html-如果方法不返回数据,则打印一个空表单元格
作者：互联网
我在部分视图中有以下代码,以显示每个评估期间学生的平均分数：
<tr class="<%= cycle("odd", "even", name: "students")%>">
<td>
<%= link_to "#{student.name}",
student_path({student_group_id: student.student_group_id, id: student.id})%>
</td>
<% student.eval_count.times do |i| %>
<td class="center"><%= student.avg_for_eval(i) %></td>
<% end %>
<td class="center"><%= student.avg unless student.avg.nan? %></td>
</tr>
我在student.rb中使用此方法来生成平均分数,但起初我无法弄清楚为什么在没有数据的情况下无法生成空行.
def evals
evals = self.evaluations.order("eval_number").group_by(&:eval_number)
end
def eval_number_set(index)
numbers = Evaluation.where('student_id = ?', self.id).uniq.pluck(:eval_number)
numbers[index]
end
def avg_for_eval(i)
scores = []
evals = self.evals.select { |k, v| k == self.eval_number_set(i) }.values.first
for eval in evals
scores << eval.score
end
evals.empty? #(scores.sum.to_f / scores.size).round(2)
end
我将方法的最后一行更改为evals.empty？如上图所示,并在浏览器中找到了它：
然后我意识到,由于选择了任何评估,该学生都不在场(对于新学生而言)不是我用来生成代码的数据集的一部分.
正如您在图中看到的那样,问题在于,只有最后一次评估的数据的两个学生的数据没有放在正确的列中-一切都移到了左侧,因为没有< td>由视图代码生成.
那么问题是,如何才能重写方法代码,以便获得相同的输出,但是否为scores.empty？该方法插入“”或“无数据”或其他占位符,以便< td>在视图中打印？
更新
我了解现在情况正在好转.该方法：
def eval_number_set(index)
numbers = Evaluation.where('student_id = ?', self.id).uniq.pluck(:eval_number)
numbers[index]
end
正在返回每个学生的评估编号-因此,在下一个方法中使用这些编号时
def avg_for_eval(i)
scores = []
evals = self.evals.select { |k, v| k == self.eval_number_set(i) }.values.first
for eval in evals
scores << eval.score
end
evals.empty? #(scores.sum.to_f / scores.size).round(2)
end
它只能拉取学生参加的评估.回到画板…
更新2
我将类方法更改如下：
#returns all 'eval_number's for a given group of students
#as the first student will have been present for all evaluations
def eval_number_set(index)
numbers = self.student_group.students.first.evals.keys
# numbers = Evaluation.where('student_id = ?', self.id).uniq.pluck(:eval_number)
numbers[index]
end
#attempts to match the first present 'eval_number' for a given student against
#the first number in the set of all 'eval_number's and react accordingly
def avg_for_eval(i)
scores = []
if self.evals.keys[i] == self.eval_number_set(i)
"match"
else
"no_match"
end
end
这与参加每次评估的所有学生相匹配,但与错过评估的学生没有任何匹配.我将代码更改为以下内容
def avg_for_eval(i)
scores = []
if self.evals.keys[i] == self.eval_number_set(i)
"#{self.evals.keys[i]} vs #{self.eval_number_set(i)}"
else
"#{self.evals.keys[i]} vs #{self.eval_number_set(i)}"
end
end
并在浏览器中返回以下内容：
因此,我尝试在if语句中添加一个计数器,以便如果该语句匹配,它将增加尝试的键,否则将保留在同一键上：
def avg_for_eval(i)
scores = []
key_match = 0
if self.evals.keys[key_match] == self.eval_number_set(i)
"#{self.evals.keys[i]} vs #{self.eval_number_set(i)}"
key_match += 1
else
"#{self.evals.keys[i]} vs #{self.eval_number_set(i)}"
end
end
产生这个：
并且很有道理-计数增加由于调用方式而没有任何实际效果.但是尽管我认为我现在已经接近想要的效果,但我不确定如何实现它！
更新3
更近一步…我已经进行了一些更改,以便可以反向读取按键,因此可以读取所有数据-但我仍然希望将数据打印在右边,将最新的数据打印出来(请参见下图) ).现在,模型方法代码如下所示：
def eval_number_set(index)
numbers = self.student_group.students.first.evals.keys.reverse
# numbers = Evaluation.where('student_id = ?', self.id).uniq.pluck(:eval_number)
numbers[index]
end
def avg_for_eval(i)
scores = []
eval_number = self.eval_number_set(i)
if self.evals.keys.reverse[i] == eval_number
for eval in self.evals.values[i]
scores << eval.score if self.evals.values[i]
end
scores
else
"no data"
end
end
它返回以下内容,我已对其进行了注释,以便您可以清楚地了解我正在寻找的内容：
更新4
调用student.evals返回以下内容(对于ID为32的学生)：
{29=>[
#<Evaluation id: 1949, score: 3, created_at: "2013-08-28 09:44:32", updated_at: "2013-08-28 09:44:32", student_id: 32, goal_id: 63, eval_number: 29>,
#<Evaluation id: 1950, score: 4, created_at: "2013-08-28 09:44:32", updated_at: "2013-08-28 09:44:32", student_id: 32, goal_id: 64, eval_number: 29>,
#<Evaluation id: 1951, score: 5, created_at: "2013-08-28 09:44:32", updated_at: "2013-08-28 09:44:32", student_id: 32, goal_id: 65, eval_number: 29>],
30=>[
#<Evaluation id: 1957, score: 3, created_at: "2013-08-28 09:44:43", updated_at: "2013-08-28 09:44:43", student_id: 32, goal_id: 65, eval_number: 30>,
#<Evaluation id: 1956, score: 2, created_at: "2013-08-28 09:44:43", updated_at: "2013-08-28 09:44:43", student_id: 32, goal_id: 64, eval_number: 30>,
#<Evaluation id: 1955, score: 1, created_at: "2013-08-28 09:44:43", updated_at: "2013-08-28 09:44:43", student_id: 32, goal_id: 63, eval_number: 30>],
31=>[
#<Evaluation id: 1968, score: 2, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 70, eval_number: 31>,
#<Evaluation id: 1967, score: 2, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 69, eval_number: 31>,
#<Evaluation id: 1966, score: 1, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 68, eval_number: 31>,
#<Evaluation id: 1965, score: 1, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 67, eval_number: 31>,
#<Evaluation id: 1964, score: 1, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 66, eval_number: 31>,
#<Evaluation id: 1963, score: 3, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 65, eval_number: 31>,
#<Evaluation id: 1962, score: 3, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 64, eval_number: 31>,
#<Evaluation id: 1961, score: 3, created_at: "2013-08-28 11:26:56", updated_at: "2013-08-28 11:26:56", student_id: 32, goal_id: 63, eval_number: 31>],
32=>[
#<Evaluation id: 1983, score: 3, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 69, eval_number: 32>,
#<Evaluation id: 1982, score: 2, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 68, eval_number: 32>,
#<Evaluation id: 1981, score: 3, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 67, eval_number: 32>,
#<Evaluation id: 1980, score: 4, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 66, eval_number: 32>,
#<Evaluation id: 1979, score: 4, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 65, eval_number: 32>,
#<Evaluation id: 1978, score: 3, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 64, eval_number: 32>,
#<Evaluation id: 1977, score: 3, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 63, eval_number: 32>,
#<Evaluation id: 1984, score: 3, created_at: "2013-08-29 19:31:48", updated_at: "2013-08-29 19:31:48", student_id: 32, goal_id: 70, eval_number: 32>]
}
解决方法:
更新
您应该更改eval_count方法或使用常量值制作4个单元格.例如,
4.times do |i|
<td class="center"><%= student.avg_for_eval(i) %></td>
end
更新结束
您可以尝试以下方法：
def avg_for_eval(i)
scores = Array.new(4)
evals = self.evals.select { |k, v| k == self.eval_number_set(i) }.values.first
evals.each_with_index do |eval, i|
scores[i] = eval.score
end
(scores.sum.to_f / scores.size).round(2)
end
注意分数数组的初始化.在此方法中,最初创建了[nil,nil,nil,nil].对于只有两个分数的学生,分数将等于[first_score,second_score,nil,nil].
这可能导致错误：方法的结果将是
(first_score + second_score) / 4
标签：html,ruby-on-rails-3,ruby-on-rails,mysql 来源： https://codeday.me/bug/20191122/2062865.html