幫助For循環。值重複

Question

$teams = array(1, 2, 3, 4, 5, 6, 7, 8); 
$game1 = array(2, 4, 6, 8); 
$game2 = array(); 

如果teams[x]不game1然後插入game2

for($i = 0; $i < count($teams); $i++){ 
    for($j = 0; $j < count($game1); $j++){ 
     if($teams[$i] == $game1[$j]){ 
      break; 
     } else { 
      array_push($game2, $teams[$i]); 
     } 
    } 
} 

for ($i = 0; $i < count($game2); $i++) { 
    echo $game2[$i]; 
    echo ", "; 
} 

林希望得到的結果是：

1, 3, 5, 7, 

然而，即時得到：

1, 1, 1, 1, 3, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 8, 

我怎樣才能改善這一點？謝謝

Answer 1

其他人已經回答如何使用array_diff。

之所以現有的循環不起作用：

if($teams[$i] == $game1[$j]){ 
     // this is correct, if item is found..you don't add. 
     break; 
    } else { 
     // this is incorrect!! we cannot say at this point that its not present. 
     // even if it's not present you need to add it just once. But now you are 
     // adding once for every test. 
     array_push($game2, $teams[$i]); 
    } 

您可以使用一個標誌，修復現有的代碼爲：

for($i = 0; $i < count($teams); $i++){ 
    $found = false; // assume its not present. 
    for($j = 0; $j < count($game1); $j++){ 
     if($teams[$i] == $game1[$j]){ 
      $found = true; // if present, set the flag and break. 
      break; 
     } 
    } 
    if(!$found) { // if flag is not set...add. 
     array_push($game2, $teams[$i]); 
    } 
} 

Answer 2

你的循環不起作用，因爲每一個元素時從$teams不等於$game1中的元素，它將$teams元素添加到$game2。這意味着每個元素被多次添加到$game2。

使用array_diff代替：

// Find elements from 'teams' that are not present in 'game1' 
$game2 = array_diff($teams, $game1); 

Answer 3

您可以使用PHP的array_diff()：

$teams = array(1, 2, 3, 4, 5, 6, 7, 8); 
$game1 = array(2, 4, 6, 8); 
$game2 = array_diff($teams,$game1); 

// $game2: 
Array 
(
    [0] => 1 
    [2] => 3 
    [4] => 5 
    [6] => 7 
)