String.replaceAll () takes an argument of regex?

Similar Questions

• String replaceAll() help

      //kill characters method used to remove unwanted characters
      //from strings, takes in unedited string, returns formatted string
      private String killChars(String fixThis )
      {
          return fixThis.replaceAll("\\W", "");
      }//end

  Hey guys and girls.
  
  I have this method in one of my applications and I do not exactly get the results I want. Is there a way to manipulate the replaceAll function arguments so that it will replace all non alphanumeric except for spaces? In fact, I want to keep white space.
  
  Thank you
  
  Kyle

  I think you can use "[^ \\s\\w]", which is anything other than whitespace or Word because. try it, see if it does what it takes.

• How to pass strings of path as arguments to applescript?

  Since InDesign Extend script, I have the code snippet:

  var docFolder = currentDoc.filePath.getRelativeURI("/"); "

  scriptPath var = (new File($.fileName).parent.getRelativeURI("/" "));

  var script = new file (scriptPath + "/ RunFileMaker.scpt");

  argArray var = new Array();

  argArray.push (docFolder);

  If {(script.exists)

  app.doScript (script, ScriptLanguage.applescriptLanguage, argArray);

  }

  
  

  My RunFileMaker.scpt applescript has been simplified to:

  
  

  on run {argv}

  Display alert point 1 argv

  end run

  
  

  I get the error message:

  < < script > > can make point 1 in string

  
  

  The bulk of the problem is I'm trying to pass a string path to the applescript.

  
  

  
  

  Hi Bill,

  Try this... ;-)

  on run
      display dialog item 1 of arguments
  end run
  

• problem with String.replaceAll

  Everyone does not understand why "10,00".replaceAll (".") * ","0") == '00' and not '0'?"
  
  Looks like. * must match the whole string and replace it with '0 '.
  
  With the help of 1.6.0_17

  840419 wrote:
  Everyone does not understand why "10,00".replaceAll (".") * ","0") == '00' and not '0'?"

  Looks like. * must match the whole string and replace it with '0 '.

  He does.

      public static void main(String[] arg) {
          //System.out.println("10.00".replaceAll(".*", "0"));
          Pattern p = Pattern.compile(".*");
          Matcher m = p.matcher("10.00");
          while(m.find()) {
              // the first find() greedily matches everything
              // the second find() matches a 0(zero)-length charsequence at the end
              System.out.printf("[%s]\n",m.group());
          }
      }
  

  You want to

      public static void main(String[] arg) {
          //System.out.println("10.00".replaceAll(".+", "0"));
          Pattern p = Pattern.compile(".+");
          Matcher m = p.matcher("10.00");
          while(m.find()) {
              System.out.printf("[%s]\n",m.group());
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• interface not implemented

  I am really fedUp with code that promise to implement an interface

  but in fact is not, and yet always compiles!

  in the following code:

  ArrayList < String > words =...

  words.replaceAll (String::toUpperCase);

  Why line 2 compile sucssfuly?

  ArrayList.replaceAll takes a < E > UnaryOperator

  UnaryOperator < T > {}

  T apply (T, t);

  }

  While String.toUpperCase () implements the UnaryOperator < E > interface, String.toUpperCase () takes no arguments

  ==================================================================================

  in the following code from Oracle tutorials collections: aggregation operations:

  // a class to store total value and values count to get the average
  class Averager implements IntConsumer {
      int total = 0;
      int count = 0;
  
      public double average() {
          if(count > 0) return (double)total/count; else return 0;
      }
  
      public void accept(int newVal) {
          total += newVal;
          count++;
      }
  
      public void compine(Averager other) {
          total += other.total;
          count += other.count;
      }
  }
  
  // Code in the main class
  ArrayList<Integer> allVals = new ArrayList<Integer>();
  allVals.addAll(Arrays.asList(2, 3, 5, 7, 1, 89, 4, 5));
  
  Averager averageCollect =
      allVals.stream()
               .collect(Averager::new, Averager::accept, Averager::compine);
  System.out.println(averageCollect.average());
  
  
  

  Why the code at line 25 compile sucssfuly?

  Averager.Accept and Averager.compine does not implement BiConsumer < T, U > Interface

  Interface BiConsumer < T, U > {}

  void accept (T, t, U u);

  }

  They both take a single argument?

  OK, let's look at in detail in the example

  List words = ... ;
  words.replaceAll(String::toUpperCase);
  

  As you noted, the replaceAll parameter is UnaryOperator, and String::toUpperCase does not implement this interface. So, how does this work?

  First, observe that String::toUpperCase is a reference of toUpperCase() method, which is an instance of string method. You can rewrite the reference method as follows, without changing the meaning:

  words.replaceAll((String s) -> s.toUpperCase());
  

  I stated the lambda parameter type just to make it clear that it's a chain, but it could just as easily be written as: s-> s.toUpperCase ().

  Observe that this lambda expression takes one parameter, a string, and returns a value, a string.

  Now let's look at List.replaceAll (). It take a UnaryOperatorparameter. It is an instance method, and the variable E is the type of the list parameter. Words being list, then words.replaceAll () will have a UnaryOperator as its parameter.

  Now let's look at UnaryOperator. This must be a functional interface, since we provide a lambda here. However, UnaryOperator himself is not all abstract methods. He inherited his unique abstract of its superinterface function method. If we look at thefunction, it has a single abstract method

  R apply(T t)
  

  which means that it takes a parameter of type T and returns a value of type R. Foregoing, we know we need a UnaryOperator, which extends thefunction, and the types of the abstract method will be

  String apply(String t)
  

  It is a method that takes a parameter, a string, and returns a value, a string. But look! That's exactly what the lambda expression

  (String s) -> s.toUpperCase()
  

  designated operational entities and the lambda expression is equivalent to

  String::toUpperCase
  

  It is so true that the String::toUpperCase, as a method, declare that it implements UnaryOperator. But we must look at the "forms" of the methods. String::toUpperCase is a reference to a method that takes a string is a parameter and returns a string. And UnaryOperator has one abstract method that takes a string as a parameter and returns a string. That's why String::toUpperCase works in a context where a UnaryOperator is necessary.

  A wrinkle here is that the toUpperCase() method is an instance method. It seems that it takes no arguments. However, as an instance method, it must be called on an object, the "receiver". The receiver is as a first argument implicit in a method; It's just that from a syntactic point of view, it occurs in a different place. Take the example of Averager:

  class Averager {
      void accept(Integer newVal) { ... }
      void combine(Averager other) { ... }
  }
  

  (I've changed around int to take into account the autoboxing).

  Here, accept() and combine() are instance methods, so they must be called on an instance of Averager. Which is like an additional implicit for each argument. Thus, accept() takes two arguments: an Averager and an integer and returns no value (empty). For example, Averager.accept () corresponds to BiConsumer, since his abstract method is accepted (Averager t, all over u).

  (Note that it is completely irrelevant that Averager.accept has the same method name as BiConsumer.accept).

  In addition, combine() takes two arguments, an Averager (receiver) and an another Averager (the parameter as 'other'). Similarly, combine() corresponds to BiConsumer.

• Storage of functions with arguments as String

  I need to store a function with arguments in a string.

  Without arguments, you simply have to do sth. Like this:

  functionName:String = "Hello";

  This [FunctionName])

  Function Hello (): void {}

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  But as soon as you enter arguments it becomes difficult.

  I came up with this:

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  but it is not very flexible, very sturdy and seems unnecessarily complicated.

  Anyone know a faster way to achieve this?

  function parseF(s:String):void {}

  var a: Array = s.split ("("); ")

  var arg:Array = a [1].split(",");

  This [a [0]] (arg [0], arg [1] .substr (0, -1));

  }

• IFmsNotifyAction: the long string arguments get corrupted

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  Field of FmsVariant;
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  pAction-> setClientId (field);
  field.setString ("myfunc");
  pAction-> setMethodName (field);
  field.setString("012345678901234567890123456789012"); 33 characters
  pAction-> addParam (field);
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  In the main.asc app, I just added:
  
  Client.prototype.myFunc (ARG) {trace (arg) ;}
  

• split a string dynamically

  NodeList list2 = doc.getElementsByTagName("URL");
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• Regex expression for <>media

  Hello
  
  I have data within the brackets <>
  and would need a regular expression to get the same correspondent.
  
  by example-data - < 111-1-111 > would just 111-1-111
  
  
  I tried to use the String.replaceall ("<","" "); featured, but can't figure out how to pass in the regular expression to check the closing ' > ' support.
  
  any suggestions on this would be very useful
  
  
  THX

  You are not one of those who wants to solve all the problems with a regex right? In the example you give simply stripping off the first and the last character (which can be done with a call unique substring()) already nets you what you want.

  In any case regex also works:

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  But without the T-I had to put this because otherwise I could not the forum to render the characters LT and GT: s

• Cannot pass arguments to a Java exe file

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  Published by: kooldba on February 12, 2012 14:25
  
  Published by: kooldba on February 12, 2012 14:26

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  I get the error while executing the following query.
  
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  Hello

  Daniel wrote:
  I get the error while executing the following query.

  The error message really means what he says.
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• I want to convert the double data type to string type

  I want to convert the double data type to string, have idea about that

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  Use f

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  Example: string (314)
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  http://w3schools.com/XPath/xpath_functions.asp#string

  See you soon,.
  Vlad