0

我有一个 grails 域类，它看起来像这样：

class Answer {
String response
Integer score
Date created = new Date()
}

如何进行查询以返回一周中每一天的平均分数？， 像这样 ：

average for Mondays = 25
average for Tuesdays = 20
....
average for sundays = 10
4

3 回答 3

2

要获得星期天的平均值，首先要获取星期天的日期。然后使用条件查询来查找平均值。请参见下面的示例：

dateOfSunday = ..
def c = Answer.createCriteria()
def avgScore = c.get {
  projections {
    avg "score"
  }
 lt("created",dateOfSunday.plus(1) )
 gt("created",dateOfSunday.minus(1) )
}

希望你有这个想法...

编辑：

好的，如果您想要所有星期天的平均值，那么您可以在域类中使用公式，例如：

class Answer {
 String response
 Integer score
 Date created = new Date()
 static mapping = {
    dayOfWeek formula: 'DAYOFWEEK(created)'
   }
}

然后查询将是：

def c = Answer.createCriteria()
def avgScore = c.get {
  projections {
    avg "score"
  }
 eq("dayOfWeek ",7)
}

编辑：

@dmahapatro，感谢您指出。

所有平均分数也可以使用单个 db hit 获得。见下文：

def results = Answer.createCriteria().list() {
  projections {
    groupProperty('dayOfWeek')
    avg "score"
  }
}

虽然它涉及的代码较少，但它有一个缺点，它使用了 db 函数，这会使查询变慢。您可以使用 dmahapatro 的建议。

于 2013-05-24T10:45:30.030 回答
1

这是@lucke84 答案的补充，是对他的方法的一个小修改（以防我对他的答案的编辑被遗漏）

def summation = [
'monday': [],
'tuesday': [], 
'wednesday': [],
'thursday': [],
'friday': [],
'saturday': [],
'sunday': []
]


def answers = Answer.list() 
answers.each {
    def dateCreated = it.created[Calendar.DAY_OF_WEEK]

    switch (dateCreated) {
        case Calendar.MONDAY:
            summation.monday << it.score     
            break
        case Calendar.TUESDAY:
            summation.tuesday << it.score
            break
        case Calendar.WEDNESDAY:
            summation.wednesday << it.score
            break
        case Calendar.THURSDAY:
            summation.thursday << it.score
            break
        case Calendar.FRIDAY:
            summation.friday << it.score   
            break
        case Calendar.SATURDAY:
            summation.saturday << it.score
            break
        case Calendar.SUNDAY:
            summation.sunday << it.score
            break
    } ​ 
}

def averages = [:]
averages.monday = (summation['monday'].sum())?.divide(summation['monday'].size())
//So on and So forth.

记下可能会影响性能.list()的所有行。Answer在这种情况下，您可以根据需要使用findAll和传递参数。pagination

于 2013-05-24T14:19:03.900 回答
1

我不确定我是否有您的问题和您的域模型，但也许这份草稿可能会有所帮助。

def summations = [
    'monday': [],
    'tuesday': [], 
    'wednesday': [],
    'thursday': [],
    'friday': [],
    'saturday': [],
    'sunday': []
]

def answers = Answer.list() 
answers.each {
    def dateCreated = it.created[Calendar.DAY_OF_WEEK]

    switch (dateCreated) {
        case Calendar.MONDAY:
            summations['monday'] << it.score     
            break
        case Calendar.TUESDAY:
            summations['tuesday'] << it.score
            break
        case Calendar.WEDNESDAY:
            summations['wednesday'] << it.score
            break
        case Calendar.THURSDAY:
            summations['thursday'] << it.score
            break
        case Calendar.FRIDAY:
            summations['friday'] << it.score   
            break
        case Calendar.SATURDAY:
            summations['saturday'] << it.score
            break
        case Calendar.SUNDAY:
            summations['sunday'] << it.score
            break
    } ​ 
}

// you can now do the average for every list in the map (sum of elements divided by number of item)

编辑：这是来自@dmahapatro：

def averages = [:]
averages.monday = (summations['monday'].sum())?.divide(summations['monday'].size())
// So on and So forth.
于 2013-05-24T13:35:46.783 回答