Regex.Replace方法

时间:2012-08-10 14:34:34

标签: regex string replace

我想使用Regex.Replace方法。我会搜索 “new long?((long)”表达式在非常长的字符串中,将进行以下更改。

在:

OracleParameter[] array = new OracleParameter[]
{
    new OracleParameter("p_op_adi", OracleDbType.Varchar2, p_op_adi, ParameterDirection.Input),
    new OracleParameter("p_tip_no", OracleDbType.Int64, p_tip_no, ParameterDirection.Input),
    new OracleParameter("p_islem", OracleDbType.Varchar2, p_islem, ParameterDirection.Input),
    new OracleParameter("p_raise", OracleDbType.Varchar2, p_raise, ParameterDirection.Input),
    new OracleParameter("p_durum", OracleDbType.Int64, null, ParameterDirection.Output),
    new OracleParameter("p_mesaj", OracleDbType.Varchar2, 4000, null, ParameterDirection.Output)
};
sMan.ExecuteSP("FRM_USER.user_yetki_kontrol_islem", array);
p_durum = ((array[4].Value != System.DBNull.Value) ? new long?((long)array[4].Value) : null);
p_mesaj = array[5].Value.ToString();

在Regex之后:

OracleParameter[] array = new OracleParameter[]
{
    new OracleParameter("p_op_adi", OracleDbType.Varchar2, p_op_adi, ParameterDirection.Input),
    new OracleParameter("p_tip_no", OracleDbType.Int64, p_tip_no, ParameterDirection.Input),
    new OracleParameter("p_islem", OracleDbType.Varchar2, p_islem, ParameterDirection.Input),
    new OracleParameter("p_raise", OracleDbType.Varchar2, p_raise, ParameterDirection.Input),
    new OracleParameter("p_durum", OracleDbType.Int64, null, ParameterDirection.Output),
    new OracleParameter("p_mesaj", OracleDbType.Varchar2, 4000, null, ParameterDirection.Output)
};
sMan.ExecuteSP("FRM_USER.user_yetki_kontrol_islem", array);
p_durum = abcde(array[4].Value);
p_mesaj = array[5].Value.ToString();

请注意,我必须捕获数组索引号4。

2 个答案:

答案 0 :(得分:1)

此代码应该完全符合您的要求:

class Program
{
    static void Main(string[] args)
    {
        string input = @"OracleParameter[] array = new OracleParameter[]
            {
                new OracleParameter('p_op_adi', OracleDbType.Varchar2, p_op_adi, ParameterDirection.Input),
                p_durum = ((array[4].Value != System.DBNull.Value) ? new long?((long)array[4].Value) : null);
                p_durum = ((array[4].Value != System.DBNull.Value) ? new long?((long)array[4].Value) : null);
            ";

        string pattern = @"^(?<varname>[^=]*)=.*new long\?\(\(long\)(?<arrayval>[^)]*).*?$";
        string replacement = "${varname} = abcde(${arrayval});";

        string output = Regex.Replace(input, pattern, replacement, RegexOptions.Multiline);
        Console.WriteLine(output);
        Console.ReadKey();
    }
}

它收集变量名称和数组值,并在替换部分中将它们用作替换。

答案 1 :(得分:0)

此正则表达式选择完整字符串(?m)^.*?new long\?\(\(long\).*?$

string res = Regex.Replace(input, @"(?m)^.*?new long\?\(\(long\).*?$", "p_durum = abcde(array[4].Value);");