Question

    public class Game

{

 public static void main(String[] args)

 {
Scanner scan = new Scanner(System.in); // Creates scanner object.
Random numb = new Random(); // Creates an instance of the random class.

String newgame = "y";

while (newgame.equalsIgnoreCase("y")) {
  int count = 0; // Placeholder for the guess counter.
  int guess = -1; // Placeholder for users guess.
  int answer = numb.nextInt(100) + 1; // Generates a random number for the game.
  int sentinel = 0; // Placeholder for players answer as to whether they want to play again or not.

  while (guess != sentinel && guess != answer) // Loop that ends when user enters a zero.
  {
    System.out.println("Enter a number between 1-100 or 0 to quit");
    guess = scan.nextInt();
    count++;

    if (guess < answer && guess > 0) {
      System.out.println("Your guess is too low, guess again");
    } else if (guess > answer) {
      System.out.println("Your guess is to high, guess again");
    }

    else if (guess == answer) {
      System.out.println();
      System.out.println("You guessed correctly, you win!!!");
      System.out.println("It took you " + count + " guesses");
    }
  }
  System.out.println();
  System.out.println("Play another game: y or n?");
  newgame = scan.next();
}
}
}

如果$url = 'http://www.example.com/'.$type.$q.'number-'.$L.','.$country.'.html或$type之后包含$q，则示例：

但是当例如$type = 1 $q = 2 http://www.example.com/1-2-number-fl,us.html不存在时：

$type

更新：如果 $ type = 1 - 或 $ q = 1 - ，那么你可以阻止这种情况发生： http://www.example.com/1--2--number-fl,us.html < /强>

Answer 1

另一种方式：

if(isset($type)) { $array[] = $type; }
if(isset($q))    { $array[] = $q;    }
$array[] = 'number';
if(isset($L))    { $array[] = $L;    }

$url = "http://www.example.com/".implode('-', $array).",$country.html";

检查变量是否设置为isset()
如果是，请将它们添加到数组
由于number是您想要的固定字符串，因此将其添加到数组
implode()加入-的数组，以便在网址

您可以将isset()替换为!empty()，但0被视为空，因此无法在网址中使用。

Answer 2

如果两者都存在则使用isset（）然后构建字符串

if (isset($type, $q)) {
  $result = $type .'-'. $q;
{

Answer 3

检查$ type和$ q是否为空是很重要的 - 仅仅检查它们是否已设置是不够的。您还需要覆盖值为整数0的情况，因为PHP将0视为空字符串，这会导致空（0）返回true。

你可以选择这样的东西：

if ( ((int) $type !== 0 && !empty( $type )) && ((int) $q !== 0 && !empty( $q ))) {
    $result = $type . '-' . $q;
}
elsif ((int) $type !== 0 && !empty( $type )){
    $result = $type;
}
else {
    $result = $q;
}

$url = "http://www.example.com/{$result}-number-{$L}{$country}.html";

Answer 4

首先检查变量是否已设置且不为空。为每个变量添加“ - ”。

$url = 'http://www.example.com/'.$type.$q.'number-'.$L.','.$country.'.html';
// declare a variable
$result = '';
// check for the variable type and add '-'
if(isset($type) && ($type!='')){
    $result=trim($type,'-').'-';
}
// check for the variable q and add '-'
if(isset($q) && ($q!='')){
$result.=trim($q,'-').'-';
}

$url = 'http://www.example.com/'.$result.'number-'.$L.','.$country.'.html';

Answer 5

检查isset（）

if(isset($type))
    http://www.example.com/1-2-number-fl,us.html
else
    http://www.example.com/2-number-fl,us.html

PHP＆＃34;如何包含＆＃39; - ＆＃39;只有当变量存在时？

5 个答案: