Asked many times, but I can't get it to work...
I have strings like:
"text!../tmp/widgets/tmp_widget_header.html"
and am trying like this to extract widget_header:
var temps[i] = "text!../tmp/widgets/tmp_widget_header.html";
var thisString = temps[i].regexp(/.*tmp_$.*\.*/) )
but that does not work.
Can someone tell me what I'm doing wrong here?
Thanks!
This prints widget_header:
var s = "text!../tmp/widgets/tmp_widget_header.html";
var matches = s.match(/tmp_(.*?)\.html/);
console.log(matches[1]);
var s = "text!../tmp/widgets/tmp_widget_header.html",
re = /\/tmp_([^.]+)\./;
var match = re.exec(s);
if (match)
alert(match[1]);
This will match:
a / character
the characters tmp_
one or more of any character that is not the . character. These are captured.
a . character
If a match was found, it will be at index 1 of the resulting Array.
In your code:
var temps[i] = "text!../tmp/widgets/tmp_widget_header.html";
var thisString = temps[i].regexp(/.*tmp_$.*\.*/) )
You are saying:
"Match any string that starts with any number of any characters, followed by "tmp_", followed by the end of input, followed by any number of periods."
.* : Any number of any character (except newline)
tmp_ : Literally "tmp_"
$ : End of input/newline - this will never be true in this position
\. : " . ", a period
\.* : Any number of periods
Plus when using the regex() function you need to pass a string, using string notation like var re = new RegExp("ab+c") or var re = new RegExp('ab+c') not in regex notation using slash. You also have either an extra, or missing parenthesis, and no characters are actually being captured.
What you want to do is:
"Find a string that preceded by the begining of input, followed by one or more of any character, followed by "tmp_"; followed by a single period, followed by one or more of any character, followed by the end of input;t that contains one or more of any character. Capture that string."
So:
var string = "text!../tmp/widgets/tmp_widget_header.html";
var re = /^.+tmp_(.+)\..+$/; //I use the simpler slash notation
var out = re.exec(string); //execute the regex
console.log(out[1]); //Note that out is an array, the first (here only) catpture sting is at index 1
This regex /^.+tmp_(.+)\..+$/ means:
^ : Match beginning of input/line
.+ : One or more of any character (except newline), "+" is one or more
tmp_ : Constant "tmp_"
\. : A single period
.+ : As above
$ : End of input/line
You could also use this as RegEx('^.+tmp_(.+)\..+$'); not that when we use RegEx(); we do not have the slash marks, instead we use quote marks (single or double will work), to pass it as a string.
Now this would also match var string = "Q%$#^%$^%$^%$^43etmp_ebeb.45t4t#$^g" and out == 'ebeb'. So depending on the specific use you may wish to replace any " . " used to signify any character (except newline) with bracketed "[ ]" character lists, as this may filter out unwanted results. You milage may vary.
For more information visit: https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions
Related
I am passing codes to an API. These codes are alphanumeric, like this one: M84.534D
I just found out that the API does not use the trailing letters. In other words, the API is expecting M84.534, no letter D at the end.
The problem I am having is that the format is not the same for the codes.
I may have M84.534DAC, or M84.534.
What I need to accomplish before sending the code is to remove any non-numeric characters from the end of the code, so in the examples:
M84.534D -> I need to pass M84.534
M84.534DAC -> I also need to pass M84.534
Is there any function or regex that will do that?
Thank you in advance to all.
You can use the regex below. It will remove anything from the end of the string that is not a number
let code = 'M84.534DAC'
console.log(code.replace(/[^0-9]+?$/, ""));
[^0-9] matches anything that is not a numer
+? Will match between 1 and unlimited times
$ Will match the end of the string
So linked together, it will match any non numbers at the end of the string, and replace them with nothing.
You could use the following expression:
\D*$
As in:
var somestring = "M84.534D".replace(/\D*$/, '');
console.log(somestring);
Explanation:
\D stands for not \d, the star * means zero or more times (greedily) and the $ anchors the expression to the end of the string.
Given your limited data sample, this simple regular expression does the trick. You just replace the match with an empty string.
I've used document.write just so we can see the results. You use this whatever way you want.
var testData = [
'M84.534D',
'M84.534DAC'
]
regex = /\D+$/
testData.forEach((item) => {
var cleanValue = item.replace(regex, '')
document.write(cleanValue + '<br>')
})
RegEx breakdown:
\D = Anything that's not a digit
+ = One or more occurrences
$ = End of line/input
I have this string :
var str = '#this #is____#a test###__'
I want to replace all the character (#,_) by (#) , so the excepted output is :
'#this #is#a test#'
Note :
I did not knew How much sequence of (#) or (_) in the string
what I try :
I try to write :
var str = '#this #is__ __#a test###__'
str = str.replace(/[#_]/g,'#')
alert(str)
But the output was :
#this #is## ###a test#####
my try online
I try to use the (*) for sequence But did not work :
var str = '#this #is__ __#a test###__'
str = str.replace(/[#_]*/g,'#')
alert(str)
so How I can get my excepted output ?
