英文:
Storing an object with key value pairs in a variable in JSONiq
问题
这是一个示例的JSON数组：
"transcript": [
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
我需要创建一个名为 $courses 的变量，其中包含所有由具有 ssn 为 82 的学生所选的课程。我应该如何遍历该数组并创建这个变量？
我希望输出的结果是 $courses:= {"CS": 211, "MTH": 125}。
我尝试在 {} 括号中编写循环，但未成功。我是 JSONiq 新手。
英文:
This is a sample JSON array
"transcript":[
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
I need to create a variable $courses which gives me all the courses taken by a student whose ssn is 82. How do I loop over this and create this variable?
I want the output to be $courses:= {"CS" : 211, "MTH" : 125}
I tried to write the loop in {} brackets but it did not work. I am new to Jsoniq
答案1
得分: 0
以下是翻译好的部分：
{
"transcript": [
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
假设此对象存储在变量 `$x` 中，您可以使用以下方式获取所需的对象：
$x.transcript[][$$.ssn eq 82]
其中 `[]` 将数组解包为对象序列，而 `[$$.ssn eq 82]` 过滤此序列，仅保留其中 ssn 为 82 的对象。
因此，完整的查询如下：
let $x := {
"transcript": [
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
return $x.transcript[][$$.ssn eq 82]
也可以使用显式的 FLWOR 子句和 where 子句的替代方式（可以根据更复杂的用例自由扩展）：
let $x := {
"transcript": [
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
for $course in $x.transcript[]
where $course.ssn eq 82
return $course
现在，我们可以使用 let 子句将这些结果绑定到变量 `$courses` 中，请注意这是一个函数式语言，这些是绑定而不是赋值。在这种简单情况下，return 子句返回了 `$courses` 的内容，但也可以对此变量执行其他操作。
英文:
First a quick comment: you need curly braces to make sure this is well-formed (JSON and JSONiq):
{
"transcript": [
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
Assuming this object is stored in a variable $x, you can obtain the desired object with
$x.transcript[][$$.ssn eq 82]
where [] unboxes the array into a sequence of objects, and [$$.ssn eq 82] filters this sequence keeping only those where ssn is 82.
So a complete query would be:
let $x := {
"transcript":[
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
return $x.transcript[][$$.ssn eq 82]
There is also the alternative with an explicit FLWOR for clause and where clause (which are extensible at will with more complex use cases):
let $x := {
"transcript":[
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
for $course in $x.transcript[]
where $course.ssn eq 82
return $course
Now, we can bind these results to a variable $courses with a let clause -- it is important to understand that this is a functional language, that is, these are bindings and not assignments. In this simple case, the return clause returns the contents of $courses, but it is also possible to do anything else you want with this variable instead.
let $courses :=
let $x := {
"transcript":[
{
"dcode": "CS",
"cno": 211,
"ssn": 82,
"grade": "A"
},
{
"dcode": "CS",
"cno": 211,
"ssn": 75,
"grade": "A"
},
{
"dcode": "MTH",
"cno": 125,
"ssn": 82,
"grade": "F"
}
]
}
for $course in $x.transcript[]
where $course.ssn eq 82
return $course
return $courses
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论