I have a string that a user can edit at any time, and a regex that is being conducted on the string, to add it to an xml and then save it but they can add '$1' to the string. I just want the text '$1' to be saved but I have to perform a regular expression on the same string that $1 is in. It replaces the $1 with a character from the regex every time.
How do I find, and replace, the $1 in this string?
Example of what is happening:
string1 = '<item id="1">i have $100</item>'
regexp = new RegExp('<item id="1"([^<]|<[^\/]|<\/[^i]|<\/i[^t]|<\/it[^e]|<\/ite[^m]|<\/item[^>])*<\/item>');
data = '<data><item id="1">i have no money</item><item id="2">i have no money</item></data>'
data = data.replace(regexp, string1);
Results
<data><item id="1">i have >00</item><item id="2">i have no money</item></data>
If you have a variable string that you want to put in your replace() call which might possibly have $N's in it, you can prevent the $N from being treated as a backreference by replacing $ with $$. Apparently, unlike other special characters in JS regex, the $ character cannot be escaped with a \ - it must be escaped with a preceding $ (go figure).
In your example, you could do the following to fix the issue:
data = data.replace(regexp, string1.replace('$', '$$$'));
This should turn any $'s into $$ in string1, preventing them from being treated as backreferences.
(Note: I found this little nugget here)
This should only happen if you have a capturing group in the regex.
If you don't want your groups to capture, then place ?: inside the start of the group.
/foo(?:bar)/
You can escape the $. Eg:
var replacement = '<item id="1">i have \\$100</item>';
Useful when you have capturing groups and need to write a $.
Related
sorry, I confused how to search in search engine about what I want to find the answer...
What I want is catch and replace a part of string with regex, but with requirements, And the requirements is not included to the result. Here for example:
I want to replace an # (at) sign with no backslash before...
If I use "string".replace(/[^\\]#/g) it will also replace the character
before at sign, except it is back slash...
more information with same destination with the example before
string.replace(/[^\\]#/g,"Replaced");
abc\# still kept as abc\# (What I want)
abc# replaced as abReplaced (I want this NOT occured)
abc# replaced as abcReplaced (What I want, the character before # not replaced, the c character)
So the conclusion is Every # sign with backslash before it will NOT be catched/Replaced, But if it has no backslash before, it will be catched/replaced with before character of the# sign not be catched/replaced
The general answer to this question in regular expressions is you would use a negative look-behind assertion. But JavaScript doesn't have lookbehind.
In your case, you can just capture the character before (if any) and re-output it as part of the replacement:
result = string.replace(/(^|[^\\])#/, "$1Replaced");
(The ^| part is to allow for no character before, e.g., beginning of string. The | is an alternation, e.g., ^ or [^\\].)
(If you want to do it globally, you'll also want the g flag.)
Also note that to use a literal backslash in a regular expression, you must escape it (with another backslash).
Examples (using g for global replace):
function test(string) {
console.log(string, "=>", string.replace(/(^|[^\\])#/g, "$1Replaced"));
}
test("abc\\#");
test("abc#");
test("#abc#abc");
Do it with a capture group and put $1 in the replace argument to reproduce that character:
var res = 'abc#'.replace(/([^\\|^])#/g, '$1Replaced');
console.log(res);
Note that you need to cover for the possibility that # is the first character of your string, and in that case your regular expression will not have a match. Solve this by matching as alternative with the start of the string (^).
You can do this:
const data = [
'test#asd.cs',
'test\\#asd.asdf',
'#test\\#asd.asdf'
]
const removeAtSymbol = str => str.replace(/([^\\])#|^#/g, '$1')
for (const item of data){
console.log(item, '->', removeAtSymbol(item))
}
I was converting normal string in to latex format.So i was created the latex code match and replace the \ single slash into \\ double slash.why the i need it Refer this link.I tried Below code :
function test(){
var tex="$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$";
var tex_form = tex.replace("/[\\\/\\\\\.\\\\]/g", "\\");
document.getElementById('demo').innerHTML=tex_form;//nothing get
}
test();
<p id="demo"></p>
Not getting any output data.But the match in this link
i wish to need replace the \ into \\
There are these issues:
The string literal has no backslashes;
The regular expression is not a regular expression;
The class in the intended regular expression cannot match sequences, only single characters;
The replacement would not add backslashes, only replace with them.
Here you find the details on each point:
1. How to Encode Backslashes in String Literals
Your tex variable has no backslashes. This is because a backslash in a string literal is not taken as a literal backslash, but as an escape for interpreting the character that follows it.
When you have "$$\left...", then the \l means "literal l", and so the content of your variable will be:
$$left...
As an l does not need to be escaped, the backslash is completely unnecessary, and these two assignments result in the same string value:
var tex="$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$";
var tex="$$left[ x=left({{11}over{2}}+{{sqrt{3271}}over{2,3^{{{3}over{2} $$";
To bring the point home, this will also represent the same value:
var tex="\$\$\l\e\f\t\[\ \x\=\l\e\f\t\(\{\{\1\1\}\o\v\e\r\{\2\}\}\+\{\{\s\q\r\t\{\3\2\7\1\}\}\o\v\e\r\{\2\,\3\^\{\{\{\3\}\o\v\e\r\{\2\}\ \$\$";
If you really want to have literal backslashes in your content (which I understand you do, as this is about LaTeX), then you need to escape each of those backslashes... with a backslash:
var tex="$$\\left[ x=\\left({{11}\\over{2}}+{{\\sqrt{3271}}\\over{2\\,3^{{{3}\\over{2} $$";
Now the content of your tex variable will be this string:
$$\left[ x=\left({{11}\over{2}}+{{\sqrt{3271}}\over{2\,3^{{{3}\over{2} $$
2. How to Code Regular Expression Literals
You are passing a string literal to the first argument of replace, while you really intend to pass a regular expression literal. You should leave out the quotes for that to happen. The / are the delimiters of a regular expression literal, not quotes:
/[\\\/\\\\\.\\\\]/g
This should not be wrapped in quotes. JavaScript understands the / delimiters as denoting a regular expression literal, including the optional modifiers at the end (like g here).
3. Classes are sets of single characters
This regular expression has unnecessary characters. The class [...] should list all individual characters you want to match. Currently you have these characters (after resolving the escapes):
\
/
\
\
.
\
\
It is overkill to have the backslash represented 5 times. Also, in JavaScript the forward slash and dot do not need to be escaped when occurring in a class. So the above regular expression is equivalent to this one:
/[\\/.]/g
Maybe this is, or is not, what you intended to match. To match several sequences of characters, you could use the | operator. This is just an example:
/\\\\|\\\/|\\\./g
... but I don't think you need this.
4. How to actually prefix with backslashes
It seems strange to me that you would want to replace a point or forward slash with a backslash. Probably you want to prefix those with a backslash. In that case make a capture group (with parentheses) and refer to it with $1 in this replace:
tex.replace(/([\\/.])/g, "\\$1");
Note again, that in the replacement string there is only one literal backslash, as the first one is an escape (see point 1 above).
why the i need it
As the question you link to says, the \ character has special meaning inside a JavaScript string literal. It represents an escape sequence.
Not getting any output data.But the match in this link
The escape sequence is processed when the string literal is parsed by the JavaScript compiler.
By the time you apply your regular expression to them, they have been consumed. The slash characters only exist in your source code, not in your data.
If you want to put a slash character in your string, then you need to write the escape sequence for it (the \\) in the source code. You can't add them back in with JavaScript afterwards.
Not sure if I understood the problem, but try this code:
var tex_form = tex.replace("/(\\)/g","\\\\");.
You need to use '(' ')' instead of '['']' to get a match for output.
I want to turn the string dkfj-dkfj-sflj into dkfj-woop$dkfj-sflj.
Here's what I've tried:
var my_string = "dkfj-dkfj-sflj";
var regex = new RegExp("(\\w+)-(\\w+)-(\\w+)", "g");
console.log(my_string.replace(regex, "$1$woop[\$$2]$3");
And my result is: dkfj-woop$2-sflj. Because the "$" is in front of the "$2" capture group, it messes up that capture group.
Assuming I want the structure of my regex and capture group string to stay the same, what's the right way to escape that "$" so it works?
That isn't how you escape a $ for replace. Backslash escaping works at the parser level, functions like replace cannot give special meaning to new escape sequences like \$ because they don't even see the \$. The string "\$" is exactly equivalent to the string "$", both produce the same string. If you wanted to pass a backslash and a dollar sign to a function, it's the backslash itself that requires escaping: "\\$".
Regardless, replace expects you to escape a $ with $$. You need "$1$woop[$$$2]$3"; a $$ for the literal $, and $2 for he capture group.
Read Specifying a string as a parameter in the replace docs.
Use $$ in the replacement part to print a literal $ symbol and you don't need to have a character class in the replacement part if your pattern was enclosed within forward slashes.
> var my_string = "dkfj-dkfj-sflj";
undefined
> my_string.replace(/(\w+)-\w+-(\w+)/, "$1-woop$$$1-$2")
'dkfj-woop$dkfj-sflj'
using http://www.regular-expressions.info/javascriptexample.html I tested the following regex
^\\{1}([0-9])+
this is designed to match a backslash and then a number.
It works there
If I then try this directly in code
var reg = /^\\{1}([0-9])+/;
reg.exec("/123")
I get no matches!
What am I doing wrong?
Update:
Regarding the update of your question. Then the regex has to be:
var reg = /^\/(\d+)/;
You have to escape the slash inside the regex with \/.
The backslash needs to be escaped in the string too:
reg.exec("\\123")
Otherwise \1 will be treated as special character.
Btw, the regular expression can be simplified:
var reg = /^\\(\d+)/;
Note that I moved the quantifier + inside the capture group, otherwise it will only capture a single digit (namely 3) and not the whole number 123.
You need to escape the backslash in your string:
"\\123"
Also, for various implementation bugs, you may want to set reg.lastIndex = 0;.
In addition, {1} is completely redundant, you can simplify your regex to /^\\(\d)+/.
One last note: (\d)+ will only capture the last digit, you may want (\d+).
I have found the following expression which is intended to modify the id of a cloned html element e.g. change contactDetails[0] to contactDetails[1]:
var nel = 1;
var s = $(this).attr(attribute);
s.replace(/([^\[]+)\[0\]/, "$1["+nel+"]");
$(this).attr(attribute, s);
I am not terribly familiar with regex, but have tried to interpret it and with the help of The Regex Coach however I am still struggling. It appears that ([^\[]+) matches one or more characters which are not '[' and \[0\]/ matches [0]. The / in the middle I interpret as an 'include both', so I don't understand why the author has even included the first expression.
I dont understand what the $1 in the replace string is and if I use the Regex Coach replace functionality if I simply use [0] as the search and 1 as the replace I get the correct result, however if I change the javascript to s.replace(/\[0\]/, "["+nel+"]"); the string s remains unchanged.
I would be grateful for any advice as to what the original author intended and help in finding a solution which will successfully replace the a number in square brackets anywhere within a search string.
Find
/ # Signifies the start of a regex expression like " for a string
([^\[]+) # Capture the character that isn't [ 1 or more times into $1
\[0\] # Find [0]
/ # Signifies the end of a regex expression
Replace
"$1[" # Insert the item captured above And [
+nel+ # New index
"]" # Close with ]
To create an expression that captures any digit, you can replace the 0 with \d+ which will match a digit 1 or more times.
s.replace(/([^\[]+)\[\d+\]/, "$1["+nel+"]");
The $1 is a backreference to the first group in the regex. Groups are the pieces inside (). So, in this case $1 will be replaced by whatever the ([^\[]+) part matched.
If the string was contactDetails[0] the resulting string would be contactDetails[1].
Note that this regex only replaces 0s inside square brackets. If you want to replace any number you will need something like:
([^\[]+)\[\d+\]
The \d matches any digit character. \d+ then becomes any sequence of at least one digit.
But your code will still not work, because Javascript strings are immutable. That means they can't be changed once created. The replace method returns a new string, instead of changing the original one. You should use:
s = s.replace(...)
looks like it replaces arrays of 0 with 1.
For example: array[0] goes to array[1]
Explanation:
([^[]+) - This part means save everything that is not a [ into variable $1
[0]/ - This part limits Part 1 to save everything up to a [0]
"$1["+nel+"]" - Print out the contents of $1 (loaded from part 1) and add the brackets with the value of nel. (in your example nel = 1)
Square braces define a set of characters to match. [abc] will match the letters a, b or c.
By adding the carat you are now specifying that you want characters not in the set. [^abc] will match any character that is not an a, b or c.
Because square braces have special meaning in RegExps you need to escape them with a slash if you want to match one. [ starts a character set, \[ matches a brace. (Same concept for closing braces.)
So, [^\[]+ captures 1 or more characters that are not [.
Wrapping that in parenthesis "captures" the matched portion of the string (in this case "contactDetails" so that you can use it in the replacement.
$1 uses the "captured" string (i.e. "contactDetails") in the replacement string.
This regex matches "something" followed by a [0].
"something" is identified by the expression [^\[]+ which matches all charactes that are not a [. You can see the () around this expression, because the match is reused with $1, later. The rest of your regex - that is \[0\] just matches the index [0]. The author had to write \[ and \] because [ and ] are special charactes for regular expressions and have to be escaped.
$1 is a reference to the value of the first paranthesis pair. In your case the value of
[^\[]+
which matches one or more characters which are not a '['
The remaining part of the regexp matches string '[0]'.
So if s is 'foobar[0]' the result will be 'foobar[1]'.
[^\[] will match any character that is not [, the '+' means one or more times. So [^[]+ will match contactDetails. The brackets will capture this for later use. The '\' is an escape symbol so the end \[0\] will match [0]. The replace string will use $1 which is what was captured in the brackets and add the new index.
Your interpretation of the regular expression is correct. It is intended to match one or more characters which are not [, followed by a literal [0]. And used in the replace method, the match would be replaced with the match of the first grouping (that’s what $1 is replaced with) together with the sequence [ followed by the value of nel and ] (that’s how "$1["+nel+"]" is to be interpreted).
And again, a simple s.replace(/\[0\]/, "["+nel+"]") does the same. Except if there is nothing in front of [0], because in that case the first regex wouldn’t find a match.