I have a regular expression ^(?=.*?[A-Za-z])\S*$ which indicates that the input should contain alphabets and can contain special characters or digits along with the alphabets. But it is not allowing white spaces since i have used \S.
Can some one suggest me a reg exp which should contain alphabets and it can contain digits or special characters and white space but alphabets are must and the last character should not end with a white space
Quite simply:
^(?=.*?[A-Za-z]).*$
Note that in JavaScript . doesn't match new lines, and there is no dot-all flag (/s). You can use something like [\s\S] instead if that is an issue:
^(?=[\s\S]*?[A-Za-z])[\s\S]*$
Since you only have a single lookahead, you can simplify the pattern to:
^.*[A-Za-z].*$
Or, even simpler:
[A-Za-z]
[A-Za-z] will match if it finds a letter anywhere in the string, you don't really need to search the rest from start to end.
To also validate the last character isn't a whitespace, it is probably easiest to use the lookahead again (as it basically means AND in regular expressions:
^(?=.*?[A-Za-z]).*\S$
.*\S$ at the end of each patten can validate that: [A-Za-z].*\S$, for example. Can you please edit the question to add this requirement?var r = /^.*[A-Za-z].*\S$/; - nothing special here. developer.mozilla.org/en/Core_JavaScript_1.5_Guide/…^(?=.*?[A-Za-z]).*\S$. I'll update the answer. You can also use [A-Za-z].*\S$|[A-Za-z]$, but it isn't as elegant as the first option.