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如何实现这个案例：我有 20 个项目的 CheckListBox：症状 1、症状 2、..、症状 20。用户可以选择多个症状。让我感到困惑的是，如何为每个症状赋予多个值。这是我的代码：

  for i := 0 to CheckListBox1.Items.Count - 1 do
  begin
    if CheckListBox1.Checked[i] = True then
    begin
      Memo1.Lines.Append(CheckListBox1.Items.Strings[i]);
      if i = 0 
        p1 := 'Disease 1';
        p2 := 'Disease 2';
        p3 := 'Disease 3';
      if i = 1 then
        p1 := 'Disease 2';
      if i = 2 then
        p1 := 'Disease 1';
      if i = 3 then
        p1 := 'Disease 3';
      if i = 4 then
        p1 := 'Disease 2';
        p2 := 'Disease 3';
      if i = 5 then
        p1 := 'Disease 1';
        p2 := 'Disease 5';
        p3 := 'Disease 6';
      if i = 6 then
        p1 := 'Disease 5';

      Memo1.Lines.Add('Disease:' + p1+', '+p2+', '+p3);
      Memo1.Lines.Add('');
    end;
  end;
end;

但结果并不如我所料。如何动态地制作 p1,p2,p3 ？

这是我在索引 2、4、6 处检查时的结果：

Symptomp 3
Disease:Disease 1, Disease 5, Disease 6

Symptomp 5
Disease:Disease 2, Disease 5, Disease 6

Symptomp 7
Disease:Disease 5, Disease 5, Disease 6
4

2 回答 2

0

您没有得到预期结果的一个可能原因是您永远不会清除p1,p2和p3变量，因此如果CheckListBox1.Checked[0]为 true则将被分配，然后 ifp2也为 true并且仍然具有上一次迭代的值，而它们可能应该是空白的。尝试类似的东西p3CheckListBox1.Checked[1]p2p3

for i := 0 to CheckListBox1.Items.Count - 1 do
  begin
    if CheckListBox1.Checked[i] = True then
    begin
      p1 := '';
      p2 := '';
      p3 := '';

      Memo1.Lines.Append(CheckListBox1.Items.Strings[i]);

      if i = 0 then begin
        p1 := 'Disease 1';
        p2 := 'Disease 2';
        p3 := 'Disease 3';
      end;
      ...    
      Memo1.Lines.Add('Disease:' + p1+', '+p2+', '+p3);
      Memo1.Lines.Add('');
    end;
  end;
end;
于 2015-07-22T07:37:59.873 回答
0

声明疾病列表和要匹配的常量字符串数组：

// List of diseases
type
  // Note: Use descriptive names instead of a numbers
  TDisease = (td1,td2,td3,{..,}tdMaxDiseases);
  TDiseaseSet = set of TDisease;

  TSymptom = (ts1,ts2,ts3,{..,}tsMaxSymptoms);

const
  // A list of disease names
  sDisease: array[TDisease] of String =
    ('Disease 1','Disease 2','Disease 3',{..,}'Disease xx');
  // An array of disease sets corresponding to each symptom
  cMyDiseaseSet : array[TSymptom] of TDiseaseSet = ([td1,td2,td3],[td3],[td1],[td2]);

set 常量数组为每个症状声明一组疾病。

要获得每个症状的结果字符串和与症状匹配的疾病集：

// A Function to retrieve the diseases from a symptom
function DiseaseFromSymptom(aSymptom: TSymptom; var diseaseSet: TDiseaseSet): String;
var
  aDisease: TDisease;
begin
  diseaseSet := cMyDiseaseSet[aSymptom];
  Result := '';
  for aDisease in diseaseSet do
    Result := Result + sDisease[aDisease] + ', ';
  SetLength(Result,Length(Result)-2);
end;

var
  diseases,diseasesSummary: TDiseaseSet;
  s: String;

  diseasesSummary := [];
  for i := 0 to CheckListBox1.Items.Count - 1 do
  begin
    if CheckListBox1.Checked[i] = True then
    begin
      s := DiseaseFromSymptom(TSymptom(i),diseases);
      Memo1.Lines.Append(CheckListBox1.Items.Strings[i]);
      Memo1.Lines.Add('Disease:' + s);
      Memo1.Lines.Add('');
      // Insert diseases
      diseasesSummary := diseasesSummary + diseases;
    end;
  end;
  // A complete set of diseases in diseasesSummary

似乎您想要一组与所有检查症状相匹配的疾病。最新更新显示了如何做到这一点。

于 2015-07-22T08:02:29.757 回答