I want to get an array of all the words with capital letters that are included in the string. But only if the line begins with "set".
For example:
- string "setUserId", result array("User", "Id")
- string "getUserId", result false
Without limitation about "set" RegEx look like /([A-Z][a-z]+)/
$str ='setUserId';
$rep_str = preg_replace('/^set/','',$str);
if($str != $rep_str) {
$array = preg_split('/(?<=[a-z])(?=[A-Z])/',$rep_str);
var_dump($array);
}
See it
Also your regex will also work.:
$str = 'setUserId';
if(preg_match('/^set/',$str) && preg_match_all('/([A-Z][a-z]*)/',$str,$match)) {
var_dump($match[1]);
}
See it
Related
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
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 need to determine whether a string begins with a number - I've tried the following to no avail:
if (matches("^[0-9].*)", upper(text))) str = "Title"""
I'm new to DXL and Regex - what am I doing wrong?
You need the caret character to indicate a match only at the start of a string. I added the plus character to match all the numbers, although you might not need it for your situation. If you're only looking for numbers at the start, and don't care if there is anything following, you don't need anymore.
string str1 = "123abc"
string str2 = "abc123"
string strgx = "^[0-9]+"
Regexp rgx = regexp2(strgx)
if(rgx(str1)) { print str1[match 0] "\n" } else { print "no match\n" }
if(rgx(str2)) { print str2[match 0] "\n" } else { print "no match\n" }
The code block above will print:
123
no match
#mrhobo is correct, you want something like this:
Regexp numReg = "^[0-9]"
if(numReg text) str = "Title"
You don't need upper since you are just looking for numbers. Also matches is more for finding the part of the string that matches the expression. If you just want to check that the string as a whole matches the expression then the code above would be more efficient.
Good luck!
At least from example I found this example should work:
Regexp plural = regexp "^([0-9].*)$"
if plural "15systems" then print "yes"
Resource:
http://www.scenarioplus.org.uk/papers/dxl_regexp/dxl_regexp.htm
FOR EXAMPLE: Given a string... "1,2,3,4"
I need to be able to remove a given number and the comma after/before depending on if the match is at the end of the string or not.
remove(2) = "1,3,4"
remove(4) = "1,2,3"
Also, I'm using javascript.
As jtdubs shows, an easy way is is to use a split function to obtain an array of elements without the commas, remove the required element from the array, and then rebuild the string with a join function.
For javascript something like this might work:
function remove(array,to_remove)
{
var elements=array.split(",");
var remove_index=elements.indexOf(to_remove);
elements.splice(remove_index,1);
var result=elements.join(",");
return result;
}
var string="1,2,3,4,5";
var newstring = remove(string,"4"); // newstring will contain "1,2,3,5"
document.write(newstring+"<br>");
newstring = remove(string,"5");
document.write(newstring+"<br>"); // will contain "1,2,3,4"
You also need to consider the behavior you want if you have repeats, say the string is "1,2,2,4" and I say "remove(2)" should it remove both instances or just the first? this function will remove only the first instance.
Just use multiple substitutions.
s/^$removed,//;
s/,$removed$//;
s/,$removed,/,/;
This will be easier than trying to invent a single replacement that handles all those cases.
string input = "1,2,3,4";
List<string> parts = new List<string>(input.Split(new char[] { ',' }));
parts.RemoveAt(2);
string output = String.Join(",", parts);
Instead of using regex, I would do something like:
- split on comma
- delete the right element
- join with comma
Here is a perl script that does the job:
#!/usr/bin/perl
use 5.10.1;
use strict;
use warnings;
my $toremove = 5;
my $string = "1,2,3,4,5";
my #tmp = split/,/, $string;
#tmp = grep{ $_ != $toremove }#tmp;
$string =join',', #tmp;
say $string;
Output:
1,2,3,4
Javascript has improved since this question was posted.
I use the following regex to remove items from a csv string
let searchStr = "359";
let regex = new RegExp("^" + searchStr + ",?|," + searchStr);
csvStr = csvStr.replace(regex, "");
If the child_id is the start, middle or end, or only item it is replaced.
If the searchStr is at the start of the csvStr it and any trailing comma is replaced. Else if the searchStr is anywhere else in the csvStr it must be preceded with a comma so the searchStr and its preceding comma are replaced by an empty string.