C# 字符串替换为字典
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C# String replace with dictionary
提问by RaYell
I have a string on which I need to do some replacements. I have a Dictionary<string, string>where I have search-replace pairs defined. I have created following extension methods to perform this operation:
我有一个字符串，我需要对其进行一些替换。我有一个Dictionary<string, string>定义了搜索替换对的地方。我创建了以下扩展方法来执行此操作：
public static string Replace(this string str, Dictionary<string, string> dict)
{
StringBuilder sb = new StringBuilder(str);
return sb.Replace(dict).ToString();
}
public static StringBuild Replace(this StringBuilder sb,
Dictionary<string, string> dict)
{
foreach (KeyValuePair<string, string> replacement in dict)
{
sb.Replace(replacement.Key, replacement.Value);
}
return sb;
}
Is there a better way of doing that?
有没有更好的方法来做到这一点？
采纳答案by Marc Gravell
If the data is tokenized (i.e. "Dear $name$, as of $date$ your balance is $amount$"), then a Regexcan be useful:
如果数据被标记化（即“亲爱的 $name$，截至 $date$，您的余额为 $amount$”），那么 aRegex可能很有用：
static readonly Regex re = new Regex(@"$(\w+)$", RegexOptions.Compiled);
static void Main() {
string input = @"Dear $name$, as of $date$ your balance is $amount$";
var args = new Dictionary<string, string>(
StringComparer.OrdinalIgnoreCase) {
{"name", "Mr Smith"},
{"date", "05 Aug 2009"},
{"amount", "GBP200"}
};
string output = re.Replace(input, match => args[match.Groups[1].Value]);
}
However, without something like this, I expect that your Replaceloop is probably about as much as you can do, without going to extreme lengths. If it isn't tokenized, perhaps profile it; is the Replaceactually a problem?
但是，如果没有这样的事情，我希望您的Replace循环可能会尽可能多，而不会太长。如果它没有被标记化，也许可以对其进行分析；是Replace实际上是一个问题吗？
回答by Jon Skeet
Seems reasonable to me, except for one thing: it's order-sensitive. For instance, take an input string of "$x $y" and a replacement dictionary of:
对我来说似乎是合理的，除了一件事：它是顺序敏感的。例如，输入一个“$x $y”字符串和一个替换字典：
"$x" => "$y"
"$y" => "foo"
The results of the replacement are either"foo foo" or "$y foo" depending on which replacement is performed first.
替换的结果是“foo foo”或“$y foo”，具体取决于首先执行哪个替换。
You could control the ordering using a List<KeyValuePair<string, string>>instead. The alternative is to walk through the string making sure you don't consume the replacements in further replace operations. That's likely to be a lot harder though.
您可以使用 aList<KeyValuePair<string, string>>来控制排序。另一种方法是遍历字符串，确保在进一步的替换操作中不会消耗替换。不过，这可能要困难得多。
回答by Allen Wang
Do this with Linq:
使用 Linq 执行此操作：
var newstr = dict.Aggregate(str, (current, value) =>
current.Replace(value.Key, value.Value));
dictis your search-replace pairs defined Dictionary object.
dict是您的搜索替换对定义的 Dictionary 对象。
stris your string which you need to do some replacements with.
str是您需要对其进行一些替换的字符串。
回答by Francis Norton
Here's a lightly re-factored version of @Marc's great answer, to make the functionality available as an extension method to Regex:
这是@Marc 很好的答案的轻微重构版本，使该功能可用作 Regex 的扩展方法：
static void Main()
{
string input = @"Dear $name$, as of $date$ your balance is $amount$";
var args = new Dictionary<string, string>(StringComparer.OrdinalIgnoreCase);
args.Add("name", "Mr Smith");
args.Add("date", "05 Aug 2009");
args.Add("amount", "GBP200");
Regex re = new Regex(@"$(\w+)$", RegexOptions.Compiled);
string output = re.replaceTokens(input, args);
// spot the LinqPad user // output.Dump();
}
public static class ReplaceTokensUsingDictionary
{
public static string replaceTokens(this Regex re, string input, IDictionary<string, string> args)
{
return re.Replace(input, match => args[match.Groups[1].Value]);
}
}
回答by DannyB
when using Marc Gravell's RegEx solution, first check if a token is available using i.e. ContainsKey, this to prevent KeyNotFoundException errors :
使用 Marc Gravell 的 RegEx 解决方案时，首先使用 ie ContainsKey 检查令牌是否可用，以防止 KeyNotFoundException 错误：
string output = re.Replace(zpl, match => { return args.ContainsKey(match.Groups[1].Value) ? arg[match.Groups[1].Value] : match.Value; });
when using the following slightly modified sample code (1st parameter has different name):
使用以下稍微修改的示例代码时（第一个参数具有不同的名称）：
var args = new Dictionary<string, string>(
StringComparer.OrdinalIgnoreCase)
{
{"nameWRONG", "Mr Smith"},
{"date", "05 Aug 2009"},
{"AMOUNT", "GBP200"}
};
this produces the following:
这会产生以下结果：
"Dear $name$, as of 05 Aug 2009 your balance is GBP200"
“亲爱的 $name$，截至 2009 年 8 月 5 日，您的余额为 200 英镑”
回答by Albeoris
Here you are:
这个给你：
public static class StringExm
{
public static String ReplaceAll(this String str, KeyValuePair<String, String>[] map)
{
if (String.IsNullOrEmpty(str))
return str;
StringBuilder result = new StringBuilder(str.Length);
StringBuilder word = new StringBuilder(str.Length);
Int32[] indices = new Int32[map.Length];
for (Int32 characterIndex = 0; characterIndex < str.Length; characterIndex++)
{
Char c = str[characterIndex];
word.Append(c);
for (var i = 0; i < map.Length; i++)
{
String old = map[i].Key;
if (word.Length - 1 != indices[i])
continue;
if (old.Length == word.Length && old[word.Length - 1] == c)
{
indices[i] = -old.Length;
continue;
}
if (old.Length > word.Length && old[word.Length - 1] == c)
{
indices[i]++;
continue;
}
indices[i] = 0;
}
Int32 length = 0, index = -1;
Boolean exists = false;
for (int i = 0; i < indices.Length; i++)
{
if (indices[i] > 0)
{
exists = true;
break;
}
if (-indices[i] > length)
{
length = -indices[i];
index = i;
}
}
if (exists)
continue;
if (index >= 0)
{
String value = map[index].Value;
word.Remove(0, length);
result.Append(value);
if (word.Length > 0)
{
characterIndex -= word.Length;
word.Length = 0;
}
}
result.Append(word);
word.Length = 0;
for (int i = 0; i < indices.Length; i++)
indices[i] = 0;
}
if (word.Length > 0)
result.Append(word);
return result.ToString();
}
}
