I have a regex ".*00$" which restricts that the input must be ending with 00. How can I improve this to add a max length of 20 also. So:
100 => valid,
00 = > invalid,
12345678900 => valid,
111111111111111111100 = > 21 digits - invalid.
Based on your title and short description this should probably work for you:
^[0-9]{1,18}00$
This will allow 3-20 digits in input that ends with 00
This can be expressed almost literally as
/^[0-9]{1,18}00$/
20 digits, the last two being zeros.
You can use {} to specify a range of characters, so
".{1,18}00"
would allow any 1-18 characters followed by 00. If you want to restrict it further, you could use
"[1-9][0-9]{0,17}00"
so that you ensure the first number is not 0, followed by 0-17 numbers and finally 00.
Try this :
^[1-9][0-9]{0,17}00$
The first [] will ensure it doesn't start by 0.
Related
I'm trying to get the regular expression to work (using jQuery) for a specific pattern I need.
I need following pattern:
First two character
s of the string need to be numbers (0-9) but maximum number is 53. for numbers below 10 a leading 0 is required
Character on position 3 needs to be a .
the next 4 characters need to be a number between 0-9, minimum number should be 2010, maximum 2050
so, Strings like 01.2020, 21.2020, or 45.2020 have to match but 54.2020 or 04.2051 must not.
I tried to write the regex without the min and max requirement first and I'm testing the string using regex101.com but I'm unable to get it to work.
acording to the definition /^[0-9]{2}\.\d[0-9]{4}$/ should allow me to insert the strings in the format NN.NNNN.
thankful for any input.
2 numbers from 00 to 53 can be matched using this : (?:[0-4][0-9]|5[0-3]) (00 -> 49 or 50 -> 53)
Character on position 3 needs to be a . : you've already got the \.
a number between 2010 and 2050 -> 20(?:[1-4][0-9]|50) (20 followed by either 10 -> 49 or 50)
This gives :
(?:[0-4][0-9]|5[0-3])\.20(?:[1-4][0-9]|50)
I have a source of data that was converted from an oracle database and loaded into a hadoop storage point. One of the columns was a BLOB and therefore had lots of control characters and unreadable/undetectable ascii characters outside of the available codeset. I am using Impala to write regex replace function to parse some of the unicode characters that the regex library cannot understand. I would like to remove the offending 2 character hex codes BEFORE I use the unhex query function so that I can do the rest of the regex parsing with a "clean" string.
Here's the code I've used so far, which doesn't quite work:
'[2-7]{1}([A-Fa-f]|[0-9]{1})'
I've determined that I only need to capture \u0020-\u007f - or represented in the two bit hex - 20-7f
If my string looks like this:
010A000000153020405C00000000143020405CBC000000F53320405C4C010000E12F204058540100002D01
I would like to be able to capture 2 characters at a time (e.g. 01,0A,00) evaluate whether or not that fits the acceptable range of 2 byte hex I mentioned above and return only what is acceptable.
The correct output should be:
30 20 40 5C 30 20 40 5C 33 20 40 5C 4C 2F 20 40 58 and 54
However, my expression finds the first acceptable number in my first range (5) and starts the capture from there which returns the position or indexing wrong for the rest of the string... and this is the return from my expression -
010A0000001**53**0**20****40****5C**000000001**43**0**20****40****5C**BC000000F**53****32**0**40****5C****4C**010000E1**2F****20****40****58****54**010000**2D**01
I just don't know how to evaluate only two characters at a time in a mixed-length string. And, if they don't fit the expression, iterate to the next two characters. But only in two character increments.
My example: https://regex101.com/r/BZL7t0/1
I have added a Positieve Lookbehind to it. Which starts at the beginning of the string and then matches 2 characters at the time. This ensures that the group you're matching always has groups of 2 characters before it.
Positieve Lookbehind:
(?<=^(..)*)
Updated regex:
(?<=^(..)*)([2-7]{1}[A-Fa-f0-9]{1})
Preview:
Regex101
I'm trying to validate french mobile numbers:
I have already removed all non numeric character and the eventual 00 at beginning, and rules are:
start with 06 or 07 or 09
is 10 digit long:
thus :
/^0(6|7|9)\d{8}$/
but (seems) that if countrycode (33) is present, the leading zero has to be avoided, but at this point I cannot create the right regex, since with number:
33614444444
/^(33|0)?(6|7|9)\d{8}$/
it works, but works also with
614444444
while it should not
can suggest solution?
you can do it using the regex
^(33|0)(6|7|9)\d{8}$
see the regex101 demo
Why don't you simply use /^(33|0)(6|7|9)\d{8}$/ ?
I do not think you need the quantifier ?.
When you add ? after (33|0). It implies either none of them is present or one of 33 or 0 is present. It would match all the following -
614444444 // none present
0614444444 // 0 present
33614444444 // 33 present
I have been reading the regex questions on this site but my issue seems to be a bit different. I need to match a 2 digit number, such as 23 through 75. I am doing this on an HP-UX Unix system. I found examples of 3 - 44 but or any digit number, nothing that is fixed in length, which is a bit surprising, but perhaps I am not understand the variable length example answer.
Since you're not indicating whether this is in addition to any other characters (or in the middle of a larger string), I've included the logic here to indicate what you would need to match the number portion of a string. This should get you there. We're creating a range for the second numbers we're looking for only allowing those characters. Then we're comparing it to the other ranges as an or:
(2[3456789]|[3456][0-9]|7[012345])
As oded noted you can do this as well since sub ranges are also accepted (depends on the implementation of REGEX in the application you're using):
(2[3-9]|[3-6][0-9]|7[0-5])
Based on the title you would change the last 5 to a 9 to go from 75-79:
(2[3-9]|[3-6][0-9]|7[0-9])
If you are trying to match these numbers specifically as a string (from start to end) then you would use the modifiers ^ and $ to indicate the beginning and end of the string.
There is an excellent technical reference of Regex ranges here:
http://www.regular-expressions.info/numericranges.html
If you're using something like grep and trying to match lines that contain the number with other content then you might do something like this for ranges thru 79:
grep "[^0-9]?(2[3-9]|[3-6][0-9]|7[0-9])[^0-9]?" folder
This tool is exactly what you need: Regex_For_Range
From 29 to 79: \b(2[3-9]|[3-7][0-9])\b
From 29 to 75: \b(29|[3-6][0-9]|7[0-5])\b
And just for fun, from 192 to 1742: \b(19[2-9]|[2-9][0-9]{2}|1[0-6][0-9]{2}|17[0-3][0-9]|174[0-2])\b :)
If I want 2 digit range 0-63
/^[0-9]|[0-5][0-9]|6[0-3]$/
[0-9] will allow single digit from 0 to 9
[0-5][0-9] will allow from 00 to 59
6[0-3] will allow from 60 till 63
This way you can take Regular Expression for any Two Digit Range
You have two classes of numbers you want to match:
the digit 2, followed by one of the digits between 3 and 9
one of the digits between 3 and 7, followed by any digit
Edit: Well, that's the title's range (23-79). Within your question (23-75), you have three:
the digit 2, followed by one of the digits between 3 and 9
one of the digits between 3 and 6, followed by any digit
the digit 7, followed by one of the digits between 0 and 5
Just to add to this, here is a solution for generating the string from the accepted answer in javascript. You can click "Run Code Snippet" to enter your own bounds and get your own string.
function regexRangeString(lower,upper){
let current=lower;
let nextRange=function(){
let currentString=String(current);
let len=currentString.length;
let string="";
let newUpper;
for(let digit=0;digit<len;digit++){
let index=len-digit-1;
let lower=Number(currentString[index]);
let thisString="";
for(let u=9;u>=lower;u--){
let us=currentString.substring(0,index)+u+currentString.substring(index+1,len);
if(Number(us)<=upper){
if(lower==u){
thisString=lower;
}
else{
thisString=`[${lower}-${u}]`;
}
currentString=currentString.substring(0,index)+u+currentString.substring(index+1,len);
break;
}
}
if(thisString!="[0-9]"){
string=currentString.substring(0,index)+thisString+string;
break;
}
else{
string=thisString+string
}
}
current=Number(currentString)+1;
return string
}
let string=""
while(current<upper){
string+="|"+nextRange(current);
}
string="("+string.slice(1)+")";
return string
}
let lower=prompt("Enter Lower Bound")
let upper=prompt("Enter Upper Bound")
alert(regexRangeString(lower,upper))
For example:
regexRangeString(72,189)
generates the following output string:
(7[2-9]|[8-9][0-9]|1[0-8][0-9])
This should do it:
/^([2][3-9]|[3-6][0-9]|[7][0-5])$/
^ and $ will make it strict that it will match only 2 numbers, so in case that you have i.e 234 it won't work.
I have a simple question about finding a regular expression for a given language.
I am given the language L where:
L = {w ∈ {0, 1}* : w has exactly one pair of consecutive zeros}
My first attempt at this was to try L( (0 + 1)* 00 (0 + 1)*), but I noticed the problem with that would be with where I have (0 + 1)* because if 0 is chosen, it can be zero more more of them, thus leading to more than one pair of consecutive zeros.
I also know, that the possible cases I have are, two zeros in the front, in the middle, and at the end. I just am not quite sure how to create a regular expression for that.
Any help is much appreciated.
Try this:
1* (011*)* 00 (11*0)* 1*
An explanation:
1*: any amount of leading 1’s
(011*)*: if there is a 0 before the 00, it must not be followed by another 0, thus only one or more 1’s are allowed; this pattern may be repeated any number of times
00: the two 0’s
(11*0)*: if there is a 0 after the 00, it must not preceded by another 0, thus only one or more 1’s; this pattern may be repeated any number of times
1*: any amount of trailing 1’s
The best possible answer for this problem is (1 + 01)* 00 (1 + 10)*
i believe it would be like this
((1*)(01)*))* 00 ((11*)0)*1*
The sequence:
Anything but 00, ending with 1
00
Anything but 00, starting with 1
My answer would be : (1 + 01) 00 (1 + 10)**
Explanation:
The consecutive zeros shouldn't be preceded or followed by another zero.
Hence 00 should be preceded by a 1 which can be either a 1 or 01.
It can be followed by a 1 or 10.