我有两个 php 函数,一个返回一个 JSON 数组,另一个尝试对其进行解码并访问它的内容,以便从中删除一个 URL。我的问题是我收到以下错误..
1) json_decode() 期望参数 1 是字符串
2) 第 77 行为 foreach() 提供的参数无效
这是我的代码
.php 文件 1
<?php
function getAnimation($userid, $db) {
include('connect.php');
$db = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error() . "please contact d.g.folksman@ljmu.ac.uk for technical assistance";
echo "<br>";
}
$box_num = 1;
$select = "SELECT card_id, order_num FROM decks WHERE box_num=$box_num AND id=$userid ORDER BY order_num";
$result = mysqli_query($db, $select) or die("SQL Error 1: " . mysqli_error($db));
while ($row = mysqli_fetch_array($result, MYSQL_ASSOC)) {
$users[] = array(
'card_id' => $row['card_id'],
'order_num' => $row['order_num'],
);
}
json_encode($users);
return $users;
mysqli_close($db);
}
?>
试图解码 JSON 数组并访问其内容的 php 文件 2 的片段。
include('getAnimation.php');
$animation = getAnimation($result_array[0], $db); //<< atempting to access returned array here
$obj = json_decode($animation); //FIRST ERROR HERE
foreach ($obj->card_id as $item) { // SECOND ERROR HERE
$url = ($item->card_id);
echo $url; //show me the money
}
谢谢你提供的所有帮助!