Question

我一直在编码和改进代码 jszobody 为我提供了可以在此LINK上找到的代码，它让我成为了我的参考但却导致了错误{{1} }

PHP：

Fatal error: Using $this when not in object context

预期输出：

数组（[0] =＆gt; HappyOutgoingTallHandsome 1 =＆gt; HappyOutgoingTallPlain [2] =＆gt; HappyOutgoingTallUgly [3] =＆gt; HappyOutgoingShortHandsome [4] =＆gt; HappyOutgoingShortPlain [5] =＆gt; HappyOutgoingShortUgly [ 6] =＆gt; HappyOutgoingMediumHandsome [7] =＆gt; HappyOutgoingMediumPlain [8] =＆gt; HappyOutgoingMediumUgly [9] =＆gt; HappyIntrovertedTallHandsome [10] =＆gt; HappyIntrovertedTallPlain [11] =＆gt; HappyIntrovertedTallUgly [12] =＆gt; HappyIntrovertedShortHandsome [13 ] =＆gt; HappyIntrovertedShortPlain [14] =＆gt; HappyIntrovertedShortUgly [15] =＆gt; HappyIntrovertedMediumHandsome [16] =＆gt; HappyIntrovertedMediumPlain [17] =＆gt; HappyIntrovertedMediumUgly [18] =＆gt; SadOutoveTallHandsome [19] =＆gt; SadOutgoingTallPlain [20] =＆gt; SadOutgoingTallUgly [21] =＆gt; SadOutgoingShortHandsome [22] =＆gt; SadOutgoingShortPlain [23] =＆gt; SadOutgoingShortUgly [24] =＆gt; SadOutgoingMediumHandsome [25] =＆gt; SadOutgoingMediumPlain [26] =＆gt; SadOutgoingMediumUgly [27] = ＆gt; SadIntroverted TallHandsome [28] =＆gt; SadIntrovertedTallPlain [29] =＆gt; SadIntrovertedTallUgly [30] =＆gt; SadIntrovertedShortHandsome [31] =＆gt; SadIntrovertedShortPlain [32] =＆gt; SadIntrovertedShortUgly [33] =＆gt; SadIntrovertedMediumHandsome [34] =＆gt; SadIntrovertedMediumPlain [35] =＆gt; SadIntrovertedMediumUgly [36] =＆gt; AngryOutgoingTallHandsome [37] =＆gt; AngryOutgoingTallPlain [38] =＆gt; AngryOutgoingTallUgly [39] =＆gt; AngryOutgoingShortHandsome [40] =＆gt; AngryOutgoingShortPlain [41] =＆gt; AngryOutgoingShortUgly [42] =＆gt; AngryOutgoingMediumHandsome [43] =＆gt; AngryOutgoingMediumPlain [44] =＆gt; AngryOutgoingMediumUgly [45] =＆gt; AngryIntrovertedTallHandsome [46] =＆gt; AngryIntrovertedTallPlain [47] =＆gt; AngryIntrovertedTallUgly [48] =＆gt; AngryIntrovertedShortHandsome [49] =＆gt; AngryIntrovertedShortPlain [50] =＆gt; AngryIntrovertedShortUgly [51] =＆gt; AngryIntrovertedMediumHandsome [52] =＆gt; AngryIntrovertedMediumPlain [53] =＆gt; AngryIntrovertedMediumUgly [54] =＆gt; HopefulOutgoingTallHandsome [55] =＆gt; HopefulOutgoingTallPlain [56] =＆gt; HopefulOutgoingTallUgly [57] =＆gt; HopefulOutgoingShortHandsome [58] =＆gt; HopefulOutgoingShortPlain [59] =＆gt; HopefulOutgoingShortUgly [60] =＆gt; HopefulOutgoingMediumHandsome [61] =＆gt; HopefulOutgoingMediumPlain [62] =＆gt; HopefulOutgoingMediumUgly [63] =＆gt; HopefulIntrovertedTallHandsome [64] =＆gt; HopefulIntrovertedTallPlain [65] =＆gt; HopefulIntrovertedTallUgly [66] =＆gt; HopefulIntrovertedShortHandsome [67] =＆gt; HopefulIntrovertedShortPlain [68] =＆gt; HopefulIntrovertedShortUgly [69] =＆gt; HopefulIntrovertedMediumHandsome [70] =＆gt; HopefulIntrovertedMediumPlain [71] =＆gt; HopefulIntrovertedMediumUgly [72] =＆gt; ）

我哪里出错？

Answer 1

你有很多对类变量的引用，比如：

$this->permutationsResultsArray = array();

PHP抱怨因为这个函数不是类中的方法。如果您只是删除this->，那么它将会有效：

$permutationsResultsArray = array();

此外，当你完成所有工作后，你永远不会真正返回结果..就像这样：

return $permutationsResultsArray;

虽然有问题。你正在重复使用，除了在第一轮中你没有创建那个数组，但是你的代码使用它就像定义它一样。

BTW：使用3个foreach循环可以更轻松地完成您的功能：

function getCombinations($traits)
{
    $combinations = array('');
    foreach( $traits as $trait_level ) {
        $new_combinations = array();
        foreach ( $combinations as $comb ) {
            foreach ( $trait_level as $trait ){
                $new_combinations[] = "$comb $trait";
            }
        }
        $combinations = $new_combinations;
    }
    return $combinations;
}

Answer 2

函数没有自引用$this。实际上，删除所有$this->引用并使$ tempArray按引用而不是按值传递，您的代码应该可以工作...

function getAllPossiblePermutations($mdArray,  $firstTime=true,  &$tempArray=array())

错误：不在对象上下文中时使用'$ this'

2 个答案: