excel - Call PERSONAL macro from other workbook -


i working on macro receive data other file shared template. after user runs macro within shared workbook, update workbook. want make accessible possible.

i don't want users open template , update manually. that's not point. works fine. when call macro developer -> macros -> run, data copied template correct, , update works too.

but when set shortcut in developer -> macros -> options , call macro shortcut others workbooks not personal, macro stops in middle after opening shared workbook. , doesn't update data other workbooks.

am missing here guys? ideas?

edit:

my code:

    public sub kpi()      application.screenupdating = false      kpi = "c:\kpi template.xlsx"      workbooks.open (kpi)      '****** <--- macro stops here - no error      patht = "c:\"     filenn = "kpi"     filepath = dir(patht)      while true          if instr(1, filepath, filenn) > 0              goto resume1         end if          filepath = dir      wend     resume1:      workbooks.open (patht & filepath), readonly:=true      kpi_filename = dir(kpi)      r = 8     k = 2     lday = day(date)      sh_name = workbooks(kpi_filename).sheets("check").cells(r, 2)      while workbooks(kpi_filename).sheets("check").cells(r, 2) <> ""          while workbooks(filepath).sheets(sh_name).cells(lday + 7, k) <> ""              workbooks(kpi_filename).sheets("check").cells(r, k + 1) = workbooks(filepath).sheets(sh_name).cells(lday + 7, k)          k = k + 1         loop      r = r + 2     k = 2     sh_name = workbooks(kpi_filename).sheets("check").cells(r, 2)     loop

end sub


Comments

Post a Comment

Popular posts from this blog

python - Operations inside variables -

this question has answer here: evaluating mathematical expression in string 12 answers i want make python script finds maximum possible output using 2 operations. problem says there syntax error @ 2nd last line, in b in there. how can fix it? x = 0 = int(input("1st number:")) c = int(input("2nd number:" )) e = int(input("3rd number:" )) in range(4): if == 0: b = "+" elif == 1: b = "-" elif == 2: b = "/" else: b = "*" j in range(4): if j == 0: d = "+" elif j == 1: d = "-" elif j == 2: d = "/" else: d = "*" k = b c d e print(k) if you're after variables e on single line, couldn...
Read more

Generic Map Parameter java -

i having lot of methods of type: public static list<eduusers> getdetails(class c, map<string, integer> params) { list<eduusers> details = null; session session = hibernateutil.getsessionfactory().opensession(); transaction tx = null; try { tx = session.begintransaction(); criteria cr = session.createcriteria(c); (map.entry<string, integer> entry : params.entryset()) { cr.add(restrictions.eq(entry.getkey(), entry.getvalue())); } details = cr.list(); tx.commit(); } catch (exception asd) { log.debug(asd.getmessage()); if (tx != null) { tx.commit(); } } { session.close(); } return details; } i trying have generic method them , have written far: public static <t> list<t> getdetails(class<t> c, class<t> b) { l...
Read more

arrays - What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it? -

what arrayindexoutofboundsexception mean , how rid of it? here code sample triggers exception: string[] name = {"tom", "dick", "harry"}; for(int = 0; i<=name.length; i++) { system.out.print(name[i] +'\n'); } your first port of call should documentation explains reasonably clearly: thrown indicate array has been accessed illegal index. index either negative or greater or equal size of array. so example: int[] array = new int[5]; int boom = array[10]; // throws exception as how avoid it... um, don't that. careful array indexes. one problem people run thinking arrays 1-indexed, e.g. int[] array = new int[5]; // ... populate array here ... (int index = 1; index <= array.length; index++) { system.out.println(array[index]); } that miss out first element (index 0) , throw exception when index 5. valid indexes here 0-4 inclusive. correct, idiomatic for statement here be: for (int index = 0; index <...
Read more