Php Zammad API：使用标记创建票证

对于那些不想阅读整个问题的人：
我正在寻找API请求（Zammad）中的索引，以便在创建票证时设置标记

详细信息：
我正在使用PHP向安装了Zammad的服务器发出API请求。下面显示了我通过curl发送的数据：

json_encode([
    "title" => $title,
    "group_id" => 2,
    "priority_id" => 2,
    "category" => 'Some Category',
    "state_id" => 1,
    "type" => "Some Type", 
    "customer" => $userID, 
    "article" => [
        "body" => $htmlBody,
        "type" => "note",
        "content_type" => "text/html",
    ],
    "tag_list" => [ // <-- The question is about this segment
        "The tag i want to add",
    ],
]);

最后：

...
"tag" => "The tag i want to add",

不用说，我没有成功。有时我会得到一个错误id（我假设它是因为索引不存在[谁会想到？：）]），有时我什么也得不到，Zammad只是创建了没有标签的票证。我说“有时”是什么意思？我指的是我在上面指定的尝试

我也尝试过的：

---

在网上搜索答案。接近我想要的东西是。但是我宁愿用标签创建票证，而不是仅仅为了添加标签而发出另一个请求。

我已经查看了代码内部，它是用ruby编写的。索引是“标签”，需要由

，

指定

基本上：

json_encode([
    "title" => $title,
    "group_id" => 2,
    "priority_id" => 2,
    "category" => 'Some Category',
    "state_id" => 1,
    "type" => "Some Type", 
    "customer" => $userID, 
    "article" => [
        "body" => $htmlBody,
        "type" => "note",
        "content_type" => "text/html",
    ],
    "tags" => "tag1,tag2,tag3", // or simply "tags" => "tag1"
]);

它可能会在将来帮助某人

json_encode([
    "title" => $title,
    "group_id" => 2,
    "priority_id" => 2,
    "category" => 'Some Category',
    "state_id" => 1,
    "type" => "Some Type", 
    "customer" => $userID, 
    "article" => [
        "body" => $htmlBody,
        "type" => "note",
        "content_type" => "text/html",
    ],
    "tags" => "tag1,tag2,tag3", // or simply "tags" => "tag1"
]);