Question

我不断收到通知，“注意：第33行路径名中的数组到字符串转换”。当我隐藏通知error_reporting(E_ALL ^ E_NOTICE);时，我仍然会出现“数组”这个词出现的次数与我返回的行数一样多。作为新人我似乎无法摆脱他们。代码中的错误就在这附近。

<?php
$query = 'SELECT * FROM NETWORK';

$result = mysqli_query($conn,$query);
while($r=mysqli_fetch_assoc($result)){
echo $rows["network"][] = $r;
}

print json_encode($rows);
?>

当我在浏览器上测试时，它会返回此信息。

Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
Notice: Array to string conversion in /var/www/html/dl.php on line 33
Array
{"network":[{"lid":"1","sid":"1","player_id":"1","active":"1"},
{"lid":"1","sid":"1","pid":"4","active":"1"},
{"lid":"1","sid":"2","pid":"2","active":"1"}, 
{"lid":"1","sid":"2","pid":"5","active":"1"},
{"lid":"1","sid":"3","pid":"3","active":"1"},
{"lid":"2","sid":"1","pid":"1","active":"1"},
{"lid":"2","sid":"2","pid":"2","active":"1"},
{"lid":"2","sid":"2","pid":"4","active":"1"},
{"lid":"2","sid":"3","pid":"3","active":"1"},
{"lid":"2","sid":"3","pid":"5","active":"1"}]}

总体问题是我做错了什么？但是因为我得到通知而不是错误，我认为可能没有错，但也许有更好的方法来获得我想要的最终结果，这是用json编码的数组。所有和任何帮助表示赞赏。

Answer 1

摆脱echo。 $r是一个不是字符串的数组，不能与echo一起使用：

//echo $rows["network"][] = $r;
$rows["network"][] = $r;

为什么要构建$rows但json_encode $rows3？

注意：第33行阵列上路径名中的数组转换为字符串

1 个答案: