[RegularExpression("^\\d{5}$||d{0}", ErrorMessage = "Girdiğiniz değer 5 karakter uzunluğunda olmalıdır ve rakamlardan oluşmalıdır")]
public string PostaKodu { get; set; }
When I get PostaKodu from old database , it returns value as five space character if it is null. In form view, it gives validation error. How can i add five white space character to my regular expression
You should modify your expression like use spaces like \s, this is the character for white space in regular expression. Something like \\d{5}$||d{0}||^\\s{5}$ would do the job :D
Related
i would appreciate your help on this, since i do not know which range of characters to use, or if there is a character class like [[:cntrl:]] that i have found in ruby?
by means of non printable, i mean delete all characters that are not shown in ie output, when one prints the input string. Please note, i look for a c# regex, i do not have a problem with my code
You may remove all control and other non-printable characters with
s = Regex.Replace(s, #"\p{C}+", string.Empty);
The \p{C} Unicode category class matches all control characters, even those outside the ASCII table because in .NET, Unicode category classes are Unicode-aware by default.
Breaking it down into subcategories
To only match basic control characters you may use \p{Cc}+, see 65 chars in the Other, Control Unicode category. It is equal to a [\u0000-\u0008\u000E-\u001F\u007F-\u0084\u0086-\u009F \u0009-\u000D \u0085]+ regex.
To only match 161 other format chars including the well-known soft hyphen (\u00AD), zero-width space (\u200B), zero-width non-joiner (\u200C), zero-width joiner (\u200D), left-to-right mark (\u200E) and right-to-left mark (\u200F) use \p{Cf}+. The equivalent including astral place code points is a (?:[\xAD\u0600-\u0605\u061C\u06DD\u070F\u08E2\u180E\u200B-\u200F\u202A-\u202E\u2060-\u2064\u2066-\u206F\uFEFF\uFFF9-\uFFFB]|\uD804[\uDCBD\uDCCD]|\uD80D[\uDC30-\uDC38]|\uD82F[\uDCA0-\uDCA3]|\uD834[\uDD73-\uDD7A]|\uDB40[\uDC01\uDC20-\uDC7F])+ regex.
To match 137,468 Other, Private Use control code points you may use \p{Co}+, or its equivalent including astral place code points, (?:[\uE000-\uF8FF]|[\uDB80-\uDBBE\uDBC0-\uDBFE][\uDC00-\uDFFF]|[\uDBBF\uDBFF][\uDC00-\uDFFD])+.
To match 2,048 Other, Surrogate code points that include some emojis, you may use \p{Cs}+, or [\uD800-\uDFFF]+ regex.
You can try with :
string s = "Täkörgåsmrgås";
s = Regex.Replace(s, #"[^\u0000-\u007F]+", string.Empty);
Updated answer after comments:
Documentation about non-printable character:
https://en.wikipedia.org/wiki/Control_character
Char.IsControl Method:
https://msdn.microsoft.com/en-us/library/system.char.iscontrol.aspx
Maybe you can try:
string input; // this is your input string
string output = new string(input.Where(c => !char.IsControl(c)).ToArray());
To remove all control and other non-printable characters
Regex.Replace(s, #"\p{C}+", String.Empty);
To remove the control characters only (if you don't want to remove the emojis 😎)
Regex.Replace(s, #"\p{Cc}+", String.Empty);
you can try this:
public static string TrimNonAscii(this string value)
{
string pattern = "[^ -~]*";
Regex reg_exp = new Regex(pattern);
return reg_exp.Replace(value, "");
}
i am trying to validate the input of a user to insert only alphanumeric characters including special characters like .,_ white space.
Point 1. the user cant insert only special characters like ####### or .........
Point 2. or any numeric numbers like 2222222.
it should be any valid format like. "hi i am asking a question on stack overflow.this is my 11th attempt. "
i tried these expressions but its not let me restrict user like point 1 and pont 2 ,please help.
here is my code
[RegularExpression(#"^([a-zA-Z0-9 \.\&\'\-]+)$", ErrorMessage = "Invalid User Name")]
public string UserName { get; set; }
[Required]
[StringLength(250, MinimumLength = 5, ErrorMessage = "User Description must have minimum 5 and maximum 250 characters.")]
[RegularExpression(#"^[^<>!##%/?*]+$", ErrorMessage = "Invalid User Description")]
public string Description { get; set; }
You need to use a negative lookahead at the start.
#"^(?![\W_]+$)(?!\d+$)[a-zA-Z0-9 .&',_-]+$"
DEMO
(?![\W_]+$) Negative lookahead asserts that the string won't contain only special characters.
(?!\d+$) asserts that the string won't contain only digits.
^(?=.*[a-zA-Z])[a-zA-Z0-9,_.&' -]+$
You can enforce a condition thru lookahead stating that a letter is always necessary.
I want to create a url friendly string (one that will only contain letters, numbers and hyphens) from a user input to :
remove all characters which are not a-z, 0-9, space or hyphens
replace all spaces with hyphens
replace multiple hyphens with a single hyphen
Expected outputs :
my project -> my-project
test project -> test-project
this is # long str!ng with spaces and symbo!s -> this-is-long-strng-with-spaces-and-symbos
Currently i'm doing this in 3 steps :
$identifier = preg_replace('/[^a-zA-Z0-9\-\s]+/','',strtolower($project_name)); // remove all characters which are not a-z, 0-9, space or hyphens
$identifier = preg_replace('/(\s)+/','-',strtolower($identifier)); // replace all spaces with hyphens
$identifier = preg_replace('/(\-)+/','-',strtolower($identifier)); // replace all hyphens with single hyphen
Is there a way to do this with one single regex ?
Yeah, #Jerry is correct in saying that you can't do this in one replacement as you are trying to replace a particular string with two different items (a space or dash, depending on context). I think Jerry's answer is the best way to go about this, but something else you can do is use preg_replace_callback. This allows you to evaluate an expression and act on it according to what the match was.
$string = 'my project
test project
this is # long str!ng with spaces and symbo!s';
$string = preg_replace_callback('/([^A-Z0-9]+|\s+|-+)/i', function($m){$a = '';if(preg_match('/(\s+|-+)/i', $m[1])){$a = '-';}return $a;}, $string);
print $string;
Here is what this means:
/([^A-Z0-9]+|\s+|-+)/i This looks for any one of your three quantifiers (anything that is not a number or letter, more than one space, more than one hyphen) and if it matches any of them, it passes it along to the function for evaluation.
function($m){ ... } This is the function that will evaluate the matches. $m will hold the matches that it found.
$a = ''; Set a default of an empty string for the replacement
if(preg_match('/(\s+|-+)/i', $m[1])){$a = '-';} If our match (the value stored in $m[1]) contains multiple spaces or hyphens, then set $a to a dash instead of an empty string.
return $a; Since this is a function, we will return the value and that value will be plopped into the string wherever it found a match.
Here is a working demo
I don't think there's one way of doing that, but you could reduce the number of replaces and in an extreme case, use a one liner like that:
$text=preg_replace("/[\s-]+/",'-',preg_replace("/[^a-zA-Z0-9\s-]+/",'',$text));
It first removes all non-alphanumeric/space/dash with nothing, then replaces all spaces and multiple dashes with a single one.
Since you want to replace each thing with something different, you will have to do this in multiple iterations.
Sorry D:
I tried (\s|\t).*[\b\w*\s\b], this one is almost ok but I want also except lines with #.
#Name Type Allowable values
#========================== ========= ========================================
_absolute-path-base-uri String -
add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0
As #anubhava said in his answer, it looks you just need to check for # at the beginning of the line. The regex for that is simple, but the mechanics of applying the regex varies wildly, so it would help if we knew which regex flavor/tool you're using (e.g. PHP, .NET, Notepad++, EditPad Pro, etc.). Here's a JavaScript version:
/^[^#].*$/mg
Notice the modifiers: m ("multiline") allows ^ and $ to match at line boundaries, and g ("global") allows you to find all the matches, not just the first one.
Now let's look at your regex. [\b\w*\s\b] is a character class that matches a word character (\w), a whitespace character (\s), an asterisk (*), or a backspace (\b). In other words, both * and \b lose their special meanings when the appear in a character class.
\s matches any whitespace character including \t, so (\s|\t) is needlessly redundant, and may not be needed at all. What it's actually doing in your case is matching the newline before each matched line. There's no need for that when you can use ^ in multiline mode. If you want to allow for horizontal whitespace (i.e., spaces and tabs) before the #, you can do this:
/^(?![ \t]*#).*$/mg
(?![ \t]*#) is a negative lookahead; it means "from this position, it is impossible to match zero or more tabs or spaces followed by #". Coming right after the ^ line anchor as it does, "this position" means the beginning of a line.
Try this:
^[A-z0-9_-]+\s+(.+)$
Assuming your first string will consist of only letters, numbers, underscores or hyphens, the first part will match that. Then we match whitespace, and then capture the rest. However, this is all dependent on the regular expression engine being used. Is this using language support for regexes, a specific editor, or a certain library? Which one? There isn't a standard: each regex engine works slightly differently.
Try this:
^[^#].*?(\s|\t)(?<Group>.*)$
After a match is found, the Group group will contain your string.
I would use this regex. In English, this says "First character is not a pound sign (#), then non-white space to match the first 'word', then white space, then match the whole line.
^[^#]\S*\s+(.+)$
Can I suggest another approach though? It looks like there are tabs between each field in the text, so why not just read the text line-by-line and split by tab into an array?
Here is an example in C# (untested):
using(StreamReader sr = new StreamReader("C:\\Path\\to\\file.txt"))
{
string line = sr.ReadLine();
while(!sr.EndOfStream)
{
//skip the comment lines
if(line.StartsWith("#"))
continue;
string[] fields = line.Split(new string[] {"\t"}, StringSplitOptions.RemoveEmptyEntries);
//now fields[0] contains the Name field
//fields[1] contains the Type field
//fields[2] contains the Allowable Values field
line = sr.ReadLine();
}
}
Try this code in php:
<?php
$s="#Name Type Allowable values
#========================== ========= ========================================
_absolute-path-base-uri String -
add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0 ";
$a = explode("\n", $s);
foreach($a as $str) {
preg_match('~^[^#].*$~', $str, $m);
var_dump($m);
}
?>
OUTPUT
array(0) {
}
array(0) {
}
array(1) {
[0]=>
string(79) "_absolute-path-base-uri String - "
}
array(1) {
[0]=>
string(77) "add-xml-decl Boolean y/n, yes/no, t/f, true/false, 1/0 "
}
Code is pretty simple, it just ignores matching # at the start of a line thus ingoring those lines completely.
The regular expression ^[A-Za-z](\W|\w)* matches when the user gives the first letter as white space, and the first letter should not be a digit and remaining letters may be alpha numerical. When the user gives a white space as the first character it should automatically be trimmed. How?
^\s*([A-Za-z]\w*)
Should do it. Just get group 1.
I'm not sure the language you are using, I'm going to assume C#, so here is a C# sample:
string testString = " myMatch123 not in the match";
Regex regexObj = new Regex("^\\s*([A-Za-z]\\w*)",
RegexOptions.IgnoreCase | RegexOptions.Multiline);
string result = regexObj.Match(testString).Groups[1].Value;
Console.WriteLine("-" + result + "-");
This will print
-myMatch123-
to the console window.
Is it possible to Trim() your input before giving it to your regex?
If you're looking for alpha-numerical, starting with non-numeric, you probably want:
\s*([A-Za-z][A-Za-z0-9]+)
If you allow one-character user names, change that plus to a star.