Arrays 删除mongodb阵列的一部分

嗨，我正在尝试删除第一个值之后的任何内容，例如

收藏：{

"_id' :.....
"UrlId" : "5dfc1aa2986b7c30f3398be4",
    "coverInput" : 
           [
            "https://test.com/s/files/1/00351576213050",
            "https://test.com.au"
           ],

"_id' :.....
"UrlId" : "5dfc1aa2986b7c30f3398be4",
    "coverInput" : 
           [
            "https://test.com/s/files/1/00351576213050"
           ],

}

我想用以下内容替换该文档：

收藏：{

"_id' :.....
"UrlId" : "5dfc1aa2986b7c30f3398be4",
    "coverInput" : 
           [
            "https://test.com/s/files/1/00351576213050",
            "https://test.com.au"
           ],

"_id' :.....
"UrlId" : "5dfc1aa2986b7c30f3398be4",
    "coverInput" : 
           [
            "https://test.com/s/files/1/00351576213050"
           ],

}

我相信这很简单。我尝试了这个，但出现了一个错误：

db.test.aggregate([
{
    $addFields: {
        coverInput: { $substr: [ "$coverInput", 0, { $indexOfBytes: [ "$coverInput", "," ] } ] }
    }
}
]).forEach( doc => db.test.updateOne( { _id: doc._id }, { $set: { coverInput: doc.coverInput } } ) )

---

提前谢谢

您可以使用下面的聚合

db.collection.aggregate([
  { "$project": {
    "coverInput": {
      "$slice": ["$coverInput", 1]
    }
  }}
])

或者如果要更新文档

db.test.updateOne(
  { _id: doc._id },
  [{
    $set: { coverInput: { "$slice": ["$coverInput", 1] } }
  }]
)

---

$substr

运算符用于处理字符串。必须使用聚合数组运算符来处理数组字段。