PHP循環通過對結果

Question

結合需要一些幫助，不斷來回的辦法可能做到這一點。我正在FileMaker中創建一個基本的調查屏幕，並使用PHP來獲得我的結果。我需要以可用格式整理調查數據，以便繪製數據圖表。當我使用PHP查詢FileMaker時，它會返回一個相當大的數據。下面我的代碼已經成功地輸出如下：

foreach ($data as $key => $question) 
{ 
    echo $question->getField('Question').' - '. $question->getField('Answer').'<br />'; 
} 

我的輸出

Has the noise around the surrounding are increased with the new store opening – Strongly Agree 
Has the noise around the surrounding are increased with the new store opening – Strongly Agree 
Has the noise around the surrounding are increased with the new store opening – Agree 
Has the noise around the surrounding are increased with the new store opening – Disagree 
Has the noise around the surrounding are increased with the new store opening – Strongly Disagree 
Do you think the store closing earlier at weekend would help with noise levels - Strongly Disagree 
Do you think the store closing earlier at weekend would help with noise levels - Strongly Disagree 
Do you think the store closing earlier at weekend would help with noise levels - Strongly Agree 
Do you think the store closing earlier at weekend would help with noise levels - Strongly Disagree 
Do you think the store closing earlier at weekend would help with noise levels – Disagree 

我現在需要整理這些數據，無論是作爲一個數組或JSON格式。我試圖使用下面的格式，因爲它看起來最簡單。

[Question][StronglyAgree][Agree][Disagree][StronglyDisagree] 
[Has the noise around the surrounding are increased with the new store opening][2][1][1][2] 
[Do you think the store closing earlier at weekend would help with noise levels][1][0][1][3] 
Etc…. 

我會做一個的FileMaker腳本的方法是進行循環雖然$ data數組，在當前指針到最後一個問題（最後一個指針）比較的問題。如果它的不同將問題的值放在問題數組變量中。這將獲得我所有的獨特問題。看看PHP文檔，我發現array_unique會爲我做所有這些，但我不能讓它與我的foreach（$ data as $ key => $ question）一起工作。

一旦我有我的唯一的問題，我會再循環雖然$數據對我在計算所有的非常同意，然後把該數值到第一[]針對問題查找問題又來了比較。再次，我會循環3其他每個問題的同意，不同意...

有沒有人知道的教程或答案，這將幫助我結合這些結果與問題的答案計數。不要太費心左右格式，像下面將工作

Has the noise around the surrounding are increased with the new store opening,2,1,1,2 
Or 
{「Question":"John", 「Strongly Agree」:2,「Agree」:1,「Disagree」:1,「Strongly Disagree」:2 }, 

作爲最後的手段，我想查詢數據庫4次，每次問題，但是這將是太多的電話，並殺死performance.Storing以某種方式將返回的值轉換爲格式化的變量。聽起來有點過分，但可能最終的方法可能會比循環更容易理解。

Answer 1

你爲什麼不使用的問題作爲陣列的關鍵？

/* 
* @var array $surveyResults: [ 
* "My Question" => [ 
*  "Question" => "My Question" , 
*  "Strongly Agree" => 2, 
*  ... etc ... 
* ] 
* ] 
*/ 
$surveyResults = []; 
foreach ($data as $key => $question) { 
    // if we have not processed this question before, add it to the survey results 
    if (!isset($surveyResults[$question->getField('Question')])) { 
     $surveyResults[$question->getField('Question')] = [ 
      "Question"   => $question->getField('Question'), 
      "Strongly Agree" => 0, 
      "Agree"    => 0, 
      "Disagree"   => 0, 
      "Strongly Disagree" => 0 
     ]; 
    } 
    // count answer 
    $surveyResults[$question->getField('Question')][$question->getField('Answer')]++: 
} 

echo json_encode($surveyResults); 

這應該給你喜歡的結果。