[英]PHP - How to update array object in for loop using php and mysql
I try to get a json form two different table in php/mysql.我尝试在 php/mysql 中从两个不同的表中获取一个 json。 Purpose is to print JSON out for a rest API using php/mysql.
目的是使用 php/mysql 为 rest API 打印 JSON。
Expected json is:预期的 json 是:
{
"qid":"1",
"qst":"OK",
"qoption:
[
{"id":"o1","isrt":true},
{"id":"o2","isrt":false},
{"id":"o3","isrt":false},
{"id":"o4","isrt":false}
]
},
{
"qid":"2",
"qst":"OK",
"qoption:
[
{"id":"o1","isrt":flase},
{"id":"o2","isrt":false},
{"id":"o3","isrt":true},
{"id":"o4","isrt":false}
]
}
What I had try with PHP:我在 PHP 上的尝试:
if ( isset($_GET['examsid']) && $_GET['examsid'] != "") {
$questions = array();
$conn = dbConnection();
$examsid = $_GET['examsid'];
$result = mysqli_query($conn, "SELECT `id`,`exams_id`,`title` FROM `tblquestions` WHERE exams_id='".$examsid."' ORDER BY RAND()");
if(mysqli_num_rows($result) > 0){
while($q = mysqli_fetch_array($result)){
$questions[] = $q;
}
}
foreach ($questions as $q) {
$options = array();
$oquery = mysqli_query($conn, "SELECT `id`,`question_id`,`title`,`iscorrect` FROM `tbloptions` WHERE `question_id`='".$q['id']."' ORDER BY RAND()");
while($o = mysqli_fetch_array($oquery)){
$options[] = $o;
}
array_push($q['options'], $options);
}
print_r($questions);
}
Current Output using json_encode($jsonobj):使用 json_encode($jsonobj) 的当前输出:
{
"0": "1",
"1": "1",
"2": "How many bones comprise the adult human skeleton?",
"3": "2019-04-11 11:18:44",
"4": "0000-00-00 00:00:00",
"id": "1",
"exams_id": "1",
"title": "How many bones comprise the adult human skeleton?",
"c_date": "2019-04-11 11:18:44",
"m_date": "0000-00-00 00:00:00"
},
{
"0": "3",
"1": "1",
"2": "Which of the following is the first calculating device?",
"3": "2019-04-11 11:19:56",
"4": "0000-00-00 00:00:00",
"id": "3",
"exams_id": "1",
"title": "Which of the following is the first calculating device?",
"c_date": "2019-04-11 11:19:56",
"m_date": "0000-00-00 00:00:00"
}
MySQL: Questions- MySQL:问题-
CREATE TABLE `tblquestions` (
`id` int(11) NOT NULL,
`exams_id` varchar(100) CHARACTER SET latin1 NOT NULL,
`title` varchar(500) CHARACTER SET latin1 NOT NULL,
`c_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`m_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
MySQL: Options - MySQL:选项 -
CREATE TABLE `tbloptions` (
`id` int(11) NOT NULL,
`question_id` int(100) NOT NULL,
`title` varchar(300) NOT NULL,
`iscorrect` varchar(100) CHARACTER SET latin1 NOT NULL,
`c_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`m_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
Seems its returns only question no options also pushed title twice on each questions.似乎它只返回问题没有选项也会在每个问题上两次推送标题。 Please any one help me to resolve this.
请任何人帮我解决这个问题。
You are adding the options to the $q
variable that is only visible inside the second foreach loop.您正在向
$q
变量添加选项,该变量仅在第二个 foreach 循环内可见。 Use your $questions
array instead.请改用您的
$questions
数组。 Furthermore, you need to get the current index of your question to add by adding $key => $q
in your foreach head.此外,您需要通过在 foreach 头中添加
$key => $q
来获取要添加的问题的当前索引。 Can you tell me what the output of this code is (I changed everything mentioned before)?你能告诉我这段代码的输出是什么吗(我改变了之前提到的所有内容)?
if ( isset($_GET['examsid']) && $_GET['examsid'] != "") {
$questions = array();
$conn = dbConnection();
$examsid = $_GET['examsid'];
$result = mysqli_query($conn, "SELECT * FROM `tblquestions` WHERE exams_id='".$examsid."' ORDER BY RAND()");
if(mysqli_num_rows($result) > 0){
while($q = mysqli_fetch_array($result)){
$questions[] = $q;
}
}
foreach ($questions as $key => $q) {
$options = array();
$oquery = mysqli_query($conn, "SELECT * FROM `tbloptions` WHERE `question_id`='".$q['id']."' ORDER BY RAND()");
while($o = mysqli_fetch_array($oquery)){
$options[] = $o;
}
$questions[$key]['options'] = $options;
}
print_r($questions);
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.