I would like to know how to remove '\\0' from a string. This may be very simple but it's not for me since I'm a new C# developer.
I've this code:
public static void funcTest (string sSubject, string sBody)
{
Try
{
MailMessage msg = new MailMessage(); // Set up e-mail message.
msg.To = XMLConfigReader.Email;
msg.From = XMLConfigReader.From_Email;
msg.Subject = sSubject;
msg.body="TestStrg.\r\nTest\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\r\n";
}
catch (Exception ex)
{
string sMessage = ex.Message;
log.Error(sMessage, ex);
}
}
But what I want is:
msg.body="TestStrg.\r\nTest\r\n";
So, is there a way to do this with a simple code?
It seems you just want the string.Replace
function (static method).
var cleaned = input.Replace("\0", string.Empty);
Edit: Here's the complete code, as requested:
public static void funcTest (string sSubject, string sBody)
{
try
{
MailMessage msg = new MailMessage();
msg.To = XMLConfigReader.Email;
msg.From = XMLConfigReader.From_Email;
msg.Subject = sSubject;
msg.Body = sBody.Replace("\0", string.Empty);
}
catch (Exception ex)
{
string sMessage = ex.Message;
log.Error(sMessage, ex);
}
}
我使用: something.TrimEnd('\\0')
你只需要更换它
msg.body="TestStrg.\r\nTest\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\r\n".Replace("\0", string.Empty);
这是 Bing 中的第一个结果,它没有我喜欢的方法,即
string str = "MyString\0\0\0\0".Trim('\0');
尝试这个:
msg.Body = msg.Body.Replace("\0", "");
如果您使用 LINQ,结果可能会更快
str.TakeWhile(c => c != '\0');
保持愚蠢的简单=*
return stringValue.Substring(0, stringValue.IndexOf('\0'));
msg.body = sBody.Replace("\0", "");
I know I'm late here but, while String.Replace works most of the time I have found that I like the regex version of this much better and is more reliable in most cases
using System.Text.RegularExpressions;
...
msg.body=Regex.Replace("TestStrg.\r\nTest\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\r\n","\0");
var str = "TestStrg.\r\nTest\0\0\0\0\0\0\0\0\0\r\n".Replace("\0", "");
String.Replace()
将用空字符串替换所有\\0
,从而删除它们。
这条线应该工作:
string result = Regex.Replace(input, "\0", String.Empty);
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