帮助For循环.值重复

Question

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

如果teams[x]不在game1那么插入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 ", ";
}

我期待结果是:

1, 3, 5, 7,

但是,我得到:

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

我怎么能改善这个？谢谢

Answer 1

其他人已经回答了如何使用array_diff.

您现有循环不起作用的原因:

    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]);
    }

您可以使用标志来修复现有代码:

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

你的循环不起作用,因为每次元素来自$teams不等于元素时$game1,它都会添加$teams元素$game2.这意味着每个元素都被$game2多次添加.

array_diff改为使用:

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