javascript - Hiding dynamically generated divs on 'Click Event' -


i working on project have created 4 divs dynamically, first "display:block". when click next, second div "display:block" , 1st, 3rd, 4th "display:none"; problem when click next button fast 2 times - 2 divs "display:block" , both seen on screen. have tried following code, fed of this. there way control show-hide of dynamically generated divs? 

 <script>  /* first check 1 div must show on click of 4 clickable buttons - failed .. ;( */           $(document).ready(function(){             $('span#backpage').click(function(){                 var alpha1 = $('div.ui-formwizard-content').css('display','block');                 $('div.ui-formwizard-content').css('display','none');                 alpha1.css('display','block');             });             $('span#nextpage').click(function(){                 var alpha1 = $('div.ui-formwizard-content').css('display','block');                 $('div.ui-formwizard-content').css('display','none');                 alpha1.css('display','block');             });             $('span#prevfield').click(function(){                 var alpha1 = $('div.ui-formwizard-content').css('display','block');                 $('div.ui-formwizard-content').css('display','none');                 alpha1.css('display','block');             });            $('span#nextfield').click(function(){                 var alpha1 = $('div.ui-formwizard-content').css('display','block');                 $('div.ui-formwizard-content').css('display','none');                 alpha1.css('display','block');             });  /* double check 1 div must show @ 1 time - failing */             if($('div#step_0').css('display','block')){$('div.step').css('display','none');$(this).css('display','block');}             if($('div#step_1').css('display','block')){$('div.step').css('display','none');$(this).css('display','block');}             if($('div#step_2').css('display','block')){$('div.step').css('display','none');$(this).css('display','block');}             if($('div#step_3').css('display','block')){$('div.step').css('display','none');$(this).css('display','block');}          });       </script>

if can me out of mess-of multiple display:block of dynamically genearted divs, glad! ;( have tried hide() , show() selected one. have tried still problem persists. dont know how other divs hide if new 1 gets blocked!

quick , untested. show unnecessary code. perhaps track down problem.

 $(document).ready(function(){         $('span#backpage').click(function(){             showalpha(this);         });         $('span#nextpage').click(function(){             showalpha(this);         });         $('span#prevfield').click(function(){             showalpha(this);         });        $('span#nextfield').click(function(){             showalpha(this);         });          function showalpha(_this) {             var alpha1 = $('div.ui-formwizard-content');             alpha1.show();             // correct active child here html figure out how in case             var child = ???;             $('div.step').hide();             $(child).show();         }      });

edited


Comments

Post a Comment

Popular posts from this blog

php - failed to open stream: HTTP request failed! HTTP/1.0 400 Bad Request -

this code i'm using find latitude , longitude of place: function coords($address){ echo $url="http://maps.googleapis.com/maps/api/geocode/json?address=".$address; $json = file_get_contents(urlencode($url)); $data = json_decode($json, true); print_r($data['results'][0]['geometry']['location']['lat']); print_r($data['results'][0]['geometry']['location']['lng']); } however returns warning: failed open stream: http request failed! http/1.0 400 bad request if use above code in simple procedure works fine not in function. note: have tried curl_init()... method produced same result? instead of encoding whole url encode address part. following code work fine $add='jamshoro phase 2'; $url="http://maps.googleapis.com/maps/api/geocode/json?address=".urlencode($add); $json = file_get_contents($url); $data = json_decode($json, true); print_r($data['r
Read more

java - How to filter a backspace keyboard input -

i have code gives message when key number pressed: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochararray().length - 1]; if (!(ch >= '0' && ch <= '9')) { system.out.println("the char entered not number"); t.consume(); } } }); now if press on wrong button , press backspace remove it, error message. how can add backspace input if statement? i have found answer: txtvalueofclause.addeventfilter(keyevent.key_typed, new eventhandler<keyevent>() { @override public void handle(keyevent t) { char ar[] = t.getcharacter().tochararray(); char ch = ar[t.getcharacter().tochar
Read more

java - Show Soft Keyboard when EditText Appears -

i have code in android app, alert dialog , keyboard, , when dialog shown want keyboard appear. not , not sure why? here code: final alertdialog.builder alert = new alertdialog.builder(this); alert.settitle("title"); alert.setmessage("message"); final edittext input = new edittext(this); input.setinputtype(inputtype.type_class_number); alert.setview(input); input.setonfocuschangelistener(new view.onfocuschangelistener() { @override public void onfocuschange(view v, boolean hasfocus) { if (hasfocus) { log.i(tag, "in focus"); inputmethodmanager inputmethodmanager = (inputmethodmanager) getsystemservice(context.input_method_service); inputmethodmanager.showsoftinput(input, inputmethodmanager.show_implicit); } } }); but keyboard not appear though edit text in focus, , tell appear why, doesn't come up? thanks in advance.
Read more