I have the following string:
var str = '\x27';
I have no control on it, so I cannot write it as '\\x27' for example. Whenever I print it, i get:
'
since 27 is the apostrophe. When I call .length on it, it gives me 1. This is of course correct, but how can I treat it like a not escaped string and have it print literally
\x27
and give me a length of 4?
I'm not sure if you should do what you are trying to do, but this is how it works:
var s = '\x27';
var sEncoded = '\\x' + s.charCodeAt(0).toString(16);
s is a string that contains one character, the apostrophe. The character code as a hexadecimal number is 27.
After the assignment var str = '\x27';, you can't tell where the contents of str came from. There's no way to find out whether a string literal was assigned, or whether the string literal contained an escape sequence. All you have is a string containing a single apostrophe character (Unicode code point U+0027). The original assignment could have been
var str = '\x27'; // or
var str = "'"; // or
var str = String.fromCodePoint(3 * 13);
There's simply no way to tell.
That said, your question looks like an XY problem. Why are you trying to print \x27 in the first place?
Related
I want to replace the smart quotes like ‘, ’, “ and ” to regular quotes. Also, I wanted to replace the ©, ® and ™. I used the following code. But it doesn't help.
Kindly help me to resolve this issue.
str.replace(/[“”]/g, '"');
str.replace(/[‘’]/g, "'");
Use:
str = str.replace(/[“”]/g, '"');
str = str.replace(/[‘’]/g, "'");
or to do it in one statement:
str = str.replace(/[“”]/g, '"').replace(/[‘’]/g,"'");
In JavaScript (as in many other languages) strings are immutable - string "replacement" methods actually just return the new string instead of modifying the string in place.
The MDN JavaScript reference entry for replace states:
Returns a new string with some or all matches of a pattern replaced by a replacement.
…
This method does not change the String object it is called on. It simply returns a new string.
replace return the resulting string
str = str.replace(/["']/, '');
The OP doesn't say why it isn't working, but there seems to be problems related to the encoding of the file. If I have an ANSI encoded file and I do:
var s = "“This is a test” ‘Another test’";
s = s.replace(/[“”]/g, '"').replace(/[‘’]/g,"'");
document.writeln(s);
I get:
"This is a test" "Another test"
I converted the encoding to UTF-8, fixed the smart quotes (which broke when I changed encoding), then converted back to ANSI and the problem went away.
Note that when I copied and pasted the double and single smart quotes off this page into my test document (ANSI encoded) and ran this code:
var s = "“This is a test” ‘Another test’";
for (var i = 0; i < s.length; i++) {
document.writeln(s.charAt(i) + '=' + s.charCodeAt(i));
}
I discovered that all the smart quotes showed up as ? = 63.
So, to the OP, determine where the smart quotes are originating and make sure they are the character codes you expect them to be. If they are not, consider changing the encoding of the source so they arrive as “ = 8220, ” = 8221, ‘ = 8216 and ’ = 8217. Use my loop to examine the source, if the smart quotes are showing up with any charCodeAt() values other than those I've listed, replace() will not work as written.
To replace all regular quotes with smart quotes, I am using a similar function. You must specify the CharCode as some different computers/browsers default settings may identify the plain characters differently ("",",',').
Using the CharCode with call the ASCII character, which will eliminate the room for error across different browsers, and operating systems. This is also helpful for bilingual use (accents, etc.).
To replace smart quotes with SINGLE QUOTES
function unSmartQuotify(n){
var name = n;
var apos = String.fromCharCode(39);
while (n.indexOf("'") > -1)
name = name.replace("'" , apos);
return name;
}
To find the other ASCII values you may need. Check here.
I want to replace the smart quotes like ‘, ’, “ and ” to regular quotes. Also, I wanted to replace the ©, ® and ™. I used the following code. But it doesn't help.
Kindly help me to resolve this issue.
str.replace(/[“”]/g, '"');
str.replace(/[‘’]/g, "'");
Use:
str = str.replace(/[“”]/g, '"');
str = str.replace(/[‘’]/g, "'");
or to do it in one statement:
str = str.replace(/[“”]/g, '"').replace(/[‘’]/g,"'");
In JavaScript (as in many other languages) strings are immutable - string "replacement" methods actually just return the new string instead of modifying the string in place.
The MDN JavaScript reference entry for replace states:
Returns a new string with some or all matches of a pattern replaced by a replacement.
…
This method does not change the String object it is called on. It simply returns a new string.
replace return the resulting string
str = str.replace(/["']/, '');
The OP doesn't say why it isn't working, but there seems to be problems related to the encoding of the file. If I have an ANSI encoded file and I do:
var s = "“This is a test” ‘Another test’";
s = s.replace(/[“”]/g, '"').replace(/[‘’]/g,"'");
document.writeln(s);
I get:
"This is a test" "Another test"
I converted the encoding to UTF-8, fixed the smart quotes (which broke when I changed encoding), then converted back to ANSI and the problem went away.
Note that when I copied and pasted the double and single smart quotes off this page into my test document (ANSI encoded) and ran this code:
var s = "“This is a test” ‘Another test’";
for (var i = 0; i < s.length; i++) {
document.writeln(s.charAt(i) + '=' + s.charCodeAt(i));
}
I discovered that all the smart quotes showed up as ? = 63.
So, to the OP, determine where the smart quotes are originating and make sure they are the character codes you expect them to be. If they are not, consider changing the encoding of the source so they arrive as “ = 8220, ” = 8221, ‘ = 8216 and ’ = 8217. Use my loop to examine the source, if the smart quotes are showing up with any charCodeAt() values other than those I've listed, replace() will not work as written.
To replace all regular quotes with smart quotes, I am using a similar function. You must specify the CharCode as some different computers/browsers default settings may identify the plain characters differently ("",",',').
Using the CharCode with call the ASCII character, which will eliminate the room for error across different browsers, and operating systems. This is also helpful for bilingual use (accents, etc.).
To replace smart quotes with SINGLE QUOTES
function unSmartQuotify(n){
var name = n;
var apos = String.fromCharCode(39);
while (n.indexOf("'") > -1)
name = name.replace("'" , apos);
return name;
}
To find the other ASCII values you may need. Check here.
In Javascript, i want the below original string:
I want to replace \"this\" and \"that\" words, but NOT the one "here"
.. to become like:
I want to replace ^this^ and ^that^ words, but NOT the one "here"
I tried something like:
var str = 'I want to replace \"this\" and \"that\" words, but NOT the one "here"';
str = str.replace(/\"/g,"^");
console.log( str );
Demo: JSFiddle here.
But still .. the output is:
I want to replace ^this^ and ^that^ words, but NOT the one ^here^
Which means i wanted to replace only the \" occurrences but NOT the " alone. But i cannot.
Please help.
As #adeneo's comment, your string was created wrong and not exactly like your expectation. Please try this:
var str = 'I want to replace \\"this\\" and \\"that\\" words, but not the one "here"';
str = str.replace(/\\\"/g,"^");
console.log(str);
You can use RegExp /(")/, String.prototype.lastIndexOf(), String.prototype.slice() to check if matched character is last or second to last match in input string. If true, return original match, else replace match with "^" character.
var str = `I want to replace \"this\" and \"that\" words, but NOT the one "here"`;
var res = str.replace(/(")/g, function(match, _, index) {
return index === str.lastIndexOf(match)
|| index === str.slice(0, str.lastIndexOf(match) -1)
.lastIndexOf(match)
? match
: "^"
});
console.log(res);
The problem with String.prototype.replace is that it only replaces the first occurrence without Regular Expression. To fix this, you need to add a g and the end of the RegEx, like so:
var mod = str => str.replace(/\\\"/g,'^');
mod('I want to replace \\"this\\" and \\"that\\" words, but NOT the one "here"');
A less effective but easier to understand to do what you wanted is to split the string with the delimiter and then join it with the replacement, like so:
var mod = str => str.split('\\"').join('^');
mod('I want to replace \\"this\\" and \\"that\\" words, but NOT the one "here"');
Note: You can wrap a string with either ' or ". Suppose your string contains ", i.e. a"a, you will need to put an \ in front of the " as "a"a" causes syntax error. 'a"a' won't cause syntax error as the parser knows the " is part of the string, but when you put a \ in front of " or any other special characters, it means the following character is a special character. So 'a\"a' === 'a"a' === "a\"a". If you want to store \, you will need to use \ regardless of the type of quote you use, so to store \", you will need to use '\\"', '\\\"' or "\\\"".
I'm trying to build a regular expression that parses a string and skips things in brackets.
Something like
string = "A bc defg hi [hi] jkl mnop.";
The .match() should return "hi" but not [hi]. I've spent 5 hours running through RE's but I'm throwing in the towel.
Also this is for javascript or jquery if that matters.
