I am very new to Coldfusion and not sure what the format should be to use this function correctly.
I want to convert 0000411111 to 0411111 get rid of the first three zeros
<cfset origValue = "#query.column#">
<cfset newValue = ReReplace(origValue, "0+", "", "all")>
<cfoutput>#newValue#</cfoutput>
This removes all zeros is there anyway to just keep one zero. Just curious.
Thanks in advance for your assistances.
If the string will always be 7 characters you can use
<cfset newValue = numberFormat(000411111,'0000000')>
If you don't know the length and always want to remove leading 0's and leave one at the begining you can do
<cfset newValue = '0' & int(000411111)>
If you always want to remove the first three characters, you can use the right() function:
<cfset newValue = right(query.column, len(query.column)-3>
This will return all the characters from the right side of the string without the leading three characters.
You could do it 2 different ways:
<Cfset newvalue=right(origvalue,len(origvalue)-3>
This method returns the string without the left 3 most characters
or
<Cfset newvalue=mid(origvalue,4,len(origvalue)-3>
this method starts at position 4 and grabs the rest of the string.
I think the numberFormat() answer is the best one, but other people have been suggesting using mid() and right() which I think - whilst those approaches work - are more cumbersome than you need to make it. If you simply wish to remove the first three chars of the string, there's a removeChars() function. It's unclear from your question though whether this actually achieves what you want: if it's only when the number is left-padded with too many zeros you want to do this, then the numberFormat() approach is best. If it's any three characters, then this approach is better.
newValue = removeChars(origValue, 1, 3);
The regex string you are looking for really is for matching 2 or more 0's at the start of the string, and replacing them with simply a single 0.
This gives the regex ^0+0
^ matches the start of the string, 0+ matches 1 or more 0's, 0 matches the second zero. This will mean that if there is only 1 leading zero then it won't need to do anything. Finally you only need to do this once, as you are only replacing the ones at the start of the string. This brings to the CF code
newValue = ReReplace(origValue, "^0+0", "0", "one")
This should replace multiple leading zeros with a single one, while not adding zeros where there weren't any to begin with.
As a final note, a good place to play around with regex is http://gskinner.com/RegExr/
Related
I have a string, and I want to extract, using regular expressions, groups of characters that are between the character : and the other character /.
typically, here is a string example I'm getting:
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
and so, I want to retrieved, 45.72643,4.91203 and also hereanotherdata
As they are both between characters : and /.
I tried with this syntax in a easier string where there is only 1 time the pattern,
[tt]=regexp(str,':(\w.*)/','match')
tt = ':45.72643,4.91203/'
but it works only if the pattern happens once. If I use it in string containing multiples times the pattern, I get all the string between the first : and the last /.
How can I mention that the pattern will occur multiple time, and how can I retrieve it?
Use lookaround and a lazy quantifier:
regexp(str, '(?<=:).+?(?=/)', 'match')
Example (Matlab R2016b):
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = regexp(str, '(?<=:).+?(?=/)', 'match')
result =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
In most languages this is hard to do with a single regexp. Ultimately you'll only ever get back the one string, and you want to get back multiple strings.
I've never used Matlab, so it may be possible in that language, but based on other languages, this is how I'd approach it...
I can't give you the exact code, but a search indicates that in Matlab there is a function called strsplit, example...
C = strsplit(data,':')
That should will break your original string up into an array of strings, using the ":" as the break point. You can then ignore the first array index (as it contains text before a ":"), loop the rest of the array and regexp to extract everything that comes before a "/".
So for instance...
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
Breaks down into an array with parts...
1 - 'abcd'
2 - '45.72643,4.91203/Rou'
3 - 'hereanotherdata/defgh'
Then Ignore 1, and extract everything before the "/" in 2 and 3.
As John Mawer and Adriaan mentioned, strsplit is a good place to start with. You can use it for both ':' and '/', but then you will not be able to determine where each of them started. If you do it with strsplit twice, you can know where the ':' starts :
A='abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
B=cellfun(#(x) strsplit(x,'/'),strsplit(A,':'),'uniformoutput',0);
Now B has cells that start with ':', and has two cells in each cell that contain '/' also. You can extract it with checking where B has more than one cell, and take the first of each of them:
C=cellfun(#(x) x{1},B(cellfun('length',B)>1),'uniformoutput',0)
C =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
Starting in 16b you can use extractBetween:
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = extractBetween(str,':','/')
result =
2×1 cell array
{'45.72643,4.91203'}
{'hereanotherdata' }
If all your text elements have the same number of delimiters this can be vectorized too.
I want to remove zeros in a String.
For example,
String A = AY000120
then the output should be
AY120
so basically any thing between AY and next number which is greater than 0 should be removed. Also, if any zero occurs after a number which is greater than 1 then that zero will not be deleted.
A reg ex will be very useful.
replace ^(AY)0*([1-9].*)
by \1\2
Or, if you knew your input is in fixed format AY+(zero or more 0)+(other Numbers), you can just:
replace ^AY0* by AY
Looks like you are using Java but here is a solution that I wrote in JavaScript. I think regex part should work for you
let str = "AY000120"
let result = str.replace(/(0*)(?=[1-9])/g, ''); //AY120
post any questions if you still have.
(0*) - looks for any number of 0s (greedy)
(?=[1-9]) - positive look ahead to make sure any number other than 0 exists
I'm comparing 2 URL query strings to see if they're equal; however, I want to ignore a specific query parameter (always with a numeric value) if it exists. So, these 2 query strings should be equal:
firstName=bobby&lastName=tables¶mToIgnore=2
firstName=bobby&lastName=tables¶mToIgnore=5
So, I tried to use a regex replace using the REReplaceNoCase function:
REReplaceNoCase(myQueryString, "¶mToIgnore=[0-9]*", "")
This works fine for the above example. I apply the replace to both strings and then compare. The problem is that I can't be sure that the param will be the last one in the string... the following 2 query strings should also be equal:
firstName=bobby&lastName=tables¶mToIgnore=2
paramToIgnore=5&firstName=bobby&lastName=tables
So, I changed the regex to make the preceding ampersand optional... "&?paramToIgnore=[0-9]*". But - these strings will still not be equal as I'll be left with an extra ampersand in one of the strings but not the other:
firstName=bobby&lastName=tables
&firstName=bobby&lastName=tables
Similarly, I can't just remove preceding and following ampersands ("&?paramToIgnore=[0-9]*&?") as if the query param is in the middle of the string I'll strip one ampersand too many in one string and not the other - e.g.
firstName=bobby&lastName=tables¶mToIgnore=2
firstName=bobby¶mToIgnore=5&lastName=tables
will become
firstName=bobby&lastName=tables
firstName=bobbylastName=tables
I can't seem to get my head around the logic of this... Can anyone help me out with a solution?
If you can't be sure of the order the parameters appear i would recommend, that you don't compare them by the string itsself.
I recommend splitting the string up like this:
String stringA = "firstName=bobby&lastName=tables¶mToIgnore=2";
String stringB = "firstName=bobby&lastName=tables¶mToIgnore=5";
String[] partsA = stringA.split("&");
String[] partsB = stringB.split("&");
Then go through arrays and make the paramToIgnore somehow euqal:
for(int i = 0; i < partsA.length; i++)
{
if(partsA[i].startsWith("paramToIgnore"){
partsA[i] = "IgnoreMePlease";
}
}
for(int j = 0; j < partsB.length; j++)
{
if(partsB[i].startsWith("paramToIgnore"){
partsB[i] = "IgnoreMePlease";
}
}
Then you can sort and compare the arrays to see if they are equal:
Arrays.sort(partsA);
Arrays.sort(partsB);
boolean b = Arrays.equals(partsA, partsB);
I'm pretty sure it's possible to make this more compact and give it a better performance. But with comparing strings like you do, you somehow alsways have to care about the order of your parameters.
You can use the QueryStringDeleteVar UDF on cflib to remove the query string variables you want to ignore from both strings, then compare them.
Make it in two steps:
first remove your param, as you described in example
then remove ampersand which is left at the begining or the end of query with separate regex, or any double/triple/... ampersands in the middle of the query
How about having an 'or' in the RegEx to match an ampersand at the start or the end?
¶mToIgnore=[0-9]*|paramToIgnore=[0-9]*&
Seems to do the job when testing in regexpal.com
try changing it to:
REReplaceNoCase(myQueryString, "&?paramToIgnore=[0-9]+", "")
plus instead of star should capture 1 or more of the preceding matched characters. It won't match anything but 0-9 so if there is another parameter after that it'll stop when it can't match any more digits.
Alternatively, you could use:
REReplaceNoCase(myQueryString, "&?paramToIgnore=[^&]", "")
This will match anything but an ampersand. It will cover the case if the parameter exists but there is no value; which is probably something you'd want to account for.
Say I have a CString object strMain="AAAABBCCCCCCDDBBCCCCCCDDDAA";
I also have two smaller strings, say strSmall1="BB";
strSmall2="DD";
Now, I want to replace all occurence of strings which occur between strSmall1("BB") and strSmall2("DD") in strMain, with say "KKKKKKK"
Is there a way to do it without Regex. I cannot use regex as adding another file to the project is prohibited.
