注意：数组字符串转换在 [英] Notice: Array to string conversion in

问题描述

我试图从数据库中的值，并使用SELECT它显示给用户。在线36阵列字符串转换中（路径）：不过，我不断收到这个 - 通知。
我认为@mysql_fetch_assoc（）;想解决这个问题，但我仍然得到通知。这是code的一部分，在那里我得到的错误：

I'm trying to select a value from a database and display it to the user using SELECT. However I keep getting this- Notice: Array to string conversion in (pathname) on line 36. I thought that the @mysql_fetch_assoc(); would fix this but I still get the notice. This is the part of the code where I'm getting the error:

  {
  $loggedin = 1;

  $get = @mysql_query("SELECT money FROM players WHERE username = 
 '$_SESSION[username]'");
  $money = @mysql_fetch_assoc($get);

  echo '<p id= "status">'.$_SESSION['username'].'<br>
  Money: '.$money.'.
  </p>';
  }

我是什么做错了吗？我是pretty新的PHP。

What am I doing wrong? I'm pretty new to PHP.

推荐答案

的问题是，$钱是一个数组，你像一个字符串或可以很容易地转换成字符串变量对待它。你应该这样说：

The problem is that $money is an array and you are treating it like a string or a variable which can be easily converted to string. You should say something like:

 '.... Money:'.$money['money']

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