python - How do I fully break out of a while loop nested inside a while loop? -


this question has answer here:

i wondering if can me figure out how break out of while loop(s) , continue rest of program. thanks!

import time while true:     company_name = input("\nwhat name of company? ")     if company_name == "":         time.sleep(1)         print("\nthis not eligible. please try again")     else:         while true:             verify_name = input("\nplease verify {} correct name of company \nusing yes or no: ".format(company_name))             if verify_name.lower() == "no":                 print("\nplease re-enter company name.")                 time.sleep(1)                 break             elif verify_name.lower() not in ('yes', 'y'):                 print("\nthis invalid response, please try again.")                 time.sleep(1)                 break             else:                 print("\nwelcome {}.".format(company_name))                 verify_name == true                 break         else:             break #continue rest of program

the solution below adds flag control when break out of external loop set break out each loop, , set if no break has occurred in internal loop, i.e. else statement has been reached on inner loop.

import time no_break_flag = true while no_break_flag:     no_break_flag = false     company_name = input("\nwhat name of company? ")     if company_name == "":         time.sleep(1)         print("\nthis not eligible. please try again")     else:         while true:             verify_name = input("\nplease verify {} correct name of company \nusing yes or no: ".format(company_name))             if verify_name.lower() == "no":                 print("\nplease re-enter company name.")                 time.sleep(1)                 break             elif verify_name.lower() not in ('yes', 'y'):                 print("\nthis invalid response, please try again.")                 time.sleep(1)                 break             else:                 print("\nwelcome {}.".format(company_name))                 verify_name == true                 break         else:             no_break_flag = true #continue rest of program

obviously have condition of while true on inner loop exit breaking, if had other condition break external loop if break statement reached on inner loop.


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