我在这里好好看看,但还没找到我想要的东西。我正在阅读一个ini文件,使用我的ini课程,我可以调用每个部分以及所有键和值,但我对[Races]部分有疑问。在它说IOM = 7和UK = 6的情况下,我想调用关键的IOM并循环通过比赛1-6并对该部分中的任何其他键值对执行相同的操作。这是我的代码。
List<string> PlacesList= new List<string>();
List<string> PositionsList= new List<string>();
public void btnReadini_Click(object sender, EventArgs e)
{
PlacesList = ListEntries("Places");
PositionsList = ListEntries("Positions");
}
public List<string> ListEntries(string sectionName)
{
IniFile INI = new IniFile(@"C:\Races.ini");
List<string> entries = null;
string[] Entry = INI.GetEntryKeyNames(sectionName);
if (Entry != null)
{
entries = new List<string>();
foreach (string EntName in Entry)
{
entries.Add(EntName + "=" + INI.GetEntryKeyValue(sectionName, EntName));
}
}
return entries;
}
这是我的班级:
public class IniFile
{
[DllImport("kernel32")]
static extern int GetPrivateProfileString(string Section, string Key,
string Value, StringBuilder Result, int Size, string FileName);
[DllImport("kernel32")]
static extern int GetPrivateProfileString(string Section, int Key,
string Value, [MarshalAs(UnmanagedType.LPArray)] byte[] Result,
int Size, string FileName);
[DllImport("kernel32")]
static extern int GetPrivateProfileString(int Section, string Key,
string Value, [MarshalAs(UnmanagedType.LPArray)] byte[] Result,
int Size, string FileName);
public string path;
public IniFile(string INIPath)
{
path = INIPath;
}
public string[] GetSectionNames()
{
for (int maxsize = 500; true; maxsize *= 2)
{
byte[] bytes = new byte[maxsize];
int size = GetPrivateProfileString(0, "", "", bytes, maxsize, path);
if (size < maxsize - 2)
{
string Selected = Encoding.ASCII.GetString(bytes, 0,
size - (size > 0 ? 1 : 0));
return Selected.Split(new char[] { '\0' });
}
}
}
public string[] GetEntryKeyNames(string section)
{
for (int maxsize = 500; true; maxsize *= 2)
{
byte[] bytes = new byte[maxsize];
int size = GetPrivateProfileString(section, 0, "", bytes, maxsize, path);
if (size < maxsize - 2)
{
string entries = Encoding.ASCII.GetString(bytes, 0,
size - (size > 0 ? 1 : 0));
return entries.Split(new char[] { '\0' });
}
}
}
public object GetEntryKeyValue(string section, string entry)
{
for (int maxsize = 250; true; maxsize *= 2)
{
StringBuilder result = new StringBuilder(maxsize);
int size = GetPrivateProfileString(section, entry, "",
result, maxsize, path);
if (size < maxsize - 1)
{
return result.ToString();
}
}
}
}
这是我的ini文件:
[Places]
IOM=Isle of man
UK=United Kingdom
IRE=Ireland
[Races]
IOM=7
UK=6
[Positions]
WN=Win
2nd=Second
3rd=Third
4th=Fourth
我知道我需要像这样循环比赛1-7:
For(int i = 1; i < 7) i++)
我只是不确定如何打电话给ini并做那部分。 我最终想做的是这样的事情:
foreach (IOM Isle of man)
{
for( 1 - 7)
{
foreach(Win - Fourth)
{
listBox1.Items.Add(the result of the above);
}
}
}
答案 0 :(得分:0)
这里是如何迭代所有部分和所有部分的键:
IniFile ini = new IniFile();
ini.Load("path to your .ini file ...");
foreach (IniSection section in ini.Sections)
{
foreach (IniKey key in section.Keys)
{
string sectionName = section.Name;
string keyName = key.Name;
string keyValue = key.Value;
// Do something ...
}
}
答案 1 :(得分:0)
我通过创建一个地方类来对此进行排序,在地方类中我创建了一个名为noOfRaces的方法。当我在我的代码中使用该函数时,它看起来像这样:
List<Place> placeList = new List<Place>();
public class Place
{
private string _Program;
private string _Name;
public int _NoOfRaces
{ get; set; }
public string Program
{ get { return _Program; } }
public string Name
{ get { return _Name; } }
public Place(string pKey, string pValue)
{
_Program = pKey;
_Name = pValue;
}
public void setNoOfRaces(int pRaces)
{
_NoOfRaces = pRaces;
}
}
然后在我的.cs中,我使用上面的类:
private int findNoOfRaces(Place pPlace)
{
foreach (Place program in placeList)
{
foreach (string line in raceDetails)
{
string raceNoTotal = "";
if (line.StartsWith(pTrack.Program + "="))
{
char delimiter = '=';
string value = line;
string[] raceSubstring = value.Split(delimiter);
raceNoTotal = raceSubstring[1];
for (int i = 1; i <= Convert.ToInt32(raceNoTotal); i++)
{
foreach (Pool nextPool in poolList)
{
listBox1.Items.Add(pTrack.Program + " " + " " + nextPool.poolName + " " +i);
}
}
return Convert.ToInt32(raceNoTotal);
}
}
}
return -1;
}
现在这给了我想要的输出,如下所示:
IOM Win 1
IOM second 1
IOM Third 1
IOM Fourth 1
IOM Win 2
IOM second 2
IOM Third 2
IOM Fourth 2
IOM一直到7 通过使用函数ican循环遍历ini中的所有种族。