Any help is appreciated. Also I'm working on getting my questions formatted correctly : )
EDIT:
Ok I just had a eureka moment and figured out that the original RegExp I was using actually did work. But when I was replaces the matches with the [matches] it simply replaced the first match in the string... over and over. I thought this was my regex refusing to skip the brackets but after much time of trying almost all of the solutions below, I realized that I was derping Hardcore.
When .replace was working its magic it was on the first match, so I quite simply added a space to the end of the result word as follows:
var result = string.match(regex);
var modifiedResult = '[' + result[0].toString() + ']';
string.replace(result[0].toString() + ' ', modifiedResult + ' ');
This got it to stop targeting the original word in the string and stop adding a new set of brackets to it with every match. Thank you all for your help. I am going to give answer credit to the post that prodded me in the right direction.
preprocess the target string by removing everything between brackets before trying to match your RE
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*\]/, "")
then apply your RE to tmpstring
correction: made the match for brackets eager per nhahtd comment below, and also, made the RE global
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*?\]/g, "")
You don't necessarily need regex for this. Simply use string manipulation:
var arr = string.split("[");
var final = arr[0] + arr[1].split("]")[1];
If there are multiple bracketed expressions, use a loop:
while (string.indexOf("[") != -1){
var arr = string.split("[");
string = arr[0] + arr.slice(1).join("[").split("]").slice(1).join("]");
}
Using only Regular Expressions, you can use:
hi(?!])
as an example.
Look here about negative lookahead: http://www.regular-expressions.info/lookaround.html
Unfortunately, javascript does not support negative lookbehind.
I used http://regexpal.com/ to test, abcd[hi]jkhilmnop as test data, hi(?!]) as the regex to find. It matched 'hi' without matching '[hi]'. Basically it matched the 'hi' so long as there was not a following ']' character.
This of course, can be expanded if needed. This has a benefit of not requiring any pre-processing for the string.
r"\[(.*)\]"
Just play arounds with this if you wanto to use regular expressions.
What do yo uwant to do with it? If you want to selectively replace parts like "hi" except when it's "[hi]", then I often use a system where I match what I want to avoid first and then what I want to watch; if it matches what I want to avoid then I return the match, otherwise I return the processed match.
Like this:
return string.replace(/(\[\w+\])|(\w+)/g, function(all, m1, m2) {return m1 || m2.toUpperCase()});
which, with the given string, returns:
"A BC DEFG HI [hi] JKL MNOP."
Thus: it replaces every word with uppercase (m1 is empty), except if the word is between square brackets (m1 is not empty).
This builds an array of all the strings contained in [ ]:
var regex = /\[([^\]]*)\]/;
var string = "A bc defg hi [hi] [jkl] mnop.";
var results=[], result;
while(result = regex.exec(string))
results.push(result[1]);
edit
To answer to the question, this regex returns the string less all is in [ ], and trim whitespaces:
"A bc defg [hi] mnop [jkl].".replace(/(\s{0,1})\[[^\]]*\](\s{0,1})/g,'$1')
Instead of skipping the match you can probably try something different - match everything but do not capture the string within square brackets (inclusive) with something like this:
var r = /(?:\[.*?[^\[\]]\])|(.)/g;
var result;
var str = [];
while((result = r.exec(s)) !== null){
if(result[1] !== undefined){ //true if [string] matched but not captured
str.push(result[1]);
}
}
console.log(str.join(''));
The last line will print parts of the string which do not match the [string] pattern. For example, when called with the input "A [bc] [defg] hi [hi] j[kl]u m[no]p." the code prints "A hi ju mp." with whitespaces intact.
You can try different things with this code e.g. replacing etc.
Given a string in Javacript, such as
var str = "this's kelly";
I want to replace the apostrophe (') with another character. Here is what I've tried so far:
str.replace('"', 'A');
str.replace('\'', 'A');
None of these work.
How do I do it?
Can you also please advices me with the invalid characters that when passed to the query string or URL crashes the page or produces undesired results ? e.g passing apostrophe (') produces undesired result are their any more of them.
var str = "this's kelly"
str = str.replace(/'/g, 'A');
The reason your version wasn't working is because str.replace returns the new string, without updating in place.
I've also updated it to use the regular expression version of str.replace, which when combined with the g option replaces all instances, not just the first. If you actually wanted it to just replace the first, either remove the g or do str = str.replace("'", 'A');
Do this:
str = str.replace("'","A");
str = str.replace("'", "A");
Your running the function but not assigning it to anything again so the var remains unchanged