php - 对 watch 。健康)状况

Question

 for ($i=1; $i < 10; $i++) { 
    $a = rand(1,6);
    $b = rand(1,6);
    if ($a != $b) {
        $mysqli->query("UPDATE `jpa` SET `war` = $b WHERE `id` = $a");
        $mysqli->query("UPDATE `jpa` SET `war` = $a WHERE `id` = $b");
    }
 }

有必要编制一份对 watch 。执行时得到同样的结果，如何避免这种情况？

Answer 1

从 id 数组开始。

$ids = [1, 2, 3, 4, 5, 6];

随机播放

shuffle($ids);

然后分成 2 个 block 并迭代它们，进行更新。

foreach (array_chunk($ids, 2) as $opponents) {
    if (count($opponents) == 2) {  // check this in case you have odd numbers of ids
        list($a, $b) = $opponents;
        $mysqli->query("UPDATE `jpa` SET `war` = $b WHERE `id` = $a");
        $mysqli->query("UPDATE `jpa` SET `war` = $a WHERE `id` = $b");
    }
}