查询在正确执行时未获取数据

Question

我觉得这很奇怪，但是感觉不对。 我想用来自mysql的数据填充JSON数组。 第一个查询将带入有关类别和问题的数据，然后针对每个我想得到答案的问题。 我从第一查询中获取数据，但是从第二查询中我没有。

我的代码：

<?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

我设法用以下代码修复它：

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']);
        }
}

现在，我要为数组中的每个项目使用数组时间每个项目的值填充“ question_answers”数组。

我已经试过了：

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

但这并没有给我想要的结果。

我希望结果是这样的：

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

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

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

这怎么可能。 我尝试的最后一种方法不起作用。 给我空数组。

我尝试了while循环和foreach，但还是没有。 我将不胜感激！

Answer 1

我认为问题在于您在两个查询中都使用了相同的PDO资源。 尝试在foreach循环之前添加$handler->closeCursor() 。

您可能还应该在循环中使用准备好的语句。

Answer 2

最后，我设法解决了这个问题：

我执行了第一个查询，该查询为我返回了2个数组。 使用以下代码，其中1个包含正常数据，而1个包含问题：

$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，我与另一个foreach遍历第二个数组，以执行第二个查询，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;
        }
}

然后我得到了预期的结果。 谢谢大家的宝贵时间。

希望我能帮助一个人！！！