts wrongly:
When moving from 3 to 1, it correctly goes via 2. When going from 1 to 3 however, it goes via 4.
When moving between 3 and 5, it goes via 4 in either direction instead of the shorter way via 6
What can be wrong? Here's my code (AS3):
public static function getPath(from:Point, to:Point, grid:NodeGrid):PointLine {
// get target node
var target:NodeGridNode = grid.getClosestNodeObj(to.x, to.y);
var backtrace:Map = new Map();
var openList:LinkedSet = new LinkedSet();
var closedList:LinkedSet = new LinkedSet();
// begin with first node
openList.add(grid.getClosestNodeObj(from.x, from.y));
// start A*
var curNode:NodeGridNode;
while (openList.size != 0) {
// pick a new current node
if (openList.size == 1) {
curNode = NodeGridNode(openList.first);
}
else {
// find cheapest node in open list
var minScore:Number = Number.MAX_VALUE;
var minNext:NodeGridNode;
openList.iterate(function(next:NodeGridNode, i:int):int {
var score:Number = curNode.distanceTo(next) + next.distanceTo(target);
if (score < minScore) {
minScore = score;
minNext = next;
return LinkedSet.BREAK;
}
return 0;
});
curNode = minNext;
}
// have not reached
if (curNode == target) break;
else {
// move to closed
openList.remove(curNode);
closedList.add(curNode);
// put connected nodes on open list
for each (var adjNode:NodeGridNode in curNode.connects) {
if (!openList.contains(adjNode) && !closedList.contains(adjNode)) {
openList.add(adjNode);
backtrace.put(adjNode, curNode);
}
}
}
}
// make path
var pathPoints:Vector.<Point> = new Vector.<Point>();
pathPoints.push(to);
while(curNode != null) {
pathPoints.unshift(curNode.location);
curNode = backtrace.read(curNode);
}
pathPoints.unshift(from);
return new PointLine(pathPoints);
}
NodeGridNode::distanceTo()
public function distanceTo(o:NodeGridNode):Number {
var dx:Number = location.x - o.location.x;
var dy:Number = location.y - o.location.y;
return Math.sqrt(dx*dx + dy*dy);
}
The problem I see here is the line
if (!openList.contains(adjNode) && !closedList.contains(adjNode))
It may be the case that an adjNode may be easier(shorter) to reach through the current node although it was reached from another node previously which means it is in the openList.
Found the bug:
openList.iterate(function(next:NodeGridNode, i:int):int {
var score:Number = curNode.distanceTo(next) + next.distanceTo(target);
if (score < minScore) {
minScore = score;
minNext = next;
return LinkedSet.BREAK;
}
return 0;
});
The return LinkedSet.BREAK (which acts like a break statement in a regular loop) should not be there. It causes the first node in the open list to be selected always, instead of the cheapest one.