I want to remove double quotes from a String

Question

I want to remove the    around a String   e g  if the String is   I am here  then I want to output only I am here

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A one liner for the lazy people  var str     a string    str   str replace        g

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If you have control over your data and you know your double quotes surround the string  so do not appear inside string  and the string is an integer     say with     20151212211647278    or similar this nicely removes the surrounding quotes  JSON parse  20151212211647278      Not a universal answer however slick for niche needs

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This simple code will also work  to remove for example double quote from a string surrounded with double quote    var str    remove  foo  delimiting double quotes   console log str replace         g    1

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var expressionWithoutQuotes       for var i  0  i lt length i         if expressionDiv charAt i                   expressionWithoutQuotes    expressionDiv charAt i             This may work for you

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If the string is guaranteed to have one quote  or any other single character  at beginning and end which you d like to remove   str   str slice 1  -1     slice has much less overhead than a regular expression

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If you re trying to remove the double quotes try following     var Stringstr      I am here       var mystring   String Stringstr     mystring   mystring substring 1  mystring length - 1     alert mystring

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This works      x000D   x000D  var string1     foo    x000D  var string2     bar    x000D   x000D  function removeFirstAndLastQuotes str   x000D    var firstChar   str charAt 0   x000D    var lastChar   str str length -1   x000D      double quotes x000D    if firstChar  amp  amp  lastChar     String fromCharCode 34    x000D      str   str slice 1  -1   x000D      x000D      single quotes x000D    if firstChar  amp  amp  lastChar     String fromCharCode 39    x000D      str   str slice 1  -1   x000D      x000D    return str  x000D    x000D  console log removeFirstAndLastQuotes string1    x000D  console log removeFirstAndLastQuotes string2    x000D   x000D   x000D

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str   str replace               1      This regexp will only remove the quotes if they are the first and last characters of the string  F ex    I am here     gt  I am here  replaced  I  am  here    gt  I  am  here  untouched  I am here      gt  I am here   untouched

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Please try following regex for remove double quotes from string         string    I am here        string    tr    d      print  string      exit

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If you only want to remove the boundary quotes   function stripquotes a        if  a charAt 0           amp  amp  a charAt a length-1                     return a substr 1  a length-2             return a      This approach won t touch the string if it doesn t look like  text in quotes

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To restate your problem in a way that s easier to express as a regular expression      Get the substring of characters that is contained between zero or one leading double-quotes and zero or one trailing double-quotes    Here s the regexp that does that     var regexp                    var newStr   str replace                 1      Breaking down the regexp        a greedy match for zero or one leading double-quotes     a greedy match for zero or one trailing double-quotes those two bookend a non-greedy capture of all other characters         which will contain the target as  1    This will return delimited  string  here for   str    delimited  string  here          str     delimited  string  here          str    delimited  string  here          and str     delimited  string  here

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If you want to remove all double quotes in string  use  var str     some  quoted  string    console log  str replace    g            some quoted string   Otherwise you want to remove only quotes around the string  use   var str     some  quoted  string    console log  clean   str replace        g            some  quoted  string

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Assuming  var someStr    He said  quot Hello  my name is Foo quot    console log someStr replace     quot    g         That should do the trick     if your goal is to replace all double quotes   Here s how it works      quot   is a character class  matches both single and double quotes  you can replace this with  quot  to only match double quotes     one or more quotes  chars  as defined by the preceding char-class  optional  g  the global flag  This tells JS to apply the regex to the entire string  If you omit this  you ll only replace a single char   If you re trying to remove the quotes around a given string  ie in pairs   things get a bit trickier  You ll have to use lookaround assertions  var str    remove  quot foo quot  delimiting double quotes   console log str replace   quot     quot       quot    quot  g    1       logs remove foo delimiting quotes str    remove only  quot foo quot  delimiting  quot     note trailing  quot  at the end console log str replace   quot     quot       quot    quot  g    1       logs remove only foo delimiting  quot  lt -- trailing double quote is not removed  Regex explained    quot   literal  matches any literal  quot     begin capturing group  Whatever is between the parentheses      will be captured  and can be used in the replacement value     quot     Character class  matches all chars  except  quot  1 or more times     quot    zero-width  as in not captured  positive lookahead assertion  The previous match will only be valid if it s followed by a  quot  literal    end capturing group  we ve captured everything in between the opening closing  quot   quot   another literal  cf list item one  The replacement is   1   this is a back-reference to the first captured group  being    quot      or everyting in between the double quotes  The pattern matches both the quotes and what s inbetween them  but replaces it only with what s in between the quotes  thus effectively removing them  What it does is some  quot string with quot  quotes - gt  replaces  quot string with quot  with - gt  string with  Quotes gone  job done  If the quotes are always going to be at the begining and end of the string  then you could use this  str replace    quot        quot     quot       1     or this for double and single quotes  str replace     quot           quot        quot         1     With input remove  quot foo quot  delimiting  quot   the output will remain unchanged  but change the input string to  quot remove  quot foo quot  delimiting quotes quot   and you ll end up with remove  quot foo quot  delimiting quotes as output  Explanation     quot   matches the beginning of the string   and a  quot   If the string does not start with a  quot   the expression already fails here  and nothing is replaced         quot      matches  and captures  everything  including double quotes one or more times  provided the positive lookahead is true     quot     the positive lookahead is much the same as above  only it specifies that the  quot  must be the end of the string        end   quot    matches that ending quote  but does not capture it  The replacement is done in the same way as before  we replace the match  which includes the opening and closing quotes   with everything that was inside them   You may have noticed I ve omitted the g flag  for global BTW   because since we re processing the entire string  this expression only applies once  An easier regex that does  pretty much  the same thing  there are internal difference of how the regex is compiled applied  would be  someStr replace    quot      quot      1     As before   quot  and  quot   match the delimiting quotes at the start and end of a string  and the      matches everything in between  and captures it  I ve tried this regex  along side the one above  with lookahead assertion  and  admittedly  to my surprize found this one to be slightly slower  My guess would be that the lookaround assertion causes the previous expression to fail as soon as the engine determines there is no  quot  at the end of the string  Ah well  but if this is what you want need  please do read on  However  in this last case  it s far safer  faster  more maintainable and just better to do this  if  str charAt 0        quot    amp  amp  str charAt str length -1        quot          console log str substr 1 str length -2       Here  I m checking if the first and last char in the string are double quotes  If they are  I m using substr to cut off those first and last chars  Strings are zero-indexed  so the last char is the charAt str length -1   substr expects 2 arguments  where the first is the offset from which the substring starts  the second is its length  Since we don t want the last char  anymore than we want the first  that length is str length - 2  Easy-peazy  Tips  More on lookaround assertions can be found here Regex s are very useful  and IMO fun   can be a bit baffeling at first  Here s some more details  and links to resources on the matter  If you re not very comfortable using regex s just yet  you might want to consider using  var noQuotes   someStr split   quot    join       If there s a lot of quotes in the string  this might even be faster than using regex

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If you only want to remove quotes from the beginning or the end  use the following regular expression     Hello   replace          g

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this code is very better for show number in textbox    this     your textbox               var number     this  val                number   number replace        g                   number   number toString   replace   B     d 3       d   g                      this  val number       1 234 567 890

[javascript] I want to remove double quotes from a String

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