int n; cin >> n; cin.ignore(); // are you sure you tried this? for(int i = 0; i < n; i++) { string input; getline(cin, input); cout << “Line Entered: ” << input << endl; } 0 solved How to force getline() to input one line at a time [duplicate]
That is reinventing the wheel, use str.isalpha You could use assert and AssertionError from string import ascii_letters value = None while True: try: value = input(“Give a value: “) assert all(c in ascii_letters for c in value) break except AssertionError: print(“Invalid input, try again”) print(“Valid input:”, value) Give a value: aa! Invalid input, try again … Read more
while (1) { scanf(“%d”, &n0); if (n0==0) break; … } 1 solved How to solve this code? [closed]
raw_input() takes in a line of input as a string. The .split() string method produces a list of the whitespace-separated words in a string. int() parses a string as an int. So putting that together, you can read the input you want using t = int(raw_input()) cases = [] for i in range(t): cases.append( map(int, … Read more
You can use OS redirection operators java Test < inputFile > outputFile then this public class Test { public static void main(String[] args) throws IOException { for(int i; (i = System.in.read()) != -1;) { System.out.write((byte)i); } } } will copy from inputFile to outputFile 7 solved Give standard input to program through files and take … Read more
the correct way is: int myreadfile(void) { FILE *fp; int i, n; unsigned char a[50]; if (!(fp=fopen(“x.exe”,”rb”))) return(0); while(n=fread(a,1,sizeof(a), fp)) { for (i=0; i<n; i++) printf(“%02x “,a[i]); printf(“\n”); } fclose(fp); return 1; } Note that the buffer is of type unsigned char. That is because a) you don’t know if the file is a complete … Read more
int main() { string calculateagain = “yes”; do { //… Your Code cout << “Do you want to calculate again? (yes/no) ” cin >> calculateagain; } while(calculateagain != “no”); return 0; } Important things to note: This is not checked, so user input may be invalid, but the loop will run again. You need to … Read more
BufferedWriter bw = new BufferedWriter( new OutputStreamWriter( new FileOutputStream(new File(“path of the output file”)), “UTF-8”)); Or in java 7 use Files.copy. 4 solved read special characters from a file and write in another file using java [duplicate]
This should do it: http://regexr.com?353i8 (/([\d]{18}\.)(jpg|png|gif)/g) This site will help you with your regex’ 8 solved Regex formula for a file name
// Logic One Using For Loop <?php error_reporting(0); $con=mysql_connect(‘localhost’,’root’,”) or die(‘Not Connect’); mysql_select_db(‘multiple’,$con); if (isset($_POST[“submit”])){ $field_name = $_POST[‘values’]; $values = “”; for ($i = 0; $i < sizeof($field_name); $i++) { $values .= “(‘”.$field_name[$i].”‘)”; if ($i != sizeof($field_name) – 1) { $values .= “, “; } } $sql = mysql_query(“INSERT INTO php_test VALUES (” . $values.”)”); … Read more
Go to Your example And make Ctrl-U everything you need is visible 2 solved CSS/Javascript Create Input with Border-Bottom [closed]
Best way I think double NthRoot(double value, double degree) { return pow(value, (double)(1 / degree)); } From here. 0 solved n-th root of a number [closed]
I can’t write a python average function that will average 3 input integers from user. (average has to be in float) [closed] solved I can’t write a python average function that will average 3 input integers from user. (average has to be in float) [closed]
Just you can use int(): number=int(input(‘Enter a number? ‘)) print(number) Otherwise, to be fool-proof, use try..except: while True: try: number=int(input(“Enter a number ? “)) break except ValueError: print(“Please input a vaild number !!”) solved How to convert a string to integer entered in input functon? [duplicate]
well what the comments said and change it like the following too : id = [1, 2, 3, 4, 5, 6] name = [‘sarah’, ‘john’, ‘mark’, ‘james’, ‘jack’] userid = int(input(‘enter user ID: ‘)) if userid in id: ind = id.index(userid) #notice this statement is inside the if and not outside print(name[ind]) else: print(“wrong id”) … Read more