Question

对于我所拥有的系统，它使用以下代码按位置顺序生成问题：

foreach ($questionnaire->questions->sortBy('position_number') as $question)

希望我能够做的是为每个问题ID访问以前的问题ID和下一个问题ID。因此，如果我在问题id 1上有一个位置为3，我希望能够访问位置为2且问题ID为5的问题ID 3。

Answer 1

IF如果数组是一个顺序数组，以下代码应该可以工作：

$myArray = ($questionnaire->questions->sortBy('position_number');
foreach ($myArray as $key => $value)
{
  $keyPrev = $key - 1;
  $keyNext = $key + 1;

  echo "Next value is ".$myArray[$keyNext]."<br>";
  echo "Current value is ".$value."<br>";
    //OR
  echo "Current value is ".$myArray[$key]."<br>";

  if ($key > 0)
  echo "Previous value was ".$myArray[$keyPrev]."<br>";
}

请记住在使用if ($key > 0)之前检查$keyPrev，因为在第一次迭代中，您仍然要计算$ keyPrev并且$ keyPrev将无效。

Answer 2

像往常一样使用array_keys()和foreach抓取数组的键。使用计数器帮助跟踪您所在的密钥。这甚至可以用于非数字索引。

$counter = 0;
// store it so that we can access it however we like
$questions = $questionnaire->questions->sortBy('position_number');
// get an array of keys of the $questions array. They are returned in order.
$questionKeys = array_keys($questions);
// use to check that there is a next question
$countKeys = count($keys) - 2;

foreach($questions as $question) {
  // only want to access the previous question if there is a previous question
  if ($counter > 0) {
    $previousQuestion = $questions[$questionKeys[$counter - 1]];
    // do interesting things with the previous question
  }

  // only access the next question if there is a next question
  if ($counter < $countKeys) {
    $nextQuestion = $questions[$questionKeys[$counter + 1]];
    //do interesting things
  }
  $counter++;
}

向上和向下移动问题 - PHP

2 个答案: