Question

我有一个结构为[String: [String: String]]的多维数组。我能够通过for循环访问[String: String]位，但我无法弄清楚如何访问主键（此位[String: [）。

let items = snapshot.value as! [String: [String: String]]

for item in items.values {
    if let from = item["from"] {
         self.users.append(from)   // this works
    }
}

如何到达此节点的主ID？

节点如下所示：

- table
     - "1"
        - "from": "AA"
        - "to": "BB"
     - "2"
        - "from": "AA"
        - "to": "BB"

我想获得var array = ["1", "2", "3"]

Answer 1

使用元组会很清楚试试这个：

$mail = new PHPMailer(true);
$mail->Host = "smtp.office365.com";
$mail->Port       = 587;
$mail->SMTPSecure = '';
$mail->SMTPAuth   = true;
$mail->Username = "email";   
$mail->Password = "password";
$mail->SetFrom('email', 'Name');
$mail->addReplyTo('email', 'Name');
$mail->SMTPDebug  = 2;
$mail->IsHTML(true);
$mail->MsgHTML($message);
$mail->Send();

更新

let dict = ["1":["from":"aa","to":"bb"],"2":["from":"AA","to":"BB"]]

var array = [String]()
for (_ ,value) in dict{
    if let v = value["from"] {
        array.append(v)
    }
}
print(array)

Answer 2

对于这样的数组，你有key-＆gt;值对，你可以将键值作为元素.0访问，将值作为元素.1

在操场上试试看它是否过得很快：

for firstLevel in tableArray {
    print(firstLevel.0)
    for innerLevel in firstLevel.1 {
        print(innerLevel.0)
        print(innerLevel.1)
    }
    print("\n")
}

获取多维数组的键（`[String：[String：String]]`）

2 个答案:

更新