redirect to another page after ajax function - ajax

Can anyone help me with, I am trying to create a download counter to my website.
I have a ajax script that counts up by 1 when the users clicks the download link, the issue I am having is on some browsers it goes to the download link before completing the ajax count script.
Is there a way that I can redirect to the download file once the script has completed. At the moment I have as follows
This is the link :-
<a href='downloads/".$downfile."' onclick=\"Counter('$referid');\"'>Download File</a>
This is the counter script:-
<script type="text/javascript">
function Counter(id)
{
$.get("clickcounter.php?id="+id);
{
return false;
}
}
</script>
This is the php script (clickcounter.php)
<?php
include('dbutils.php');
$referid = $_GET['id'];
$q = "SELECT * FROM downloads WHERE downid =".$referid;
$r = mysql_query($q);
while ($row = mysql_fetch_array($r))
{
$click = stripslashes(trim($row['downcount']));
$download = $row['downfile'];
}
$countup = $click + 1;
$qUpdate = "UPDATE downloads
SET downcount=$countup
WHERE downid=$referid";
$rUpdate = mysql_query($qUpdate);
?>

A few relatively small modifications should solve the problem. First, change the onclick to the following:
onclick=\"Counter('$referid', this); return false;\"
What we have done is to send in this as the second argument to the Counter function so we have a reference to the clicked link. Secondly, we have added return false, which blocks the browser from navigating to the url specified in the href.
The modified counter function looks like this:
function Counter(id, link) {
$.get("clickcounter.php?id=" + id, function() {
location.href = $(link).attr("href");
});
}
We now have a reference to the clicked link. A function has now been specified as the second argument to $.get(). This is the success-function, which is called when the ajax call has been successfully called. Inside that function we now redirect to the url specified in the href attribute on the clicked link.
I feel I should point out that the recommended way is to bind the onclick using jQuery separate from the html. The referid can be stored in a data attribute (which I chose to call data-rid):
<a href='downloads/".$downfile."' class='dl' data-rid='$referid'>Download File</a>
Then you bind the onclick for all download links (a elements with a "dl" class):
$(function() {
$("a.dl").click(function() {
var id = $(this).attr("data-rid");
var href = $(this).attr("href");
$.get("clickcounter.php?id=" + id, function() {
location.href = href;
});
return false;
});
});​
(I feel I should point out that the code has not been tested, so it's possible that a typo has snuck in somewhere)

Related

Google App Scripts Function to Open URL [duplicate]

Is there a way to write a google apps script so when ran, a second browser window opens to www.google.com (or another site of my choice)?
I am trying to come up with a work-around to my previous question here:
Can I add a hyperlink inside a message box of a Google Apps spreadsheet
This function opens a URL without requiring additional user interaction.
/**
* Open a URL in a new tab.
*/
function openUrl( url ){
var html = HtmlService.createHtmlOutput('<html><script>'
+'window.close = function(){window.setTimeout(function(){google.script.host.close()},9)};'
+'var a = document.createElement("a"); a.href="'+url+'"; a.target="_blank";'
+'if(document.createEvent){'
+' var event=document.createEvent("MouseEvents");'
+' if(navigator.userAgent.toLowerCase().indexOf("firefox")>-1){window.document.body.append(a)}'
+' event.initEvent("click",true,true); a.dispatchEvent(event);'
+'}else{ a.click() }'
+'close();'
+'</script>'
// Offer URL as clickable link in case above code fails.
+'<body style="word-break:break-word;font-family:sans-serif;">Failed to open automatically. Click here to proceed.</body>'
+'<script>google.script.host.setHeight(40);google.script.host.setWidth(410)</script>'
+'</html>')
.setWidth( 90 ).setHeight( 1 );
SpreadsheetApp.getUi().showModalDialog( html, "Opening ..." );
}
This method works by creating a temporary dialog box, so it will not work in contexts where the UI service is not accessible, such as the script editor or a custom G Sheets formula.
You can build a small UI that does the job like this :
function test(){
showURL("http://www.google.com")
}
//
function showURL(href){
var app = UiApp.createApplication().setHeight(50).setWidth(200);
app.setTitle("Show URL");
var link = app.createAnchor('open ', href).setId("link");
app.add(link);
var doc = SpreadsheetApp.getActive();
doc.show(app);
}
If you want to 'show' the URL, just change this line like this :
var link = app.createAnchor(href, href).setId("link");
EDIT : link to a demo spreadsheet in read only because too many people keep writing unwanted things on it (just make a copy to use instead).
EDIT : UiApp was deprecated by Google on 11th Dec 2014, this method could break at any time and needs updating to use HTML service instead!
EDIT :
below is an implementation using html service.
function testNew(){
showAnchor('Stackoverflow','http://stackoverflow.com/questions/tagged/google-apps-script');
}
function showAnchor(name,url) {
var html = '<html><body>'+name+'</body></html>';
var ui = HtmlService.createHtmlOutput(html)
SpreadsheetApp.getUi().showModelessDialog(ui,"demo");
}
There really isn't a need to create a custom click event as suggested in the bountied answer or to show the url as suggested in the accepted answer.
window.open(url)1 does open web pages automatically without user interaction, provided pop- up blockers are disabled(as is the case with Stephen's answer)
openUrl.html
<!DOCTYPE html>
<html>
<head>
<base target="_blank">
<script>
const url1 ='https://stackoverflow.com/a/54675103';
const winRef = window.open(url1);
winRef ? google.script.host.close() : window.alert('Allow popup to redirect you to '+url1) ;
window.onload=function(){document.getElementById('url').href = url1;}
</script>
</head>
<body>
Kindly allow pop ups</br>
Or <a id='url'>Click here </a>to continue!!!
</body>
</html>
code.gs:
function modalUrl(){
SpreadsheetApp.getUi()
.showModalDialog(
HtmlService.createHtmlOutputFromFile('openUrl').setHeight(50),
'Opening StackOverflow'
)
}
Google Apps Script will not open automatically web pages, but it could be used to display a message with links, buttons that the user could click on them to open the desired web pages or even to use the Window object and methods like addEventListener() to open URLs.
It's worth to note that UiApp is now deprecated. From Class UiApp - Google Apps Script - Google Developers
Deprecated. The UI service was deprecated on December 11, 2014. To
create user interfaces, use the HTML service instead.
The example in the HTML Service linked page is pretty simple,
Code.gs
// Use this code for Google Docs, Forms, or new Sheets.
function onOpen() {
SpreadsheetApp.getUi() // Or DocumentApp or FormApp.
.createMenu('Dialog')
.addItem('Open', 'openDialog')
.addToUi();
}
function openDialog() {
var html = HtmlService.createHtmlOutputFromFile('index')
.setSandboxMode(HtmlService.SandboxMode.IFRAME);
SpreadsheetApp.getUi() // Or DocumentApp or FormApp.
.showModalDialog(html, 'Dialog title');
}
A customized version of index.html to show two hyperlinks
<a href='http://stackoverflow.com' target='_blank'>Stack Overflow</a>
<br/>
<a href='http://meta.stackoverflow.com/' target='_blank'>Meta Stack Overflow</a>
Building of off an earlier example, I think there is a cleaner way of doing this. Create an index.html file in your project and using Stephen's code from above, just convert it into an HTML doc.
<!DOCTYPE html>
<html>
<base target="_top">
<script>
function onSuccess(url) {
var a = document.createElement("a");
a.href = url;
a.target = "_blank";
window.close = function () {
window.setTimeout(function() {
google.script.host.close();
}, 9);
};
if (document.createEvent) {
var event = document.createEvent("MouseEvents");
if (navigator.userAgent.toLowerCase().indexOf("firefox") > -1) {
window.document.body.append(a);
}
event.initEvent("click", true, true);
a.dispatchEvent(event);
} else {
a.click();
}
close();
}
function onFailure(url) {
var div = document.getElementById('failureContent');
var link = 'Process';
div.innerHtml = "Failure to open automatically: " + link;
}
google.script.run.withSuccessHandler(onSuccess).withFailureHandler(onFailure).getUrl();
</script>
<body>
<div id="failureContent"></div>
</body>
<script>
google.script.host.setHeight(40);
google.script.host.setWidth(410);
</script>
</html>
Then, in your Code.gs script, you can have something like the following,
function getUrl() {
return 'http://whatever.com';
}
function openUrl() {
var html = HtmlService.createHtmlOutputFromFile("index");
html.setWidth(90).setHeight(1);
var ui = SpreadsheetApp.getUi().showModalDialog(html, "Opening ..." );
}
I liked #Stephen M. Harris's answer, and it worked for me until recently. I'm not sure why it stopped working.
What works for me now on 2021-09-01:
function openUrl( url ){
Logger.log('openUrl. url: ' + url);
const html = `<html>
<a id='url' href="${url}">Click here</a>
<script>
var winRef = window.open("${url}");
winRef ? google.script.host.close() : window.alert('Configure browser to allow popup to redirect you to ${url}') ;
</script>
</html>`;
Logger.log('openUrl. html: ' + html);
var htmlOutput = HtmlService.createHtmlOutput(html).setWidth( 250 ).setHeight( 300 );
Logger.log('openUrl. htmlOutput: ' + htmlOutput);
SpreadsheetApp.getUi().showModalDialog( htmlOutput, `openUrl function in generic.gs is now opening a URL...` ); // https://developers.google.com/apps-script/reference/base/ui#showModalDialog(Object,String) Requires authorization with this scope: https://www.googleapis.com/auth/script.container.ui See https://developers.google.com/apps-script/concepts/scopes#setting_explicit_scopes
}
https://developers.google.com/apps-script/reference/base/ui#showModalDialog(Object,String) Requires authorization with this scope: https://www.googleapis.com/auth/script.container.ui See https://developers.google.com/apps-script/concepts/scopes#setting_explicit_scopes

Laravel render for differend controller method

I'm struggling with the render() method in Laravel 5.
When $whatever->render() is runned, it takes the controller method name as the route by default.
Example:
When i run this command in DelasController#updateFilter, the pagination route is set to whatever.com/marketplace/updateFiler?page=2, which does not make a sense to me.
Problem:
I want to keep the route as simple as whatever.com/marketplace?page=2.
Question:
Can anybody gives me a hint on how to solve this?
Thank you for your time and a discussion.
Looking forward for a reply.
I have an application in which various paginated lists are displayed in "windows" on the page and are updated via AJAX calls to the server. Here's how I did it:
Set up a route to render the whole page, something like this:
Route::get('/marketplace', function ($arguments) {
....
});
Set up a route which will return the current page of the list. For example, it might be something like this:
Route::get('/marketplace/updateFiler', function ($arguments) {
....
});
In your Javascript code for the page, you need to change the pagination links so that, instead of loading the new page with the URL for the link, it makes the AJAX request to the second route. The Javascript could look something like this:
$('ul.pagination a').on('click', function (event) {
// stop the default action
event.stopPropagation();
event.preventDefault();
// get the URL from the link
var url = $(event.currentTarget).attr('href');
// get the page number from the URL
var page = getURLParameterByName(url, 'page');
$.get('marketplace/updateFiler', { page: page }, function (data){
// do something with the response from the server
});
});
The getURLParameterByName function is simply a helper that extracts the named parameter from a URL:
var getURLParameterByName = function (url, name, defaultValue) {
// is defaultValue undefined? if so, set it to false
//
if (typeof defaultValue === "undefined") {
defaultValue = false;
}
name = name.replace(/[\[]/, "\\[").replace(/[\]]/, "\\]");
var regex = new RegExp("[\\?&]" + name + "=([^&#]*)"),
results = regex.exec(url);
return results === null ?
defaultValue :
decodeURIComponent(results[1].replace(/\+/g, " "));
};
I adapted this code from an answer I found here on Stack Overflow: https://stackoverflow.com/a/901144/2008384.

Joomla - OnMouseover show data from controller in tooltip using Qtip script

I need help to show the data (from controller) in tooltip (using qTip script) on MouseOVer event.
User mouserover on link (Multiple links like in foreach)
id send to a js function like function getData(id){}
Call the PHP function from the controller & using the variable return the data to tooltip.
Show return data into tooltip.
HTML & PHP:
foreach ($rows as $row) {
<a href="#" onmouseover="getData(<?php echo $row->id; ?>)" >Name</a>
}
JS function :
function getData(id)
{
var url='index.php?option=com_test&controller=test&task=getDetails&format=raw';
var data = 'item_id=' + id ;
var request = new Request({
url: url,
method:'post',
data: data,
async: true,
onSuccess: function(responseText)
{
// How i show the "responseText" data here in tooltip using qTip
}
}).send();
}
Controller function:
function getDetails()
{
echo $return = JRequest::getVar('item_id');
}
Would it not be easier to preload the tip in a content item view override (add it to the title= attribute), and adding the call to qTip (one per page) would do the trick.
Also, ajax would make the tooltip not very quick.
Just make sure you properly escape " or ' characters in the title attribute or you will break your markup.
edit: added code. i'm writing this here so there may be typos, but it should give you the idea:
foreach ($rows as $row) {
// let's assume $tip contains the right tip for each row:
Name
}
This first part will show the default browser tooltip.
Now you have to inject the qTip script:
$document = JFactory::getDocument();
$document->addScriptDeclaration('
jQuery("a.tipme").qTip();
');
Now how to put the value in $tip? I see you have a com_test component, you don't need to go through the controller, but can instantiate directly the model that will provide the tooltip; if it's called "tip":
$tipmodel=JModel::getInstance('tip', 'ComtestModel'); // get the model (once!)
$tip = $tipmodel->getTip($row->id);
All the variable, model, controller names in this story are fictional.

Zend Form: onchange select load another view content

In my application I have a form in controller/index that consists out of 3 select boxes. When all three boxes have a value selected I need to show additional html and extra form options in the same view based on those select values. The obvious solution seems to make an ajax call to another action that handles the database operation and creates a view and loading that view into the controller/index.phtml
I have been able to load a view of another action in the index.phtml by using:
$('#select').change(function() {
event.preventDefault();
var id = $(this).attr('id');
$('#results').show();
$('#results').load('/controller/index/' + $(this).attr('value'));
return false;
});
However I need to pass the variables of all three select boxes and for that I alternatively used:
$('#select1').change(function() {
var select1 = $('#select1').val();
var select2 = $('#select2').val();
var select3 = $('#select3').val();
$.ajax({
type: 'POST',
dataType: 'json',
url: '/controller/index/',
data: { select1: select1, select2: select2, select3: select3},
success: function(result){
var return1 = result.return1;
var return2 = result.return2;
}
});
});
The last method works in as far that I do see the variables passed in the headers and the response contains the view, but I cant fix it that just the content of the ajax view is placed within the index view. (Ofcourse by not using AjaxContent switching, the ajax view will load but that includes the complete layout as well.) Anything that I echo in the ajax action or ajax view do not show in the index view. Any pointer would be more than welcome
EDIT
the ajax action now looks like
$this->view->layout()->disableLayout();
$this->_helper->viewRenderer->setNoRender(true);
$select1 = $this->_request->getParam('select1');
$select2 = $this->_request->getParam('select2');
$select3 = $this->_request->getParam('select3');
// DO THE OTHER STUFF AND LOGIC HERE
$results = array(
'return1' => 'value1',
'return2' => 'value2'
);
$this->_response->setBody(json_encode($results));
and the controller init
public function init() {
$ajaxContext = $this->_helper->getHelper('AjaxContext');
$ajaxContext->addActionContext('ajax', 'json')->initContext();
}
So everything works, I can see the returned values in the response by using developer tool (network) in my browser, however I just do not know how I can use this to "update" the view
You can do two things:
You can enable the layout of the action you are calling via ajax. See you have disabled layout so even if the view phtml file of the ajax action contains something, it won't show. You can enable layout, use text/html dataType instead of json and show the returned HTML somewhere.
Or, in the success event of the ajax call, write javascript codes to update DOM.
Thanks #Salman for your suggestions as they lead me in the right direction and I managed to solve the problem.
I managed to pass multiple parameters with the ajax .load() call by passing them as get parameters.
The results of the ajaxAction could then be formatted in the ajax.ajax.phtml view and were consecutively
shown within the #results div that resides in the index.phtml where the select boxes are.
controller/index.phtml
<div id="results" style="display:block;">Select all three values</div>
IndexController init and ajaxAction
public function init() {
$ajaxContext = $this->_helper->getHelper('AjaxContext');
$ajaxContext->addActionContext('ajax', 'html')->initContext('html');
}
public function ajaxAction() {
$select1 = $this->_request->getQuery('select1');
$select2 = $this->_request->getQuery('select2');
$select3 = $this->_request->getQuery('select3');
$form = new Application_Form();
// Database operations and logic
$this->view->form = $form;
$this->view->array = $somearray;
}
}
jquery script in index.phtml
$(document).ready(function(){
$('.selector').change(function() {
var select1 = $('#select1').val();
var select2 = $('#select2').val();
var select3 = $('#select3').val();
if ( select1 && select2 && select3) {
$('#results').show();
$('#results').load('/controller/ajax?select1=' + select1 + '&select2=' + select2 + '&select3=' + select3);
}
});
});
controller/ajax.ajax.phtml
<?php if ( $this->array ) : ?>
<?php echo( $this->form ); ?>
<?php else: ?>
Nothing found for selected values
<?php endif ?>

How can I return an id value from a div already populated through ajax

I am having some difficulty passing a correct id function back to AJAX.
I'm creating a product bulletin generator that lets items to be added by their SKU code (which works fine). My problem is that when a bulletin is clicked on, a preview of that bulletin is loaded into a div and shows all products associated with that bulletin.
From inside those results, I am trying to add the ability to delete a product from the bulletin. The problem is that the value being passed back to AJAX belongs to the first product only. It won't send the value belonging to the particular item if it is any other item than the first one.
This is the code (belonging to main.php) that gets loaded via AJAX into a div and is looped with each product associated with a selected bulletin
echo "<form name='myDelForm'>
$news_gen_id<br>
<input type='hidden' id='delccode' value='".$news_gen_id."'>
<input type='hidden' id='deledit' value='".$edit."'>
<input type='button' onclick='ajaxDelCcode()' value='Delete' /><br></form>
</td>";
The AJAX code (on index.php, where the div that calls in main.php is also located) is this
function ajaxDelCcode(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new
ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById("ajaxMain2");
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
var deledit = document.getElementById("deledit").value;
var delccode = document.getElementById("delccode").value;
var queryString = "?delccode=" + delccode + "&deledit=" + deledit;
ajaxRequest.open("GET", "main.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
Currently, using those two pieces of code, I can successfully delete only the first product. The delccode variables do not seem to change when the products are looped (although when I echo the variables during the loop, it is definitely changing to the appropriate value...it's just not passing it correctly back to AJAX.)
I tried taking the AJAX code, putting it inside the main.php product loop, and change the function name during each loop (so ajaxDelCcode$news_gen_id() for example) and also to the button itself so that it is calling the AJAX specific to it. And it works if you are visiting main.php directly...but not from index.php after main.php has been called into the div.
I can't figure out how to pass the correct looped value from main.php within the div, back to the AJAX code on index.php
Can anyone help me with this?
Thanks,
Dustin
Instead of storing the id in the input, just pass it as an argument to the function:
function ajaxDelCcode(delccode) { ...
<input type='button' onclick='ajaxDelCcode(\"".$news_gen_id."\")' value='Delete' />
Also, I'd swap the quotes if I were you. Or better yet, instead of using echo, break the PHP code and just write HTML:
<? ... ?><input type="button" onclick="ajaxDelCcode('<?= $news_gen_id ?>')" value="Delete" /><? ... ?>
What does the code you use to delete look like? Is it in the same php file as the form you posted above? If so, is the form getting submitted to itself accidentally? Like perhaps when a user presses enter while on an input type=text control? I understand that you want to do this by ajax but I am suspecting that the form is your problem.
Seconding the jQuery comment.
Here try this
1) add jquery to your document.
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
2) give your inputs name attributes
<input type='hidden' name='delcode' id='delccode' value='".$news_gen_id."'>
<input type='hidden' name='deledit' id='deledit' value='".$edit."'>
3) Use a function something like this instead of all that code above
function ajaxDelCcode() {
$.ajax({
url: "main.php",
type: "GET",
dataType: "text",
data: $("#myDelForm").serialize(),
success: function(rText) {
$("#ajaxMain2").text(rText);
}
});
}

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