Search Results

Search found 30 results on 2 pages for 'javed abbas'.

Page 2/2 | < Previous Page | 1 2 

  • difference between calling javascript function on body load or directly from script.

    - by Abbas
    i am using a javascript where in i am creating multiple div (say 5) at runtime, using javascript function, all the divs contain some text, which is again set at runtime, now i want to disable all the divs at runtime and have the page numbers in the bottom, so that whenever user clicks on the page number only that div should get visible else other should get disable, i have created a function, which accepts parameter, as page number, i enable the div whose page number is clicked and using a for loop, i disable all the other divs, now here my problem is i have created two functions, 1st (for adding divs and disabling all the divs except 1st) and writing content to it, and other for enabling the div whose page number is clicked, and i have called the Adding div function on body onload; now first time when i run, page everthing goes well, but next time when i click on any of the page number, it just gets enabled and again that AddDiv function, runs and re-enables all the divs.. Please reply why this is happening and how should i resolve my issue... Below is my script, content for the div are coming using Json. <body onload="JsonScript();"> <script language="javascript" type="text/javascript"> function JsonScript() { var existingDiv = document.getElementById("form1"); var newAnchorDiv = document.createElement("div"); newAnchorDiv.id = "anchorDiv"; var list = { "Article": articleList }; for(var i=0; i < list.Article.length; i++) { var newDiv = document.createElement("div"); newDiv.id = "div"+(i+1); newDiv.innerHTML = list.Article[i].toString(); newAnchorDiv.innerHTML += "<a href='' onclick='displayMessage("+(i+1)+")'>"+(i+1)+"</a>&nbsp;"; existingDiv.appendChild(newDiv); existingDiv.appendChild(newAnchorDiv); } for(var j = 2; j < list.Article.length + 1; j ++) { var getDivs = document.getElementById("div"+j); getDivs.style.display = "none"; } } function displayMessage(currentId) { var list = {"Article" : articleList} document.getElementById("div"+currentId).style.display = 'block'; for(var i = 1; i < list.Article.length + 1; i++) { if (i != currentId) { document.getElementById("div"+i).style.display = 'none'; } } } </script> Thanks and Regards

    Read the article

  • Running multiple environments on one AWS EC2 instance (Elastic Beanstalk)

    - by Abbas
    I am very new to the Amazon AWS services. I was wondering if there is a way to run an instance of EC2 (say, Amazon Linux AMI) and then connect two environments to this instance. Particularly, I'd like to run a PHP and a Tomcat environment on a single EC2 instance. The problem is, every time I create a new environment in Elastic Beanstalk, it seems to create a new EC2 instance as well. Am I missing something here? I'd appreciate any hint on this.

    Read the article

  • Django QuerySet API: How do I join iexact and icontains?

    - by Zeynel
    Hello, I have this join: lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name) This is the site If you try Last Name: Abbas and First Name: Amr it tells you that amr abbas has 1 schoolmates. But if you try First name only it says that there are no lawyers in the database called amr (obviously there is). If I change (last__iexact=last_name) to (last__icontains=last_name) then leaving Last Name blank works fine and amr is found. But with last__icontains=last_name if you search for "collin" you also get "collins" and "collingwood" which is not what I want. Do you know how I can use iexact and also have it ignored if it is blank? Thanks This is the view function: def search_form(request): if request.method == 'POST': search_form = SearchForm(request.POST) if search_form.is_valid(): last_name = search_form.cleaned_data['last_name'] first_name = search_form.cleaned_data['first_name'] lawyers = Lawyer.objects.filter(last__iexact=last_name).filter(first__icontains=first_name) if len(lawyers)==0: form = SearchForm() return render_to_response('not_in_database.html', {'last': last_name, 'first': first_name, 'form': form}) if len(lawyers)>1: form = SearchForm(initial={'last_name': last_name}) return render_to_response('more_than_1_match.html', {'lawyers': lawyers, 'last': last_name, 'first': first_name, 'form': form}) q_school = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('school', flat=True) q_year = Lawyer.objects.filter(last__icontains=last_name).filter(first__icontains=first_name).values_list('year_graduated', flat=True) lawyers1 = Lawyer.objects.filter(school__iexact=q_school[0]).filter(year_graduated__icontains=q_year[0]).exclude(last__icontains=last_name) form = SearchForm() return render_to_response('search_results.html', {'lawyers': lawyers1, 'last': last_name, 'first': first_name, 'form': form}) else: form = SearchForm() return render_to_response('search_form.html', {'form': form, })

    Read the article

  • Code golf: find all anagrams

    - by Charles Ma
    An word is an anagram if the letters in that word can be re-arranged to form a different word. Task: Find all sets of anagrams given a word list Input: a list of words from stdin with each word separated by a new line e.g. A A's AOL AOL's Aachen Aachen's Aaliyah Aaliyah's Aaron Aaron's Abbas Abbasid Abbasid's Output: All sets of anagrams, with each set separated by a separate line Example run: ./anagram < words marcos caroms macros lump's plum's dewar's wader's postman tampons dent tend macho mocha stoker's stroke's hops posh shop chasity scythia ... I have a 149 char perl solution which I'll post as soon as a few more people post :) Have fun!

    Read the article

< Previous Page | 1 2