Hi I would like to build a function which will replace all the special character ie: * & ! # # $ % with "//(with whatever the match will be)
so I string like "1*234#" will become 1//*/234//#"
is there a replace function in swift to do that?
String.replacingOccurrences which can be used like so:
let replacements = ["!" : "-exclamation-", "." : "-period-"]
var stringToModify = "hello! This is a string."
replacements.keys.forEach { stringToModify = stringToModify.replacingOccurrences(of: $0, with: replacements[$0]!)}
print(stringToModify)
output: hello -exclamation- This is a string -period-
There is also an overload with more options, incase you want to do stuff like case insensitive compare. https://developer.apple.com/reference/foundation/nsstring/1416484-replacingoccurrences
Related
I'm trying to implement the escape character functionality in a macro generator I'm writing in Dart. For example, I would like the program to grab all the occurrences of '¶m' in my string and replace it with 'John', unless the '&' character is preceded with the escape character '\'. Example: "My name is ¶m and my parameter is called \¶m." -> "My name is John and my parameter is called ¶m". What would be the regular expression to catch all the substrings that contain the '&', then my parameter's name, and without the preceding '\'?
It's possible to match that, even avoiding escapes of backslashes, as:
var re = RegExp(r"(?<!(?:^|[^\\])(?:\\{2})*\\)&\w+");
This uses negative lookbehind to find a & followed by word-characters, and not preceded by an odd number of backslashes.
More likely, you want to also recognize double-backslashes and convert them to single-backslashes. That's actually easier if you try to find all matches, because then you know all preceding double-backslashes are part of an earlier match:
var re = RegExp(r"\\\\|(?<!\\)&\w+");
This, when used as re.allMatches will find all occurrences of \\ and &word where the latter is not preceded by an odd number of backslashes.
var _re = RegExp(r"\\\\|(?<!\\)&(\w+)");
String template(String input, Map<String, String> values) {
return input.replaceAllMapped(_re, (m) {
var match = m[0]!;
if (match == r"\\") return r"\";
var replacement = values[m[1]!];
if (replacement != null) return replacement;
// do nothing for undefined words.
return match;
});
}
(You might also want to allow something like &{foo} if parameters can occur next to other characters, like &{amount)USD).
To keep the character before ¶m when it matches a non-backslash character you need to use so called capturing groups. These are are subexpressions of a regular expression inside parentheses. To use capturing groups in Dard you need to use the method replaceAllMapped. We also have the case when the template starts with ¶m and in this case we match at the beginning of the string instead.
Try this:
void main() {
final template = 'My name is ¶m and my parameter is called \\¶m.';
final populatedTemplate = template.replaceAllMapped(RegExp(r'(^|[^\\])¶m\b'), (match) {
return '${match.group(1)}John';
});
final result = populatedTemplate.replaceAll(RegExp(r'\\¶m\b'), 'John');
print(result);
}
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
In my simulink i have a propagate signal which look like this:
<foo_boo>
and at source
foo_boo
i would like to build a regular expression the return from
<foo_boo>
simply foo_boo and from foo_boo i would like to get foo_boo.
In other words, i would like a regular expression that remove '>' and '<' from my string and the string can include [a-zA-Z_0-9] chars.
Pretty easy. Use regexprep to search for symbols that contain < or > in your input string and replace them with nothing. In other words:
out = regexprep(in, '<|>', '');
in would be the string you want to operate on (i.e. <foo_boo>) and out contains the processed string.
Example:
in = '<foo_boo>';
out = regexprep(in, '<|>', '')
out =
foo_boo
Since I think logical indexing is the answer to most things MATLAB (the other being bsxfun), I throw this in:
str = '<foo_boo>';
str( (str=='<') | (str=='>') ) = [];
seems there's no need to use regex:
str = '<foo_boo>'
str([strfind(str,'<'),strfind(str,'>')]) = []
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
I have a list of several phrases in the following format
thisIsAnExampleSentance
hereIsAnotherExampleWithMoreWordsInIt
and I'm trying to end up with
This Is An Example Sentance
Here Is Another Example With More Words In It
Each phrase has the white space condensed and the first letter is forced to lowercase.
Can I use regex to add a space before each A-Z and have the first letter of the phrase be capitalized?
I thought of doing something like
([a-z]+)([A-Z])([a-z]+)([A-Z])([a-z]+) // etc
$1 $2$3 $4$5 // etc
but on 50 records of varying length, my idea is a poor solution. Is there a way to regex in a way that will be more dynamic? Thanks
A Java fragment I use looks like this (now revised):
result = source.replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
result = result.substring(0, 1).toUpperCase() + result.substring(1);
This, by the way, converts the string givenProductUPCSymbol into Given Product UPC Symbol - make sure this is fine with the way you use this type of thing
Finally, a single line version could be:
result = source.substring(0, 1).toUpperCase() + source(1).replaceAll("(?<=^|[a-z])([A-Z])|([A-Z])(?=[a-z])", " $1$2");
Also, in an Example similar to one given in the question comments, the string hiMyNameIsBobAndIWantAPuppy will be changed to Hi My Name Is Bob And I Want A Puppy
For the space problem it's easy if your language supports zero-width-look-behind
var result = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "(?<=[a-z])([A-Z])", " $1");
or even if it doesn't support them
var result2 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "([a-z])([A-Z])", "$1 $2");
I'm using C#, but the regexes should be usable in any language that support the replace using the $1...$n .
But for the lower-to-upper case you can't do it directly in Regex. You can get the first character through a regex like: ^[a-z] but you can't convet it.
For example in C# you could do
var result4 = Regex.Replace(result, "^([a-z])", m =>
{
return m.ToString().ToUpperInvariant();
});
using a match evaluator to change the input string.
You could then even fuse the two together
var result4 = Regex.Replace(#"thisIsAnExampleSentanceHereIsAnotherExampleWithMoreWordsInIt", "^([a-z])|([a-z])([A-Z])", m =>
{
if (m.Groups[1].Success)
{
return m.ToString().ToUpperInvariant();
}
else
{
return m.Groups[2].ToString() + " " + m.Groups[3].ToString();
}
});
A Perl example with unicode character support:
s/\p{Lu}/ $&/g;
s/^./\U$&/;