ios - different methods on one standard headCell in func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell{ -

-

i have 2 tableviews in same view controller: "studentstable" , "professorstable". wanted create standard head cell both tables. subclassed uitableviewcell , created headcell type. assigned 1 headcell both if indexpath.row==0. went smooth, however, there 1 problem. whenever click "addingbutton" want "addstudent" function called if on "studentstable" , "addprofessor" if on "professorstable". can create 2 headcells, 1 each of tables wonder if there more professional way of handling writing less code. here code suppose may work if right "if" condition given. in advance

 func tableview(tableview: uitableview, cellforrowatindexpath indexpath: nsindexpath) -> uitableviewcell{    var actiontype:string=string()    if(/*what condition should put?*/)     {actiontype="addstudent"}    else    {actiontype="addprofessor"}       if(indexpath.row==0)            {                var theheadcell : headcell = tableview.dequeuereusablecellwithidentifier("headcell") headcell             // cell setup               if(actiontype != "")                 {                    addbutton.addtarget(self, action: nsselectorfromstring(actiontype), forcontrolevents: uicontrolevents.touchupinside)                 }               return theheadcell            }      else      {          if (tableview==self.studentstable)              {                  var cell : uitableviewcell = tableview.dequeuereusablecellwithidentifier("studentscell") uitableviewcell                    // cell setup                   return cell              }         else             {                     var cell : uitableviewcell = tableview.dequeuereusablecellwithidentifier("professorscell") uitableviewcell                  // cell setup                 return cell            }     }         } 

if tableview == self.studentstable     {actiontype = "addstudent"} else if tableview == self.professorstable    {actiontype = "addprofessor"}

Comments

Post a Comment

Popular posts from this blog

php - Admin SDK -- get information about the group -

-
please, help. https://www.googleapis.com/admin/directory/v1/groups/106274944935548788283 sending request obtain information group id 1062 ... here link method in google https://developers.google.com/admin-sdk/directory/v1/guides/manage-groups#get_group in response error - 401 unauthorized. oauth authenticate via user has received rights necessary request, included in console admin sdk. authorization successful, there token. not can not understand. if not hard - response. requested information: file_get_contents("https://www.googleapis.com/admin/directory/v1/groups/1062749‌​44935548788283?key=aizasybruhiomdf9nlx9lx3imxwesqmcrnhh4l4"); id group - "106274944935548788283" key applications in case - "aizasybruhiomdf9nlx9lx3imxwesqmcrnhh4l4" error: failed open stream: http request failed! http/1.0 401 unauthorized
Read more

dns - How To Use Custom Nameserver On Free Cloudflare? -

-
i have problem. i own small hosting company. trying add cloudflare site free cdn. problem when sign have change nameservers nameservers. have nameserver dns set through godaddy, how add them cloudflare? main site uses company's nameservers whenever add them site goes offline. when go cloudflare's dns records click on dropdown, click ns, , there's 2 boxes 'name' , 'nameserver'. mean input 'ns1' 'name' , 'ns1.domain.com' 'nameserver'? input server's ip addresses? is there way of accomplishing free version of cloudflare? that impossible. option have custom name servers available on business , enterprise. see https://www.cloudflare.com/plans . also, might mistaken. must change name servers in domain register settings, not on cloudflare.
Read more

Python Error - TypeError: input expected at most 1 arguments, got 3 -

-
can explain why can not use your_name in goal variable? my_name = "bryson" my_age = 29 your_name = input ("what name? ") your_age = input ("what age? ") print ("my name is", my_name,", , am", my_age, "years old.") print ("your name is", your_name,", , are", your_age,".") print("thank buying book,", your_name,"!") goal = input ("what favorite part of book,", your_name, "?") print("awesome!") the error is: goal = input ("what favorite part of book,", your_name, "?") typeerror: input expected @ 1 arguments, got 3 you got error because in fact gave 3 arguments input function when expecting 1 (namely, string prompt). in input ("what favorite part of book,", your_name, "?") ---------------------------------------- , ---------, --- the underlined parts comma-separated arguments: st...
Read more