查詢在正確執行時未獲取數據

Question

我覺得這很奇怪，但是感覺不對。 我想用來自mysql的數據填充JSON數組。 第一個查詢將帶入有關類別和問題的數據，然后針對每個我想得到答案的問題。 我從第一查詢中獲取數據，但是從第二查詢中我沒有。

我的代碼：

<?php
error_reporting(E_ALL ^ E_NOTICE);
ini_set("default_charset", "UTF-8");
header('Content-type: text/html; charset=UTF-8');
try {
    $handler = new PDO('mysql:host=localhost;dbname=database', 'root', '');
    $handler->setAttribute(PDO::MYSQL_ATTR_INIT_COMMAND, "SET NAMES utf8");
    $handler->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $handler->exec("SET CHARACTER SET 'utf8'"); 
} catch (Exception $e) {
    echo $e->getMessage();
    die();
}

$query = $handler->query('SELECT DISTINCT c.cat_name, c.cat_id, q.question FROM `categories` c
LEFT JOIN `questions` q ON c.cat_id = q.cat_id WHERE c.cat_id = 1');
$records = array();

$records = $query->fetchAll(PDO::FETCH_ASSOC);
echo "<pre>";
print_r($records);
echo "</pre>";
$answers = array();
foreach($records as $k => $v){
    $ques = $v['question'];
    $ques = trim($ques);
    $qu = $handler->query("SELECT a.answer, a.iscorrect FROM `answers` a INNER JOIN `questions` q ON a.quest_id = q.q_id WHERE q.question = '".$ques."' ");
    echo "SELECT a.answer, a.iscorrect FROM `answers` a INNER JOIN `questions` q ON a.quest_id = q.q_id WHERE q.question = '".$ques."'<br>";
    $answers = $qu->fetch(PDO::FETCH_BOTH);
    /*$answers = $qu->fetchAll(PDO::FETCH_ASSOC);
    foreach ($answers as $key => $value) {
        echo "Key: " . $key . " Value: " . $value;
    }
    //$answersR = $qu->fetchAll(PDO::FETCH_ASSOC);*/
    echo "<pre>";
    print_r($answers);
    echo "</pre>";
}
$j['quiz'] = $json;
echo json_encode($j);
/*$json[] = array(
        "category_name" => $v['cat_name'], "category_id" => $v['cat_id'],  "question_name" => $v['question'],
        "answers" => array(
            "answer" => $answers['answer'],
            "iscorrect" => $answers['iscorrect']
    ));*/
?>

更新

我設法用以下代碼修復它：

foreach($records as $k => $v){
    $a[] = array("category_name" => $v['cat_name'], "category_id" => $v['cat_id'],  "question_name" => $v['question'], "question_answers" => array() );
    $normal[] = $v['question'];
}
foreach ($normal as $key => $value) {
        $ques = $value;
        $qu = $handler->query("SELECT a.answer, a.iscorrect FROM `answers` a INNER JOIN `questions` q ON a.quest_id = q.q_id WHERE q.question = '".$ques."' ");
        $ans = $qu->fetchAll(PDO::FETCH_ASSOC);
        foreach ($ans as $key => $value) {
            $times[] = array('answer' => $value['answer'], 'iscorrect' => $value['iscorrect']);
        }
}

現在，我要為數組中的每個項目使用數組時間每個項目的值填充“ question_answers”數組。

我已經試過了：

foreach ($times as $w => $e) {
    $a['question_answers'][]  = array("answer" => $e['answer'], "iscorrect" => $e['iscorrect']);
}

但這並沒有給我想要的結果。

我希望結果是這樣的：

"category_name" => categoryname,
"category_id" => categoryid,
"question_name" => questionname,
"question_answers" =>[
    "answer" => answer1,
    "iscorrect" => yes,

    "answer" => answer2,
    "iscorrect" => no,

    "answer" => answer3,
    "iscorrect" => no,
]

這怎么可能。 我嘗試的最后一種方法不起作用。 給我空數組。

我嘗試了while循環和foreach，但還是沒有。 我將不勝感激！

Answer 1

我認為問題在於您在兩個查詢中都使用了相同的PDO資源。 嘗試在foreach循環之前添加$handler->closeCursor() 。

您可能還應該在循環中使用准備好的語句。

Answer 2

最后，我設法解決了這個問題：

我執行了第一個查詢，該查詢為我返回了2個數組。 使用以下代碼，其中1個包含正常數據，而1個包含問題：

$query = $handler->query('SELECT DISTINCT c.cat_name, c.cat_id, q.question FROM `categories` c
INNER JOIN `questions` q ON c.cat_id = q.cat_id WHERE c.cat_id = 1');
$records = array();
$records = $query->fetchAll(PDO::FETCH_ASSOC);
$a = array();
$ans = array();
foreach($records as $k => $v){
    $first[] = array("category_name" => $v['cat_name'], "category_id" => $v['cat_id'],  "question_name" => $v['question'], "question_answers" => array());
    $second[] = $v['question'];
}

然后，我與另一個foreach遍歷第二個數組，以執行第二個查詢，如下所示：

foreach ($second $key => $value) {
        $ques = $value;
        $qu = $handler->query("SELECT a.answer, a.iscorrect FROM `answers` a INNER JOIN `questions` q ON a.quest_id = q.q_id WHERE q.question = '".$ques."' ");
        $third = $qu->fetchAll(PDO::FETCH_ASSOC);
        foreach ($first as $k => $v) {
            $first[$key]['question_answers'] = $third;
        }
}

然后我得到了預期的結果。 謝謝大家的寶貴時間。

希望我能幫助一個人！！！