Can anyone tell me how to remove unwanted characters from username field.
ex: i:0#.w|abcventures\sreekiran.k I need to remove the characters which are generated before abcventures\sreekiran.k.
I have used translate() to eliminate those characters, but, it is removing i & w characters also, from the username.
To expand on the current answers, I think they meant to use substring-after(). This will do the trick:
substring-after(userName(), "i:0#.w|")
To expand on Mekalikot's answer -
substring-before(userName(), "i:0#.w|")
Will return just the string after the "i:0#.w|" from the user name function. If the user name function does not include that string, however, the formula will return nothing. Since you'll get a different value from the user name function in preview vs. the browser, you'll probably need to test this with the published form.
Use substring-before to remove those characters.
Related
When adding data with the block call.StorageValue, the string is saved in firebase with "/" before and after the string,
There does not seem to be any block to remove it, How can I do it?
It's a normal firebase function that allows to separate the values and read them as such.
Example :
on Firebase, "\"English-EN\"" is a single value sent from the app as English-EN
and "[\"863674037411046\",\"863674037411046\",\"863674037411046\",\"863674037411046\"]" is a list of numbers sent as 863674037411046.
Try to retrieve the value with a button and to a simple label and you should see that it's displayed without the extra characters.
Source:check my app "harpokrates". I've made it as a firebase DB management demo and it uses nothing else. All values are stored as you describe and are retrieved just fine, without extra symbols or any need to trim the text.
ps:However if you do have extra symbols at some point, check your use of lists and lists of lists that might generate excessive "\" if you made a mistake somewhere. You can also use the "trim" or "split text" blocks but that would be bad practice. Finding the code error is best.
This is likely an escape character that escapes the special character " (quotation marks). This is common practice to use \ as an escape character to indicate that the next character has special meaning, in this case it is not the start or end of a string but actually part of it.
As such you can't actually remove it (just the escape character) and should consider how you got a quotation mark in the string to begin with.
You should however be able to remove the entire quotation mark \"
I have done a validations using MVC4 Data Annotations.
And i restricted the Special characters too.
But i want to allow Spaces between each words.
I tried this:
[Required(ErrorMessage = "Category Name should not be empty")]
[RegularExpression(#"^[a-zA-Z0-9]+$", ErrorMessage = "Special character should not be entered")]
public virtual string Name { get; set; }
Right now system is not allowing me to Insert. Because i have entered a space. Its considering the space as special char.
How do i allow space? the same time the above validation should work.
Myself found the answer.
Instead of #"^[a-zA-Z0-9]+$" we need to add like this #"^[a-zA-Z0-9'' ']+$".
''Single Space' in the Expression
'' '
For more information I'd suggesting reading - Using Data Annotations to Customize Data Classes (MSDN)
I just came across your answer and just wanted to mention that the single quotes are not needed. You just need the space inside the brackets.
#"^[a-zA-Z0-9 ]+$"
we need to add like this
Change symbol "+" to "*" (after close square brackets)
Correct answer is :
#"^[a-zA-Z0-9]*$"
What is the default regular expression used by asp.net create user wizard?
According to the MSDN documentation, it should be something like this:
Regular Expression: #\"(?:.{7,})(?=(.*\d){1,})(?=(.*\W){1,})
Validation Message: Your password must be 7 characters long, and contain at least one number and one special character.
However, it does not work as it does not accept something like 3edc£edc, which is actually accepted when using the default create user wizard.
Any idea about how can I get this regular expression?
The error is in the ?: in (?:.{7,})(?=(.*\d){1,})(?=(.*\W){1,}) that is "consuming" the fist seven characters or more characters. It should be ?= OR you can invert the order: (?=(.*\d){1,})(?=(.*\W){1,})(?:.{7,})
Just change the order
^(?=(.*\d))(?=(.*\W)).{7,}
I additionally removed the {1,} and anchored it to the start of the string and you don't need a group around the last part
See it here on Regexr
I have a TextBox, and I want to force the user to type an email format in this field like (example#mail.com) ?
I don't want to use FilteredTextBoxExtender or the RegularExpressionValidator.
I want to do it manualy.
Use the MailAddress class of System.Net.Mail. If what you pass into it is not a valid email address it will fail.
Example:
How do I validate email address formatting with the .NET Framework?
You are really going to reinvent the wheel.
But if it is your wish, you have to use string manipulation functions built in to the String object.
First do a check whether there is a in # symbol in the text.
Use String.Contains to check that.
Or you can use String.IndexOf to check whether the # symbol is present, and if present which index is it present. (considering the string as an array of characters)
And then check whether there are any (and how many) characters present before the # symbol.
If the symbol # symbol was in the 4th index, then you know there are 3 characters before etc.
There's plethora of functions for the String object. You may have to use Length function and String.SubString to retrieve parts of the string.
Get the indexes of the # symbol and the . symbol and check whether there are at least 3 characters in between.
I really cant seem to think of all the possibilities but first list down all the possibilities and check them one by one.
You can also use the Contains method to check whether illegal characters are present :)
EDIT: String.LastIndexOf will return the last index where a specified character was found ;)
And you count and check whether the # symbol was found more than once etc
String.IndexOfAny(Char[])
String.IndexOfAny Method (Char[], Int32)
String.IndexOfAny Method (Char[], Int32, Int32)
This is the best way I found on internet.
Regex.IsMatch(YourStringEmail, "^(?("")("".+?""#)|(([0-9a-zA-Z]((\.(?!\.))|[-!#\$%&'\*\+/=\?\^`\{\}\|~\w])*)(?<=[0-9a-zA-Z])#))" + _
"(?(\[)(\[(\d{1,3}\.){3}\d{1,3}\])|(([0-9a-zA-Z][-\w]*[0-9a-zA-Z]\.)+[a-zA-Z]{2,6}))$")
Thank you.
Looks like a simple task - get a regex that tests a string for particular length:
^.{1,500}$
But if a string has "\r\n" than the above match always fails!
How should the correct regex look like to accept new line characters as part of the string?
I have a <asp:TextBox TextMode="Multiline"> and use a RegularExpressionValidator to check the length of what user types in.
Thank you,
Andrey
You could use the RegexOptions.Singleline option when validating input. This treats the input as a single line statement, and parses it as such.
Otherwise you could give the following expression a try:
^(.|\s){1,500}$
This should work in multiline inputs.
Can you strip the line breaks before checking the length of the string? That'd be easy to do when validating server-side. (In .net you could use a custom validator for that)
From a UX perspective, though, I'd implement a client-side 'character counter' as well. There's plenty to be found. jQuery has a few options. Then you can implement the custom validator to only run server-side, and then use the character counter as your client-side validation. Much nicer for the user to see how many characters they have left WHILE they are typing.
The inability to set the RegexOptions is screwing you up here. Since this is in a RegularExpressionValidator, you could try setting the options in the regular expression itself.
I think this should work:
(?s)^.{1,500}$
The (?s) part turns on the Singleline option which will allow the dot to match every character including line feeds. For what it's worth, the article here also lists the other RegexOptions and the notation needed to set them as an inline statement.