Is there a way in VC++/MFC to do it? Or any easy algorithm you can point me to?
int length = strMain.GetLength();
int begin = strMain.Find(strSmall1, 0) + strSmall1.GetLength();
int end = strMain.Find(strSmall2, 0);
CStringT left = strMain.Left(begin);
CStringT right = strMain.Right(length - end);
strMain = left + "KKKKKKK" + right
The easiest way is probably to handle the replacement recursively. Search for the starting delimiter and the ending delimiter. If you find them, put together a new string consisting of the string up to the starting delimiter, followed by the replacement string, followed by the return from recursively doing the replacement in the remainder of the string following the ending delimiter.
That, of course, assumes you want to replace all the occurrences in the main string -- if you only want to replace the first one, John Weldon's solution (for one example) will work quite nicely.
psudocode:
loop over string
if curlocation matches string strsmall1 save index break
loop over remaining string
replace till curlocation matches string strsmall2
Extra credit:
What will the next assignment be?
My answer:
Speed it up by jumping the length of strsmall1 and strsmall2 in loop iterations
What is the regular expression to validate a comma delimited list like this one:
12365, 45236, 458, 1, 99996332, ......
I suggest you to do in the following way:
(\d+)(,\s*\d+)*
which would work for a list containing 1 or more elements.
This regex extracts an element from a comma separated list, regardless of contents:
(.+?)(?:,|$)
If you just replace the comma with something else, it should work for any delimiter.
It depends a bit on your exact requirements. I'm assuming: all numbers, any length, numbers cannot have leading zeros nor contain commas or decimal points. individual numbers always separated by a comma then a space, and the last number does NOT have a comma and space after it. Any of these being wrong would simplify the solution.
([1-9][0-9]*,[ ])*[1-9][0-9]*
Here's how I built that mentally:
[0-9] any digit.
[1-9][0-9]* leading non-zero digit followed by any number of digits
[1-9][0-9]*, as above, followed by a comma
[1-9][0-9]*[ ] as above, followed by a space
([1-9][0-9]*[ ])* as above, repeated 0 or more times
([1-9][0-9]*[ ])*[1-9][0-9]* as above, with a final number that doesn't have a comma.
Match duplicate comma-delimited items:
(?<=,|^)([^,]*)(,\1)+(?=,|$)
Reference.
This regex can be used to split the values of a comma delimitted list. List elements may be quoted, unquoted or empty. Commas inside a pair of quotation marks are not matched.
,(?!(?<=(?:^|,)\s*"(?:[^"]|""|\\")*,)(?:[^"]|""|\\")*"\s*(?:,|$))
Reference.
/^\d+(?:, ?\d+)*$/
i used this for a list of items that had to be alphanumeric without underscores at the front of each item.
^(([0-9a-zA-Z][0-9a-zA-Z_]*)([,][0-9a-zA-Z][0-9a-zA-Z_]*)*)$
You might want to specify language just to be safe, but
(\d+, ?)+(\d+)?
ought to work
I had a slightly different requirement, to parse an encoded dictionary/hashtable with escaped commas, like this:
"1=This is something, 2=This is something,,with an escaped comma, 3=This is something else"
I think this is an elegant solution, with a trick that avoids a lot of regex complexity:
if (string.IsNullOrEmpty(encodedValues))
{
return null;
}
else
{
var retVal = new Dictionary<int, string>();
var reFields = new Regex(#"([0-9]+)\=(([A-Za-z0-9\s]|(,,))+),");
foreach (Match match in reFields.Matches(encodedValues + ","))
{
var id = match.Groups[1].Value;
var value = match.Groups[2].Value;
retVal[int.Parse(id)] = value.Replace(",,", ",");
}
return retVal;
}
I think it can be adapted to the original question with an expression like #"([0-9]+),\s?" and parse on Groups[0].
I hope it's helpful to somebody and thanks for the tips on getting it close to there, especially Asaph!
In JavaScript, use split to help out, and catch any negative digits as well:
'-1,2,-3'.match(/(-?\d+)(,\s*-?\d+)*/)[0].split(',');
// ["-1", "2", "-3"]
// may need trimming if digits are space-separated
The following will match any comma delimited word/digit/space combination
(((.)*,)*)(.)*
Why don't you work with groups:
^(\d+(, )?)+$
If you had a more complicated regex, i.e: for valid urls rather than just numbers. You could do the following where you loop through each element and test each of them individually against your regex:
const validRelativeUrlRegex = /^(^$|(?!.*(\W\W))\/[a-zA-Z0-9\/-]+[^\W_]$)/;
const relativeUrls = "/url1,/url-2,url3";
const startsWithComma = relativeUrls.startsWith(",");
const endsWithComma = relativeUrls.endsWith(",");
const areAllURLsValid = relativeUrls
.split(",")
.every(url => validRelativeUrlRegex.test(url));
const isValid = areAllURLsValid && !endsWithComma && !startsWithComma