php 注意：数组到字符串的转换
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Notice: Array to string conversion in
提问by Sunden
I'm trying to select a value from a database and display it to the user using SELECT. However I keep getting this error:
我正在尝试从数据库中选择一个值并使用 SELECT 将其显示给用户。但是我不断收到此错误：
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:
我以为@mysql_fetch_assoc();会解决这个问题，但我仍然收到通知。这是我收到错误的代码部分：
{
$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>';
}
What am I doing wrong? I'm pretty new to PHP.
我究竟做错了什么？我对 PHP 很陌生。
采纳答案by Razvan
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:
问题在于 $money 是一个数组，您将其视为字符串或可以轻松转换为字符串的变量。你应该这样说：
'.... Money:'.$money['money']
回答by user1046243
Even simpler:
更简单：
$get = @mysql_query("SELECT money FROM players WHERE username = '" . $_SESSION['username'] . "'");
note the quotes around usernamein the $_SESSIONreference.
注意周围的引号的用户名在$ _SESSION参考。
回答by Learner
mysql_fetch_assoc returns an array so you can not echo an array, need to print_r() otherwise particular string $money['money'].
mysql_fetch_assoc 返回一个数组，所以你不能回显一个数组，需要 print_r() 否则特定的字符串 $money['money']。
回答by Learner
You cannot echo an array. Must use print_r instead.
您不能回显数组。必须改用 print_r。
<?php
$result = $conn->query("Select * from tbl");
$row = $result->fetch_assoc();
print_r ($row);
?>
回答by Ernesto Jr Montejo
One of reasons why you will get this Notice: Array to string conversion in… is that you are combining group of arrays. Example, sorting out several first and last names.
您将收到此通知的原因之一：数组到字符串的转换在...中是您正在组合一组数组。例如，整理出几个名字和姓氏。
To echo elements of array properly, you can use the function, implode(separator, array)Example:
要正确回显数组元素，您可以使用该函数，implode(separator, array)例如：
implode(' ', $var)
result:
结果：
first name[1], last name[1]
first name[2], last name[2]
More examples from W3C.
来自W3C 的更多示例。
回答by user5336032
Store the Value of $_SESSION['username'] into a variable such as $username
将 $_SESSION['username'] 的值存储到一个变量中，例如 $username
$username=$_SESSION['username'];
$get = @mysql_query("SELECT money FROM players WHERE username =
'$username'");
it should work!
它应该工作！
