Question

我想将我的通知插入静态，但我正在检查变量类型是否为空，但此时插入显示Message: Illegal string offset 'notificationCount'和Undefined index: type。我正在尝试动态制作我的数组，但它似乎不起作用。

public function addNotification($message, $product_id, $type = ''){
    $types = array('new' => 0, 'pending' => 1, 'low stock' => 3);
    if (isset($types[$type]) === false) {
        throw new \InvalidArgumentException('Value for third parameter must be one of new, pending, or low stock.');
    }
    $type = $types[$type];
    $time = time();
    $query = "SELECT COUNT(*) AS notificationCount FROM storelte_notifications WHERE product_id = ? AND type = ? ";
    $previousNotification = $this->db->query($query, array($product_id, $type));
    if ($previousNotification[0]['notificationCount'] == 0) {
        $sql = "INSERT INTO storelte_notifications (message,type,product_id,created_at) VALUES(?, ?, ?, ?)";
        $this->db->query($sql, array($message, $type, $product_id, $time));
        try {
            if ($this->db->query($sql)) {
                return true;
            }else{
                return false;
            }
        } catch (Exception $e) {

        }
    }else{
        return true;
    }
}

控制器

public function add(){
        $this->notification->addNotification('low stock',4228,'type');
    }

Answer 1

$sql_prev_notification是您在此处创建的字符串：

$sql_prev_notification = "SELECT COUNT(*) AS notificationCount FROM storelte_notifications WHERE product_id = ? AND type = ? ";

您可以使用它来执行查询：

$this->db->query($sql_prev_notification, array($product_id, $type));

但您尚未将查询的返回结果分配给任何内容。

$sql_prev_notification仍然是一个字符串，所以当你这样做时：

if ($sql_prev_notification[0]['notificationCount'] == 0) {

$sql_prev_notification[0]指的是字符串中的第一个字母（S），它显然不是一个数组，因此

非法字符串偏移＆＃39; notificationCount＆＃39;

你可能想要更像的东西：

$sql = "SELECT COUNT(*) AS notificationCount FROM storelte_notifications WHERE product_id = ? AND type = ? ";
$sql_prev_notification = $this->db->query($sql, array($product_id, $type));
if ($sql_prev_notification[0]['notificationCount'] == 0) {

虽然在引用结果中的特定项目之前，您还应检查您的查询是否实际返回任何内容。

Answer 2

恐慌的答案（https://stackoverflow.com/a/43529550/4132627）是解决非法字符串偏移的正确答案，请继续使用。

对于错误Undefined index: type，您似乎正在传递字符串＆＃34;键入＆＃34;作为变量$type的值。然后它使用该值作为$types数组的键，但$types数组没有索引＆＃34;类型＆＃34; - 它的索引是＆＃34;新＆＃34;，＆＃34;待定＆＃34;和＆＃34;低库存＆＃34;。

要解决此问题，您必须通过＆＃34; new＆＃34;，＆＃34; pending＆＃34;或＆＃34; low stock＆＃34;作为addNotification函数的第三个参数：

$this->notification->addNotification('low stock',4228,'new');
//or
$this->notification->addNotification('low stock',4228,'pending');
//of
$this->notification->addNotification('low stock',4228,'low stock');

您还应检查传递的密钥是否有效，否则您将继续收到此通知。实际上，传递错误的值可能会导致代码不规则地运行，在这种情况下抛出异常可能是一个好主意：

if (isset($types[$type]) === false) {
    throw new \InvalidArgumentException('Value for third parameter must be one of new, pending, or low stock.');
}

$type = $types[$type];

Answer 3

你可以试试这个：

public function addNotification( $message, $product_id, $type = '' ){
    $types = array( 'new' => 0, 'pending' => 1, 'low stock' => 3);
    if (isset($types[$type]) === false) {
        throw new \InvalidArgumentException('Value for third parameter must be one of new, pending, or low stock.');
    }
    $type = $types[$type];
    $time = time();
    $this->db->select_sum('notificationCount');
    $this->db->where(['product_id' => $product_id, 'type' => $type]);
    $query = $this->db->get();
    $previousNotification = $query->row_array()
    if ( $previousNotification['notificationCount'] == 0 ) {
        $this->db->trans_start();
        $this->db->insert(storelte_notifications, ['message' => $message, 'type' => $type, 'product_id' => $product_id, 'created_at' => $time]);
        $this->db->trans_complete();
        return $this->db->trans_status();
   } else {
       return true;
   }
}

codeigniter中的数组问题

3 个答案: