Your problem comes from a bad use of switch See Switch/Case documentation. Your menu switch/case will triggered every cases switch(1) { case 1 : cout << ‘1’; // prints “1”, case 2 : cout << ‘2’; // then prints “2”…
I figured it out. string = ‘123456789’ splitted = [] prev = 0 while True: n = random.randint(1,3) splitted.append(string[prev:prev+n]) prev = prev + n if prev >= len(string)-1: break print splitted 0 solved Split string by n when n is…
Try the regular expression ^\\d{10,15}$ Here \d is a predefined character class for digits{10, 15} quantifier stands for a repeat of 10 to 15 times of the previous pattern Ex: String input = “1234567890”; Pattern pattern = Pattern.compile(“^\\d{10,15}$”); if (pattern.matcher(input).find())…
If you have char arr[10] = “1234567890” as per your question, note that it is not a string since it has no room for the NUL terminator (would need size 11, or left empty [] for the compiler to count…
To format a number using fixed-point notation, you can simply use the toFixed method: (10.8).toFixed(2); // “10.80” var num = 2.4; alert(num.toFixed(2)); // “2.40” Note that toFixed() returns a string. IMPORTANT: Note that toFixed does not round 90% of the…
This looks a bit like minimised code, where function call are often shortened by creating lookup tables, so you can save some bits on long function calls like Element.dispatchEvent and instead minimise it to e[l(p)]. The variable _0x76ed32 holds the…
In your html view page some change echo “<td><a href=””‘>Delete</a></td>”; like bellow: <?php while($row = mysql_fetch_array($result)) { echo “<tr>”; echo “<td>” . $row[‘name’] . “</td>”; echo “<td>” . $row[‘id’] . “</td>”; echo “<td>” . $row[‘rollnumber’] . “</td>”; echo “<td>” .…
Looking at your code you are trying to print the element itself. System.out.println(FatherNum); If you want to print the value of that field , you can do it by : System.out.println(FatherNum.getAttribute(“value”)); OR String Fathernum= FatherNum.getAttribute(“value”); System.out.println(Fathernum); See the doc for…
Fairly simple: use two loops, one for counting up and one for counting down. Print literal “0” between the two. #include <stdio.h> int main() { for (int i = 0; i < 10; i++) { for (int j = 10…
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