Javascript - Replacing the escape character in a string literal

I am trying to replace the backslash (escape) character in a Javascript string literal.

I need to replace it with a double backslash so that I can then do a redirect:

var newpath = 'file:///C:\funstuff\buildtools\viewer.html'.replace(/\\/g,"\\");
window.location = newpath;

However, it seems to have no result.

I don't have the option of properly escaping the backslashes before they are handled by Javascript.

How can I replace (\) with (\\) so that Javascript will be happy?

Thanks, Derek

Answers:

Answer

If it's a literal, you need to escape the backslashes before Javascript sees them; there's no way around that.

var newpath = 'file:///C:\\funstuff\\buildtools\\viewer.html';
window.location = newpath;

If newpath is getting its value from somewhere else, and really does contain single backslashes, you don't need to double them up; but if you really wanted to for some reason, don't forget to escape the backslashes in the replace() call:

newpath.replace(/\\/g,"\\\\");

Why do you not have the option of properly escaping the backslashes before they are handled by Javascript? If the problem is that your Javascript source is being generated from some other scripting language that itself uses \ as an escape character, just add a level of escaping:

var newpath = 'file:///C:\\\\funstuff\\\\buildtools\\\\viewer.html';
Answer

To better demonstrate and understand the string-escaping behavior, take the following example:

You can see what the string looks like in memory after being parsed by the JS-engine by splitting the string, thus also offering potential (ugly) solutions around this issue:

'file:///C:\funstuff\buildtools\viewer.html'.split('')
//>
 ["f", "i", "l", "e", ":", "/", "/", "/", "C", ":", "", "u", "n", "s", "t", "u",
  "f", "f", "", "u", "i", "l", "d", "t", "o", "o", "l", "s", "", "i", "e", "w",
  "e", "r", ".", "h", "t", "m", "l"]

'file:///C:\funstuff\buildtools\viewer.html'.split('').map( function(e){
     return e.charCodeAt()
});
//>
[102, 105, 108, 101, 58, 47, 47, 47, 67, 58, 12, 117, 110, 115, 116, 117, 102,
 102, 8, 117, 105, 108, 100, 116, 111, 111, 108, 115, 11, 105, 101, 119, 101, 
 114, 46, 104, 116, 109, 108]
//>in Hex values by applying .toString(16)
["66", "69", "6c", "65", "3a", "2f", "2f", "2f", "43", "3a", "c", "75", "6e", 
 "73", "74", "75", "66", "66", "8", "75", "69", "6c", "64", "74", "6f", "6f", 
 "6c", "73", "b", "69", "65", "77", "65", "72", "2e", "68", "74", "6d", "6c"]

Basically the single backslash escapes the following character, thus giving rise to unexpected results, if the escaping-context is not heeded.

Solution:

Through a look-up-table, you can restore many errantly escaped characters if they lie outside the printable ASCII character range of \x20-\x7F. For the example above for instance, 12 or \x0c [ 12..toString(16) ] would become '\\'+'v', and so on.

PS: Be aware that a loss of information occured, and you are trying to restore information through contextual- or meta- information, meaning in your case that the string is in the printable ASCII range.

Please share any implementations with the community. Cheers!

Answer

You should be replacing with "\\\\" because "\\" is escaping into a single \ thus no change.

Answer

<< Escape Characters Replacement>>

import java.util.Scanner;

public class Example7 {

  public static void main(String[] args) {
    Scanner in=new Scanner(System.in);
    System.out.println("Please enter a sentence: ");
    String a=in.nextLine();
    //System.out.println("the word had enter: "+a);

    String Str1 = a.replace("\\n", "(new_line)");
    //System.out.println(Str1);
    String Str2 = Str1.replace("\\t", "(tab)");
    //System.out.println(Str2);
    String Str3 = Str2.replace("\\t", "(tab)");
    String Str4 = Str3.replace("\\\\", "(comment_line)");
    String Str5 = Str4.replace(":)", "(smile) ");
    System.out.println("The new sentence:"  +Str5);
  }
}

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