Question

我有一个json列，我想收集id到id2名称的新列。 这是json专栏的一行

[{"id":"26","answer":[{"option":"3","text":"HIGH"}],"type":"a"},
 {"id":"30","answer":[{"option":"3","text":"LOW"}],"type":"b"},
 {"id":"31","answer":[{"option":"3","text":"LOW"}],"type":"c"}]

我的意思是如何将26,30,31 40,40,10 52,12,12添加到新列。我的表名是user_survey_start我的json列名是survey_answers而我的新ID列是id2

<?php
$con=mysqli_connect("localhost","root","","arrayy");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT `survey_answers` FROM `user_survey_start`";

if ($result=mysqli_query($con,$sql)) {
    while ($row = mysqli_fetch_row($result)) {
        $json = $row[0];
        $jason_array  = json_decode($json,true);

        // id   
        $id = array();
        foreach ($jason_array as $data){
            $id[] = $data['id'];
            // here code to insert/update values to db column
        }
        echo implode(',',$id)."</br>";
    }
}
mysqli_close($con);
?>

Answer 1

首先，您必须更新表架构，因此它将有一个名为id2的新列。

ALTER TABLE `user_survey_start` ADD `id2` VARCHAR(255) NOT NULL

如果您希望此列可选，则可以从NOT删除NOT NULL。

然后，您只需将这些ID从JSON插入新创建的id2列。

$id = implode(',', $id);
mysqli_query($con, "UPDATE `user_survey_start` SET `id2` = $id WHERE `id` = {$row['id']}");

Answer 2

解决此代码

         $id = array(); 
            foreach ($jason_array as $data) {
                            $id[] = $data['id'];
                        }
                        $ids= implode(',',$id);
                        $sql1="update user_survey_start set id2='$ids' where id_s=".$row[1];//run update sql
                        echo $sql1."<br>";
                        mysqli_query($con,$sql1);               
        // awnser2          
        $answers = array();
        foreach ($jason_array as $data) {
                foreach($data['answer'] as $ans){
                $answers[] =$ans['text'] ;
                }
        }
        $answers= implode(',',$answers);
        $sql3="update user_survey_start set awnser2='$answers' where id_s=".$row[1];//run update sql
        echo $sql3."<br>";
        mysqli_query($con,$sql3);                                                                           
      }
}

通过PHP将json对象的列插入新列

2 个答案: