Regular expression for a word.

Question

I am making a string parsing problem in PHP. I want a regular expression for a word in a dictionary. The word can only contain AZ and az. I came up with this regex, but that doesn't seem to work. Can someone help.

$regexp = "/[Az]+/"; if(preg_match($regexp,$buffer)){ print $buffer . "<BR>"; } 

Adding rkt comment for eadibility

currently using this regex

 $regexp = "/[A-Za-z]+/"; 

but still many irrelevant words are printed, for example.

 a.new,#quickbar a.new{color:#ba0000} enwiki:resourceloader:filter:minify-css:5:f2a9127573a22335c2a9102b208c73e7 wgNamespaceNumber=0; wgAction="view"; wgPageName="Roger_Federer"; wgMainPageTitle="‌​Main Page"; wgWikimediaMobileUrl="http:\/\/en.m.wikipedia.org\/wiki"; document.writeln("\x3cdiv id=\"localNotice\"\x3e\x3cp\x3e\x3c/p\x3e\n\x3c/div\x3e"); 

Answer 1

If you want to combine only the whole word ...

 /\b[AZ]+\b/i 

Answer 2

Besides the problem that Limo Wang Kenobi noticed, your regular expression does not do what you think.

 /[A-Za-z]+/ 

All this checks that the letter, in any case, appears somewhere in the input. If your input looks like this:

 1111112222222333333333A333333444444555555567 

He will still fit.

What you are looking for is a regular expression:

 /^[A-Za-z]+$/ 

This will correspond to the beginning of the line, then one or more letters and, finally, the end of the line. Now there is no place for anything but letters!

Answer 3

The regular expression should be /\b[A-Za-z]+\b/ , it was believed that for words to match in the dictionary, you probably want something more restrictive than this, because words usually don't have caps in any odd place.

Answer 4

your regular expression should be

 $regexp = "/[A-Za-z]+/"; 

Answer 5

You can check my answer on How do you separate a javascript string with spaces and punctuation? , which contains JavaScript, which can be used to break English text into an array of words and punctuation marks, where words can be hyphenated (clearly defined) or compound (will not).