How to distinguish between a SHA1 string and a date-time string? - php

Basically I have two options:
$one = "fdfeb16f096983ada02db49d46a8154475d700ae";
$two = "2011-12-28 05:20:01";
I need some sort of regex, so that I can detect wether the string follows the pattern in $one, or the pattern in $two
Detect if the string is sha1 or datetime.
What would be the best way to determine this?
Thanks

If you are ABSOLUTE sure those are the ONLY two options, I would go with strlen, and not some kind of marvelous regexp.
Even if those are not the only two options (user messed up), I would still go with strlen, and then check specifically for each format, if it is what you expect it to be.

I don't know how to apply regexes in PHP, but assuming the actual regexes are enough...
/^[a-f0-9]{40}$/
Match 40 characters of consisting of a-f or 0-9. The ^ and $ match the beginning and end of the string, so nothing else can be in it.
/^\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2}$/
Match strings with the date pattern you have.

if (preg_match($one, $string) {
echo "$string matches $one";
} else if (preg_match($two, $string) {
echo "$string matches $two";
}

Try using the preg_match() function.
http://php.net/manual/en/function.preg-match.php

Conditionals with regex to match each of the options. For the second case:
if(preg_match('|(\d{4}-\d{2}-\d{2}\s\d{2}:\d{2}:\d{2}|',$two)
{
echo "Matching two";
}
I don't see a clear pattern for number one, but you could do elseif(s) to detect other potential cases.

Related

PHP preg_match regular expression for find date in string

I try to make system that can detect date in some string, here is the code :
$string = "02/04/16 10:08:42";
$pattern = "/\<(0?[1-9]|[12][0-9]|3[01])\/\.- \/\.- \d{2}\>/";
$found = preg_match($pattern, $string);
if ($found) {
echo ('The pattern matches the string');
} else {
echo ('No match');
}
The result i found is "No Match", i don't think that i used correct regex for the pattern. Can somebody tell me what i must to do to fix this code
First of all, remove all gibberish from the pattern. This is the part you'll need to work on:
(/0?[1-9]|[12][0-9]|3[01]/)
(As you said, you need the date only, not the datetime).
The main problem with the pattern, that you are using the logical OR operators (|) at the delimiters. If the delimiters are slashes, then you need to replace the tube characters with escaped slashes (/). Note that you need to escape them, because the parser will not take them as control characters. Like this: \/.
Now, you need to solve some logical tasks here, to match the numbers correctly and you're good to go.
(I'm not gonna solve the homework for you :) )
These articles will help you to solve the problem tough:
Character classes
Repetition opetors
Special characters
Pipe character (alternation operator)
Good luck!
In your comment you say you are looking for yyyy, but the example says yy.
I made a code for yy because that is what you gave us, you can easily change the 2 to a 4 and it's for yyyy.
preg_match("/((0|1|2|3)[0-9])\/\d{2}\/\d{2}/", $string, $output_array);
Echo $output_array[1]; // date
Edit:
If you use this pattern it will match the time too, thus make it harder to match wrong.
((0|1|2|3)[0-9])/\d{2}/\d{2}\s+\d{2}:\d{2}:\d{2}
http://www.phpliveregex.com/p/fjP
Edit2:
Also, you can skip one line of code.
You first preg_match to $found and then do an if $found.
This works too:
If(preg_match($pattern, $string, $found))}{
Echo $found[1];
}Else{
Echo "nothing found";
}
With pattern and string as refered to above.
As you can see the found variable is in the preg_match as the output, thus if there is a match the if will be true.

Replace from one custom string to another custom string

How can I replace a string starting with 'a' and ending with 'z'?
basically I want to be able to do the same thing as str_replace but be indifferent to the values in between two strings in a 'haystack'.
Is there a built in function for this? If not, how would i go about efficiently making a function that accomplishes it?
That can be done with Regular Expression (RegEx for short).
Here is a simple example:
$string = 'coolAfrackZInLife';
$replacement = 'Stuff';
$result = preg_replace('/A.*Z/', $replacement, $string);
echo $result;
The above example will return coolStuffInLife
A little explanation on the givven RegEx /A.*Z/:
- The slashes indicate the beginning and end of the Regex;
- A and Z are the start and end characters between which you need to replace;
- . matches any single charecter
- * Zero or more of the given character (in our case - all of them)
- You can optionally want to use + instead of * which will match only if there is something in between
Take a look at Rubular.com for a simple way to test your RegExs. It also provides short RegEx reference
$string = "I really want to replace aFGHJKz with booo";
$new_string = preg_replace('/a[a-zA-z]+z/', 'boo', $string);
echo $new_string;
Be wary of the regex, are you wanting to find the first z or last z? Is it only letters that can be between? Alphanumeric? There are various scenarios you'd need to explain before I could expand on the regex.
use preg_replace so you can use regex patterns.

Regex to validate username

I'm trying to understand what's wrong with this regex pattern:
'/^[a-z0-9-_\.]*[a-z0-9]+[a-z0-9-_\.]*{4,20}$/i'
What I'm trying to do is to validate the username. Allowed chars are alphanumeric, dash, underscore, and dot. The restriction I'm trying to implement is to have at least one alphanumeric character so the user will not be allowed to have a nickname like this one: _-_.
The function I'm using right now is:
function validate($pattern, $string){
return (bool) preg_match($pattern, $string);
}
Thanks.
EDIT
As #mario said, yes,t here is a problem with *{4,20}.
What I tried to do now is to add ( ) but this isn't working as excepted:
'/^([a-z0-9-_\.]*[a-z0-9]+[a-z0-9-_\.]*){4,20}$/i'
Now it matches 'aa--aa' but it doesn't match 'aa--' and '--aa'.
Any other suggestions?
EDIT
Maybe someone wants to deny not nice looking usernames like "_..-a".
This regex will deny to have consecutive non alphanumeric chars:
/^(?=.{4,20}$)[a-z0-9]{0,1}([a-z0-9._-][a-z0-9]+)*[a-z0-9.-_]{0,1}$/i
In this case _-this-is-me-_ will not match, but _this-is-me_ will match.
Have a nice day and thanks to all :)
Don't try to cram it all into one regex. Make your life simpler and use a two step-approach:
return (bool)
preg_match('/^[a-z0-9_.-]{4,20}$/', $s) && preg_match('/\w/', $s);
The mistake in your regex probably was the mixup of * and {n,m}. You can have only one of those quantifiers, not *{4,20} both after another.
Very well, here is the cumbersome solution to what you want:
preg_match('/^(?=.{4})(?!.{21})[\w.-]*[a-z][\w-.]*$/i', $s)
The assertions assert the length, and the second part ensures that at least one letter is present.
Try this one instead:
'/[a-z0-9-_\.]*[a-z0-9]{1,20}[a-z0-9-_\.]*$/i'
Its probably just a matter if finetuning, you could try something like this:
if (preg_match('/^[a-zA-Z0-9]+[_.-]{0,1}[a-zA-Z0-9]+$/m', $subject)) {
# Successful match
} else {
# Match attempt failed
}
Matches:
a_b <- you might not want this.
ysername
Username
1254_2367
fg3123as
Non-Matches:
l__asfg
AHA_ar3f!
sAD_ASF_#"#T_
"#%"&#"E
__-.asd
username
1___
Non-matches you might want to be matches:
1_5_2
this_is_my_name
It is clear to me that you should split this into two checks!
Firstly check that they are using all valid characters. If they're not, then you can tell them that they are using invalid characters.
Then check that they have at least one alpha-numeric character. If they're not, then you can tell them that they must.
Two distinct advantages here: more meaningful feedback to the user and cleaner code to read and maintain.
Here is a simple, single regex solution (verbose):
$re = '/ # Match password having at least one alphanum.
^ # Anchor to start of string.
(?=.*?[A-Za-z0-9]) # At least one alphanum.
[\w\-.]{4,20} # Match from 4 to 20 valid chars.
\z # Anchor to end of string.
/x';
In Action (short form):
function validate($string){
$re = '/^(?=.*?[A-Za-z0-9])[\w\-.]{4,20}\z/';
return (bool) preg_match($re, $string);
}
Try this:
^[a-zA-Z][-\w.]{0,22}([a-zA-Z\d]|(?<![-.])_)$
From related question: Create one RegEx to validate a username
^[A-Za-z][A-Za-z0-9]*(?=.{3,31}$)[a-z0-9]{0,1}([a-z0-9._-][a-z0-9]+)*[a-z0-9.-_]{0,1}$
This will Validate the username
start with an alpha
accept underscore dash and dots
no spaces allowed
Why don't you make it simpler like this?
^[a-zA-Z][a-zA-Z0-9\._-]{3,9}
First letter should be Alphabetical.
then followed by character or symbols you allowed
length of the word should be between 4,10 (as explicitly force the first word)

Check for word in string

What is the best way to search for a word in a string
preg_match("/word/",$string)
stripos("word",$string)
Or is there a better way
One benefit to using regexp for this job is the ability to use \b (Regexp word boundary) in the regexp, and other random derivations. If you are only looking for that sequence of letters in a string stripos is likely to be a little better.
$tests = array("word", "worded", "This also has the word.", "Words are not the same", "Word capitalized should match");
foreach ($tests as $string)
{
echo "Testing \"$string\": Regexp:";
echo preg_match("/\bword\b/i", $string) ? "Matched" : "Failed";
echo " stripos:";
echo stripos("word", $string) >= 0 ? "Matched": "Failed";
echo "\n";
}
Results:
Testing "word": Regexp:Matched stripos:Matched
Testing "worded": Regexp:Failed stripos:Matched
Testing "This also has the word.": Regexp:Matched stripos:Matched
Testing "Words are not the same": Regexp:Failed stripos:Matched
Testing "Word capitalized should match": Regexp:Matched stripos:Matched
Like it says in the Notes for preg_match:
Do not use preg_match() if you only want to check if one string is contained in another string. Use strpos() or strstr() instead as they will be faster.
If you are simply looking for a substring, stripos() or strpos() and friends are much better than using the preg family of functions.
For simple string matching the PHP string functions offer more performance. Regex is more heavyweight and therefore has lower performance.
Having said that, in most cases, the performance difference is small enough to go unnoticed, unless you're looping over an array with hundreds of thousands of elements or more.
Of course, as soon as you start needing "cleverer" matching, regex becomes the only game in town.
There is also substr_count($haystack, $needle) which just returns the number of substring occurences. With the added bonus of not having to worry about 0 equating to false like stripos() if the first occurrence is at position 0. Although that's not a problem if you use strict equality.
http://php.net/manual/en/function.substr-count.php

PHP Regular Expression. Check if String contains ONLY letters

In PHP, how do I check if a String contains only letters? I want to write an if statement that will return false if there is (white space, number, symbol) or anything else other than a-z and A-Z.
My string must contain ONLY letters.
I thought I could do it this way, but I'm doing it wrong:
if( ereg("[a-zA-Z]+", $myString))
return true;
else
return false;
How do I find out if myString contains only letters?
Yeah this works fine. Thanks
if(myString.matches("^[a-zA-Z]+$"))
Never heard of ereg, but I'd guess that it will match on substrings.
In that case, you want to include anchors on either end of your regexp so as to force a match on the whole string:
"^[a-zA-Z]+$"
Also, you could simplify your function to read
return ereg("^[a-zA-Z]+$", $myString);
because the if to return true or false from what's already a boolean is redundant.
Alternatively, you could match on any character that's not a letter, and return the complement of the result:
return !ereg("[^a-zA-Z]", $myString);
Note the ^ at the beginning of the character set, which inverts it. Also note that you no longer need the + after it, as a single "bad" character will cause a match.
Finally... this advice is for Java because you have a Java tag on your question. But the $ in $myString makes it look like you're dealing with, maybe Perl or PHP? Some clarification might help.
Your code looks like PHP. It would return true if the string has a letter in it. To make sure the string has only letters you need to use the start and end anchors:
In Java you can make use of the matches method of the String class:
boolean hasOnlyLetters(String str) {
return str.matches("^[a-zA-Z]+$");
}
In PHP the function ereg is deprecated now. You need to use the preg_match as replacement. The PHP equivalent of the above function is:
function hasOnlyLetters($str) {
return preg_match('/^[a-z]+$/i',$str);
}
I'm going to be different and use Character.isLetter definition of what is a letter.
if (myString.matches("\\p{javaLetter}*"))
Note that this matches more than just [A-Za-z]*.
A character is considered to be a letter if its general category type, provided by Character.getType(ch), is any of the following: UPPERCASE_LETTER, LOWERCASE_LETTER, TITLECASE_LETTER, MODIFIER_LETTER, OTHER_LETTER
Not all letters have case. Many characters are letters but are neither uppercase nor lowercase nor titlecase.
The \p{javaXXX} character classes is defined in Pattern API.
Alternatively, try checking if it contains anything other than letters: [^A-Za-z]
The easiest way to do a "is ALL characters of a given type" is to check if ANY character is NOT of the type.
So if \W denotes a non-character, then just check for one of those.

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