通过php解析json数组和合并对象

Question

我有一个带有 id type answer 的 json 数组，我正在合并 id、type 和 answer 并将它们放在 id2 列中，那为什么我不能在输出中获取 $answers 值？

[{"id":"38","answer":[{"option":"3","text":"HIGH"}],"type":"a"},
 {"id":"39","answer":[{"option":"3","text":"LOW"}],"type":"b"},
 {"id":"40","answer":["Hello Word"],"type":"c"}] 

这是我的代码：

<?php
$con=mysqli_connect("localhost","root","","arrayok");
mysqli_set_charset($con,"utf8");

// Check connection
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";
if ($result=mysqli_query($con,$sql)){
    while ($row = mysqli_fetch_row($result)){
        $json = $row[0];
        if(!is_null($json)){                          

        $json = preg_replace("!\r?\n!", "", $json);  
        $jason_array = json_decode($json,true);

    // id2 
            $id = array();
            foreach ($jason_array as $data) {
            if (array_key_exists('id', $data)) {
            if (array_key_exists('type', $data)) {  
            if (array_key_exists('answer', $data)) { 
                foreach($data['answer'] as $ans){
                $answers[] = isset($ans['text']) ? $ans['text'] : $ans;  
                }

            $id[] = ' ID='.$data['id'].', TYPE='.$data['type'].', AWNSER='.$answers;
            }
            }
            }
            }
            // lets check first your $types variable has value or not?
             $ids= implode(',',$id); /// implode yes if you got values
            $sql1="update user_survey_start set id2='$ids' where us_id=".$row[1];//run update sql
            echo $sql1."<br>";
            mysqli_query($con,$sql1);

        }
    }
}
mysqli_close($con);
?>

这是我的输出：

update user_survey_start set id2=' ID=38, TYPE=a, AWNSER=Array,

我得到了Notice: Array to string conversion in C:\wamp64\www\json\awnser.php on line 29
我想拥有 $answers 的价值

Answer 1

自己解决 因为我不能把 Awnser 直接放到id[]，所以我把 Awnser 变成了这样：

$id[] = ' ID='.$data['id'].', TYPE='.$data['type'];
$id[] = isset($ans['text']) ? ' AWNSER='.$ans['text'] : ' AWNSER='.$ans;

这是我的代码：

<?php
$con=mysqli_connect("localhost","root","","arrayok");
mysqli_set_charset($con,"utf8");

// Check connection
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";
if ($result=mysqli_query($con,$sql)){
    while ($row = mysqli_fetch_row($result)){
        $json = $row[0];
        if(!is_null($json)){                          

        $json = preg_replace("!\r?\n!", "", $json);  
        $jason_array = json_decode($json,true);

    // id2 
            $id = array();
            foreach ($jason_array as $data) {
            if (array_key_exists('id', $data)) {
            if (array_key_exists('type', $data)) {    
            if (array_key_exists('answer', $data)) { 
                foreach($data['answer'] as $ans){
                $id[] = ' ID='.$data['id'].', TYPE='.$data['type'];
                $id[] = isset($ans['text']) ? ' AWNSER='.$ans['text'] : ' AWNSER='.$ans;
                }



            }
            }
            }
            }
            // lets check first your $types variable has value or not?
             $ids= implode(',',$id); /// implode yes if you got values
            $sql1="update user_survey_start set id2='$ids' where us_id=".$row[1];//run update sql
            echo $sql1."<br>";
            mysqli_query($con,$sql1);

        }
    }
}
mysqli_close($con);
?>