解析JSON數組，並通過PHP

Question

合併對象我有idtypeanswer我合併id，type和answer在一起，並把它們放在id2列JSON數組，爲什麼我不能得到的輸出$答案值？

[{"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"}] 

這是我的代碼：

<?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); 
?> 

這是我的輸出：

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

和我Notice: Array to string conversion in C:\wamp64\www\json\awnser.php on line 29
我希望有$價值解答了

Answer 1

自己解決 因爲我不能把Awnser直接爲id[]，我抽放工藝Awnser這樣的：

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

，這是我的代碼：

<?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); 
?>