Question

你好，我有像这样的JSON行：

//THis is one of My Rows
     [{"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"},
        {"id":"32","answer":[{"option":"3","text":"HIGH"}],"type":"d"},
        {"id":"33","answer":[{"option":"3","text":"HIGH"}],"type":"e"},
        {"id":"34","answer":[{"option":"3","text":"HIGH"}],"type":"f"},
        {"id":"40","answer":["Number 3"],"type":"g"}]

我如何回应id answer type，
和db Name: array Table Name: user_survey_start JSON row Name: survey_answers，这是我的代码：

    <?php
    $con=mysqli_connect("localhost","root","","array");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

  //  $sql="SELECT  survey_answers->"$.id" AS `twitter` FROM user_survey_start";  
    $sql="SELECT  survey_answers FROM user_survey_start";


    if ($result=mysqli_query($con,$sql))
      {
      // Fetch one and one row
      while ($row=mysqli_fetch_row($result))
        {
        printf ("%s \n",$row[0]);
        }
      // Free result set
      mysqli_free_result($result);
    }

    mysqli_close($con);
    ?>

Answer 1

你可以使用json_decode（）函数将你的json转换成php array（）。之后，您可以根据需要迭代所有数据。

Answer 2

首先，我需要选择我的主键

$sql="SELECT `survey_answers`,us_id FROM `user_survey_start`";

然后将值放入ID

$id[] = $data['id'];

之后

$sql1="update user_survey_start set id2='$ids' where us_id=".$row[1];//run update sql
echo $sql1."<br>";

在PHP的PHPMyAdmin列中回显json数组

2 个答案: