What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
Related
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
I have a string like '[1]-[2]-[3],[4]-[5],[6,7,8],[9]' or '[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]', I'd like the Pattern to get the list result, but don't know how to figure out the pattern. Basically the comma is the split, but [6,7,8] itself contains the comma as well.
the string: [1]-[2]-[3],[4]-[5],[6,7,8],[9]
the result:
[1]-[2]-[3]
[4]-[5]
[6,7,8]
[9]
or
the string: [Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]
the result:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
,(?=\[)
This pattern splits on any comma that is followed by a bracket, but keeps the bracket within the result text.
The (?=*stuff*) is known as a "lookahead assertion". It acts as a condition for the match but is not itself part of the match.
In C# code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
foreach(String s in Regex.Split(inputstring, #",(?=\[)"))
System.Console.Out.WriteLine(s);
In Java code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile(",(?=\\[)"));
for(String s : p.split(inputstring))
System.out.println(s);
Either produces:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
Although I believe the best approach here is to use split (as presented by #j__m's answer), here's an approach that uses matching rather than splitting.
Regex:
(\[.*?\](?!-))
Example usage:
String input = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile("(\\[.*?\\](?!-))");
Matcher m = p.matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Resulting output:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
An answer that doesn't use regular expressions (if that's worth something in ease of understanding what's going on) is:
substitute "]#[" for "],["
split on "#"
I have a string something like "[aaa][vad][adf]", i would like to use regex to capture the data in[], and chars in [] can be char and number and no length limit. I am regex noob, can anyone help me on this?
Thanks.
You can try something like this:
var data = "[asd][dfhg][asfsa243]";
var re = new Regex(#"\[([^\]]*)\]");
var matches = re.Matches(data);
for (int i = 0; i < matches.Count; i++ )
{
var m = matches[i];
Console.WriteLine(m.Groups[1]);
}
Console.ReadLine();
This outputs:
asd
dfhg
asfsa243
The regular expression \[([^\]]*)\] matches zero or more occurrences of a character that isn't the ] character and which is between a pair of square brackets ([ and ])
This regex might work for you:
\[(\w+)\]
That if you by char refers to word character
Here is my suggestion:
\[(\w+)\]
The charachter will be provided in the first group of the match, like this:
var regex = new Regex(#"\[([\w\d]*)\]");
MatchCollection matchCollection = regex.Matches(stringToTest);
foreach (Match match in matchCollection)
{
Debug.Print(match.Groups[0].Value);
}
Here are some good resources for building regex's
http://www.regexlib.com
http://regexpal.com/
Here is the regex for Alphanumeric:
^[a-zA-Z0-9]+$
Here's how to ensure there are 3:
/^([a-zA-Z0-9]){3}$/
You could try -
[^\[\]\W]+
var matches = System.Text.RegularExpressions.Regex.Matches("[aaa][vad][adf]",#"[^\[\]\W]+").Cast<Match>().Select(m => m.Value);
That should ignore any '[', ']' and non-word characters and would return 'aaa', 'vad' and 'adf' from your example string.