Question

你对JSON有点新意，所以请忍受我的愚蠢......我正在开发一个聊天应用程序，我试图在我的工作PHP脚本上使用json ..我已经尝试过了远在我执行数据库查询后，我将结果编码为JSON然后我使用json_decode然后将其插入我的foreach语句这是包含json_encode和json_decode的PHP代码

    <?php
session_start();
$username=$_SESSION['username'];
$user_id = $_SESSION['id'];
  require "config.php";
  $con = new PDO("mysql:host=".db_host.";dbname=chat_db",db_username,db_password);
  $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
  $sql="SELECT id,firstname,lastname,status,flag,flag2 FROM users WHERE id != :uid";
  $stmt=$con->prepare($sql);
  $stmt->bindValue("uid", $user_id, PDO::PARAM_STR);
  $stmt->execute();

  $sql2="UPDATE users SET status= 1 WHERE status = 0 AND id = :uid";
  $stmt2=$con->prepare($sql2);
  $stmt2->bindValue("uid", $user_id, PDO::PARAM_STR);
  $stmt2->execute();
$rows =$stmt->fetchAll();
$str = json_encode($rows);
$str2 = json_decode($str);
  foreach($str2 as $row){
  $uid = $row['id'];

  $firstname = $row['firstname'];
  $lastname = $row['lastname'];
  $sts = $row['status'];
  $flg = $row['flag'];
  $flg2 = $row['flag2'];
?>
<li>
  <a href="#" data-usernem="<?php echo $username; ?>" data-suid="<?php echo $user_id; ?>" data-userid="<?php echo $uid; ?>" data-role="button"><?php echo $firstname . PHP_EOL . $lastname; ?> </a>

  <span class="typing-stats idle" data-type-status="<?php echo $flg2; ?>"> Typing... </span>
  <span class="mail msg" data-flag="<?php echo $flg; ?>"> <a href="#" data-role="button"><i class="fa fa-envelope-o"></i></a> </span>
  <span class="bullet" data-status="<?php echo $sts; ?>"> &nbsp; </span>
</li>
<?php
  }
?>

如果此脚本执行，则会给我一个错误Fatal error: Cannot use object of type stdClass as array in D:\site_deploy\chat\includes\loadusers.php on line 22

我正确使用json_encode和json_decode吗？我很确定我在json_decode

下的php手册上写了什么

我试过var_dump它，它给了我编码的数据集FYI

Answer 1

使用参数true将其转换为数组。

$str2 = json_decode($str, true);

JSON编码结果来自数据库然后解码给我错误致命错误：不能使用stdClass类型的对象作为数组

1 个答案: