C# 解析C中的复杂字符串。
我目前正在维护一个用C编写的应用程序。我需要编写一个新功能,我遇到了麻烦 我正在从数据库中提取数据,用于用VB6编写的另一个应用程序。有一个字段用于保存树列表的数据。我需要在我的应用程序中重新创建此树。数据如下所示:C# 解析C中的复杂字符串。,c#,parsing,vb6,C#,Parsing,Vb6,我目前正在维护一个用C编写的应用程序。我需要编写一个新功能,我遇到了麻烦 我正在从数据库中提取数据,用于用VB6编写的另一个应用程序。有一个字段用于保存树列表的数据。我需要在我的应用程序中重新创建此树。数据如下所示: { { table1~col_b^table1~colc^= }| { { table1~col_b^table2~col_b^= }|{ table2~col
{
{
table1~col_b^table1~colc^=
}|
{
{
table1~col_b^table2~col_b^=
}|{
table2~col_a^table3~cola^=
}|AND
}|OR
}
OR
-- table1~col_b^table1~colc^=
-- AND
---- table1~col_b^table2~col_b^=
---- table2~col_a^table3~cola^=
我甚至不知道从哪里开始。我需要完成的就是这个。{}表示一个表达式,|单独的表达式。基本上,这棵树应该是这样的:
{
{
table1~col_b^table1~colc^=
}|
{
{
table1~col_b^table2~col_b^=
}|{
table2~col_a^table3~cola^=
}|AND
}|OR
}
OR
-- table1~col_b^table1~colc^=
-- AND
---- table1~col_b^table2~col_b^=
---- table2~col_a^table3~cola^=
解决这个问题的任何帮助或指导都将是伟大的
谢谢
道格我建议你什么时候定义语法。它允许非常容易地编写解析器 这是代码的粗略语法:
[Language("ExpressionEvaluator", "1.0", "Multi-line expression evaluator")]
public class ExpressionEvaluatorGrammar : Grammar
{
public ExpressionEvaluatorGrammar()
{
// 1. Terminals
var identifier = new RegexBasedTerminal("identifier", "[a-z\\d_^~]+");
// 2. Non-terminals
var root = new NonTerminal("root");
var block = new NonTerminal("block");
var expression = new NonTerminal("expression");
var expressions = new NonTerminal("expressions");
var prop = new NonTerminal("prop");
var op = new NonTerminal("op");
// 3. BNF rules
op.Rule = ToTerm("OR") | "AND";
prop.Rule = identifier + "=" ;
expression.Rule = "{" + (prop | block) + "}" + "|" ;
expressions.Rule = MakeStarRule(expressions, expression);
block.Rule = expressions + op;
root.Rule = "{" + block +"}";
Root = root;
//automatically add NewLine before EOF so that our BNF rules work correctly when there's no final line break in source
this.LanguageFlags = LanguageFlags.NewLineBeforeEOF;
}
}
} //namespace
它解析得很好,您所需要做的就是添加AST树并使用它
当你定义语法时,我建议你看看。它允许非常容易地编写解析器 这是代码的粗略语法:
[Language("ExpressionEvaluator", "1.0", "Multi-line expression evaluator")]
public class ExpressionEvaluatorGrammar : Grammar
{
public ExpressionEvaluatorGrammar()
{
// 1. Terminals
var identifier = new RegexBasedTerminal("identifier", "[a-z\\d_^~]+");
// 2. Non-terminals
var root = new NonTerminal("root");
var block = new NonTerminal("block");
var expression = new NonTerminal("expression");
var expressions = new NonTerminal("expressions");
var prop = new NonTerminal("prop");
var op = new NonTerminal("op");
// 3. BNF rules
op.Rule = ToTerm("OR") | "AND";
prop.Rule = identifier + "=" ;
expression.Rule = "{" + (prop | block) + "}" + "|" ;
expressions.Rule = MakeStarRule(expressions, expression);
block.Rule = expressions + op;
root.Rule = "{" + block +"}";
Root = root;
//automatically add NewLine before EOF so that our BNF rules work correctly when there's no final line break in source
this.LanguageFlags = LanguageFlags.NewLineBeforeEOF;
}
}
} //namespace
它解析得很好,您所需要做的就是添加AST树并使用它
如果没有任何附加库,在纯C代码中将其解析为表达式树并不难:
class TreeNode
{
private enum ParseState
{
Operator,
Expression
}
public static TreeNode ParseTree(string treeData)
{
Stack<TreeNode> parsed = new Stack<TreeNode>();
StringBuilder nodeData = new StringBuilder();
ParseState state = ParseState.Operator;
for (int charIndex = 0; charIndex < treeData.Length; charIndex++)
{
switch (treeData[charIndex])
{
case '{':
nodeData.Clear();
state = ParseState.Expression;
break;
case '\t':
case ' ':
case '\r':
case '\n':
case '|':
// ignore whitespace and |
break;
case '}':
{
if (state == ParseState.Expression)
{
state = ParseState.Operator;
parsed.Push(new TreeNodeData(nodeData.ToString()));
}
else // Operator
{
TreeNodeOperators op = (TreeNodeOperators)(Enum.Parse(typeof(TreeNodeOperators), nodeData.ToString()));
TreeNodeExpression exp = new TreeNodeExpression();
exp.Operator = op;
exp.Right = parsed.Pop();
exp.Left = parsed.Pop();
parsed.Push(exp);
}
nodeData.Clear();
}
break;
default:
nodeData.Append(treeData[charIndex]);
break;
}
}
return parsed.Pop();
}
}
enum TreeNodeOperators
{
AND,
OR
}
class TreeNodeExpression : TreeNode
{
public TreeNodeOperators Operator {get; set;}
public TreeNode Left { get; set; }
public TreeNode Right { get; set; }
}
class TreeNodeData : TreeNode
{
public string Data {get; set;}
public TreeNodeData(string data)
{
Data = data;
}
}
如果没有任何附加库,在纯C代码中将其解析为表达式树并不难:
class TreeNode
{
private enum ParseState
{
Operator,
Expression
}
public static TreeNode ParseTree(string treeData)
{
Stack<TreeNode> parsed = new Stack<TreeNode>();
StringBuilder nodeData = new StringBuilder();
ParseState state = ParseState.Operator;
for (int charIndex = 0; charIndex < treeData.Length; charIndex++)
{
switch (treeData[charIndex])
{
case '{':
nodeData.Clear();
state = ParseState.Expression;
break;
case '\t':
case ' ':
case '\r':
case '\n':
case '|':
// ignore whitespace and |
break;
case '}':
{
if (state == ParseState.Expression)
{
state = ParseState.Operator;
parsed.Push(new TreeNodeData(nodeData.ToString()));
}
else // Operator
{
TreeNodeOperators op = (TreeNodeOperators)(Enum.Parse(typeof(TreeNodeOperators), nodeData.ToString()));
TreeNodeExpression exp = new TreeNodeExpression();
exp.Operator = op;
exp.Right = parsed.Pop();
exp.Left = parsed.Pop();
parsed.Push(exp);
}
nodeData.Clear();
}
break;
default:
nodeData.Append(treeData[charIndex]);
break;
}
}
return parsed.Pop();
}
}
enum TreeNodeOperators
{
AND,
OR
}
class TreeNodeExpression : TreeNode
{
public TreeNodeOperators Operator {get; set;}
public TreeNode Left { get; set; }
public TreeNode Right { get; set; }
}
class TreeNodeData : TreeNode
{
public string Data {get; set;}
public TreeNodeData(string data)
{
Data = data;
}
}
您可以使用regexp进行标记化,使用堆栈进行递归解析,如下所示
internal class Node
{
public string Terminal { get; set; }
public List<Node> Operands { get; set; }
}
internal static readonly Regex TokensPattern = new Regex(@"(?<ws>\s+)|{\s*(?<value>[^\s}]+)\s*}|(?<token>OR|AND|.)", RegexOptions.Compiled);
static Node parseData(string str)
{
// init stack
var stack = new Stack<Node>();
stack.Push(new Node() { Operands = new List<Node>() });
// define parser
var parser = new Dictionary<string, Action<string>>();
parser.Add("{", _ => stack.Push(new Node() { Operands = new List<Node>() }));
parser.Add("}", _ => { var top = stack.Pop(); stack.Peek().Operands.Add(top); });
parser.Add("|", _ => { });
parser.Add("AND", _ => stack.Peek().Terminal = "AND");
parser.Add("OR", _ => stack.Peek().Terminal = "OR");
parser.Add("", value => stack.Peek().Operands.Add(new Node { Terminal = value }));
// execute parser
TokensPattern.Matches(str).Cast<Match>()
.Where(m => string.IsNullOrEmpty(m.Groups["ws"].Value))
.Count(m => { parser[m.Groups["token"].Value](m.Groups["value"].Value); return false; });
// return top of the tree
return stack.Peek().Operands[0];
}
static void Main(string[] args)
{
const string str = @"{
{
table1~col_b^table1~colc^=
}|
{
{
table1~col_b^table2~col_b^=
}|{
table2~col_a^table3~cola^=
}|{cccc}|AND
}|OR
}";
// print tree function
Action<int, Node> dump = null;
dump = new Action<int, Node>((level, node) =>
{
Console.WriteLine("{0}{1}", new string(' ', level * 2), node.Terminal);
if (node.Operands != null)
node.Operands.ForEach(el => dump(level + 1, el));
});
dump(0, parseData(str));
}
您可以使用regexp进行标记化,使用堆栈进行递归解析,如下所示
internal class Node
{
public string Terminal { get; set; }
public List<Node> Operands { get; set; }
}
internal static readonly Regex TokensPattern = new Regex(@"(?<ws>\s+)|{\s*(?<value>[^\s}]+)\s*}|(?<token>OR|AND|.)", RegexOptions.Compiled);
static Node parseData(string str)
{
// init stack
var stack = new Stack<Node>();
stack.Push(new Node() { Operands = new List<Node>() });
// define parser
var parser = new Dictionary<string, Action<string>>();
parser.Add("{", _ => stack.Push(new Node() { Operands = new List<Node>() }));
parser.Add("}", _ => { var top = stack.Pop(); stack.Peek().Operands.Add(top); });
parser.Add("|", _ => { });
parser.Add("AND", _ => stack.Peek().Terminal = "AND");
parser.Add("OR", _ => stack.Peek().Terminal = "OR");
parser.Add("", value => stack.Peek().Operands.Add(new Node { Terminal = value }));
// execute parser
TokensPattern.Matches(str).Cast<Match>()
.Where(m => string.IsNullOrEmpty(m.Groups["ws"].Value))
.Count(m => { parser[m.Groups["token"].Value](m.Groups["value"].Value); return false; });
// return top of the tree
return stack.Peek().Operands[0];
}
static void Main(string[] args)
{
const string str = @"{
{
table1~col_b^table1~colc^=
}|
{
{
table1~col_b^table2~col_b^=
}|{
table2~col_a^table3~cola^=
}|{cccc}|AND
}|OR
}";
// print tree function
Action<int, Node> dump = null;
dump = new Action<int, Node>((level, node) =>
{
Console.WriteLine("{0}{1}", new string(' ', level * 2), node.Terminal);
if (node.Operands != null)
node.Operands.ForEach(el => dump(level + 1, el));
});
dump(0, parseData(str));
}
如果需要起点,请通过详细定义语法来检查REGEXStart。基于一个或多个示例字符串的解析器往往会导致很多不愉快。请参阅我的编辑。如果需要起点,请通过详细定义语法检查REGEXStart。基于一个或多个示例字符串的解析器往往会导致很多不愉快。请参阅我的编辑。您假设和/或是二进制运算符,这很可能不是事实。是的,此代码基于有限的可用信息进行了一些假设,但调整以适应不符合这些假设的情况并不困难。您假设和/或是二进制运算符,这很可能不是实际情况。是的,此代码确实基于有限的可用信息做出了一些假设,但调整以适应不符合这些假设的情况并不困难。