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Javascript match a string which start with a specific char from the set of chars and end with same char using Regular Expression

What is the best regular expression to check if a string is a valid URL?Regular expression to match a line that doesn't contain a wordHow do you access the matched groups in a JavaScript regular expression?Regular expression to stop at first matchHow to match “anything up until this sequence of characters” in a regular expression?javascript to match string start with “ and end with ”; and anything in between?JavaScript Regex finding all substrings matching the starting and ending patternmatching words starting and ending with same letterregex matching for string with character limit, specific start characters, and terminationMatch the same start and end character of a string with Regex

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

2

I need to achieve a particular task in javascript, in which my goal is to match the string which starts with a char from a specific set of characters like vowel and ends with the same character where the length of the string is greater than three.

so far I have done the following code that starts and ends with the same character but doesn't know how to specify it that the first char is from the specific set of character:

function regexVar() 

 var re = /(.).*1/
 return re;


console.log("obcdo".test(s));

let suppose the specific set of chars is the vowel

(a, e, i, o, u)

in this case:

abcd ----> false

obcdo ----> true

ixyz ----> false

javascript regex

share|improve this question

edited Mar 23 at 10:09

Ali Azim

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

add a comment | 

2

I need to achieve a particular task in javascript, in which my goal is to match the string which starts with a char from a specific set of characters like vowel and ends with the same character where the length of the string is greater than three.

so far I have done the following code that starts and ends with the same character but doesn't know how to specify it that the first char is from the specific set of character:

function regexVar() 

 var re = /(.).*1/
 return re;


console.log("obcdo".test(s));

let suppose the specific set of chars is the vowel

(a, e, i, o, u)

in this case:

abcd ----> false

obcdo ----> true

ixyz ----> false

javascript regex

share|improve this question

edited Mar 23 at 10:09

Ali Azim

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

add a comment | 

I need to achieve a particular task in javascript, in which my goal is to match the string which starts with a char from a specific set of characters like vowel and ends with the same character where the length of the string is greater than three.

so far I have done the following code that starts and ends with the same character but doesn't know how to specify it that the first char is from the specific set of character:

function regexVar() 

 var re = /(.).*1/
 return re;


console.log("obcdo".test(s));

let suppose the specific set of chars is the vowel

(a, e, i, o, u)

in this case:

abcd ----> false

obcdo ----> true

ixyz ----> false

javascript regex

share|improve this question

edited Mar 23 at 10:09

Ali Azim

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

function regexVar() 

 var re = /(.).*1/
 return re;


console.log("obcdo".test(s));

share|improve this question

edited Mar 23 at 10:09

Ali Azim

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

share|improve this question

edited Mar 23 at 10:09

Ali Azim

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

asked Mar 23 at 9:52

Ali AzimAli Azim

13313

2 Answers
2

active

oldest

votes

3

You need to use a character set to ensure the captured character is one of the ones you want, backreference the first captured group at the end of the pattern, not the third group (your pattern doesn't have 3 capture groups), use ^ and $ to anchor the pattern to the start and end of the string, and repeat with 2, rather than * to make sure the whole string is at least 4 characters long:

/^([aeiou]).+1$/

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

share|improve this answer

edited Mar 23 at 10:01

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

add a comment | 

0

You can use this pattern

/^([aeiou]).+1$/i

• 
  ^ - Start of string


• 
  ([aeiou]) - Matches a,e,i,o,u any one of that. (group 1)


• 
  .+ - Match anything except new line.


• 
  1 - Match group 1


• 
  $ - End of string

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

share|improve this answer

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please explain what i is used for in this RegExp?
  
  – Ali Azim
  Mar 23 at 10:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AliAzim i flag is used for case insensitivity. so if you use i flag, regex will treat uppercase and lowercase letters as same character. else it will treat them as two different characters
  
  – Code Maniac
  Mar 23 at 10:05
  
  

  
  
  
  

add a comment | 

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3

You need to use a character set to ensure the captured character is one of the ones you want, backreference the first captured group at the end of the pattern, not the third group (your pattern doesn't have 3 capture groups), use ^ and $ to anchor the pattern to the start and end of the string, and repeat with 2, rather than * to make sure the whole string is at least 4 characters long:

/^([aeiou]).+1$/

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

share|improve this answer

edited Mar 23 at 10:01

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

add a comment | 

3

You need to use a character set to ensure the captured character is one of the ones you want, backreference the first captured group at the end of the pattern, not the third group (your pattern doesn't have 3 capture groups), use ^ and $ to anchor the pattern to the start and end of the string, and repeat with 2, rather than * to make sure the whole string is at least 4 characters long:

/^([aeiou]).+1$/

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

share|improve this answer

edited Mar 23 at 10:01

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

add a comment | 

You need to use a character set to ensure the captured character is one of the ones you want, backreference the first captured group at the end of the pattern, not the third group (your pattern doesn't have 3 capture groups), use ^ and $ to anchor the pattern to the start and end of the string, and repeat with 2, rather than * to make sure the whole string is at least 4 characters long:

/^([aeiou]).+1$/

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

share|improve this answer

edited Mar 23 at 10:01

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

/^([aeiou]).+1$/

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

const re = /^([aeiou]).2,1$/
console.log(
 re.test('abcd'),
 re.test('obcdo'),
 re.test('ixyz')
);

share|improve this answer

edited Mar 23 at 10:01

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

answered Mar 23 at 9:55

CertainPerformanceCertainPerformance

105k166797

0

You can use this pattern

/^([aeiou]).+1$/i

• 
  ^ - Start of string


• 
  ([aeiou]) - Matches a,e,i,o,u any one of that. (group 1)


• 
  .+ - Match anything except new line.


• 
  1 - Match group 1


• 
  $ - End of string

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

share|improve this answer

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please explain what i is used for in this RegExp?
  
  – Ali Azim
  Mar 23 at 10:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AliAzim i flag is used for case insensitivity. so if you use i flag, regex will treat uppercase and lowercase letters as same character. else it will treat them as two different characters
  
  – Code Maniac
  Mar 23 at 10:05
  
  

  
  
  
  

add a comment | 

0

You can use this pattern

/^([aeiou]).+1$/i

• 
  ^ - Start of string


• 
  ([aeiou]) - Matches a,e,i,o,u any one of that. (group 1)


• 
  .+ - Match anything except new line.


• 
  1 - Match group 1


• 
  $ - End of string

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

share|improve this answer

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  can you please explain what i is used for in this RegExp?
  
  – Ali Azim
  Mar 23 at 10:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @AliAzim i flag is used for case insensitivity. so if you use i flag, regex will treat uppercase and lowercase letters as same character. else it will treat them as two different characters
  
  – Code Maniac
  Mar 23 at 10:05
  
  

  
  
  
  

add a comment | 

You can use this pattern

/^([aeiou]).+1$/i

• 
  ^ - Start of string


• 
  ([aeiou]) - Matches a,e,i,o,u any one of that. (group 1)


• 
  .+ - Match anything except new line.


• 
  1 - Match group 1


• 
  $ - End of string

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

share|improve this answer

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

/^([aeiou]).+1$/i

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

let startAndEnd = (str) =>
 return /^([aeiou]).+1$/i.test(str)


console.log(startAndEnd(`ixyz`))
console.log(startAndEnd(`abcd`))
console.log(startAndEnd(`obcdo`))
 

share|improve this answer

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034

answered Mar 23 at 9:57

Code ManiacCode Maniac

13.2k21034