A well written RegEx can handle your problem rather easily.
Quoting Mohit's answer to have a startpoint:
var str = '#this #is__ __#a test###__';
var formattedStr = str.replace(/[#_,]+/g, '#');
console.log( formattedStr );
Line 2:
Put in formattedStr the result of the replace method on str.
How does replace work? The first parameter is a string or a RegEx.
Note: RegExps in Javascripts are Objects of type RegExp, not strings. So writing
/yourRegex/
or
New RegExp('/yourRegex/')
is equivalent syntax.
Now let's discuss this particular RegEx itself.
The leading and trailing slashes are used to surround the pattern, and the g at the end means "globally" - do not stop after the first match.
The square parentheses describe a set of characters who can be supplied to match the pattern, while the + sign means "1 or more of this group".
Basically, ### will match, but also # or #####_# will, because _ and # belong to the same set.
A preferred behavior would be given by using (#|_)+
This means "# or _, then, once found one, keep looking forward for more or the chosen pattern".
So ___ would match, as well as #### would, but __## would be 2 distinct match groups (the former being __, the latter ##).
Another problem is not knowing wheter to replace the pattern found with a _ or a #.
Luckily, using parentheses allows us to use a thing called capturing groups. You basically can store any pattern you found in temporary variabiles, that can be used in the replace pattern.
Invoking them is easy, propend $ to the position of the matched (pattern).
/(foo)textnotgetting(bar)captured(baz)/ for example would fill the capturing groups "variables" this way:
$1 = foo
$2 = bar
$3 = baz
In our case, we want to replace 1+ characters with the first occurrence only, and the + sign is not included in the parentheses!
So we can simply
str.replace("/(#|_)+/g", "$1");
In order to make it work.
Have a nice day!
Your regex replaces single instance of any matched character with character that you specified i.e. #. You need to add modifier + to tell it that any number of consecutive matching characters (_,#) should be replaced instead of each character individually. + modifier means that 1 or more occurrences of specified pattern is matched in one go. You can read more about modifiers from this page:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp
var str = '#this #is__ __#a test###__';
var formattedStr = str.replace(/[#_,]+/g, '#');
console.log( formattedStr );
You should use the + to match one-or-more occurrences of the previous group.
var str = '#this #is__ __#a test###__'
str = str.replace(/[#_]+/g,'#')
alert(str)
i need to replace a part of a string in Javascript
The following example should clarify what i mean
var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";
var strDesiredResult = "asd[595442/A][30333][0]";
Basically it means the second area within the brackets should get replaced with another string
How to do that?
What i did so far is something like this :
var str = "asd[595442/A][30327][0]";
var regex = /asd\[(.*)\]\[(.*)\]\[(.*)\]/;
var arrMatches = regex.exec(str);
The string appears in arrMatches[2] correctly, and i could replace this. But what happens if in arrMatches[1] is the same string ?
Because it should only replace the value in the second bracket area.
You may use a regex that will match the first [....] followed with [ and capture that part into a group (that you will be able to refer to via a backreference), and then match 1+ chars other than ] to replace them with your replacement:
var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";
console.log(str.replace(/(\[[^\]]*]\[)[^\]]*/, "$1" + strToReplace));
var strDesiredResult = "asd[595442/A][30333][0]";
console.log(strDesiredResult);
The /(\[[^\]]*]\[)[^\]]*/ has no gmodifier, it will be looking for one match only.
Since regex engine searches a string for a match from left to right, you will get the first match from the left.
The \[[^\]]*]\[ matches [, then any 0+ chars other than ] and then ][. The (...) forms a capturing group #1, it will remember the value that you will be able to get into the replacement with $1 backreference. [^\]]* matches 0+ chars other than ] and this will be replaced.
Details:
( - a capturing group start
\[ - a literal [ symbol (if unescaped, it starts a character class)
[^\]]* - a negated character class that matches zero or more (due to the * quantifier)
] - a literal ] (outside a character class, it does not have to be escaped)
\[ - a literal [
) - end of capturing group #1 (its value can be accessed with $1 backreference from the replacement pattern)
[^\]]* - 0+ (as the * quantifier matches zero or more occurrences, replace with + if you need to only match where there is 1 or more occurrences) chars other than ] (inside a character class in JS regex, ] must be escaped in any position).
Use this pattern:
'asd[595442/A][30327][0]'.replace(/^(asd\[[^\[\]]+\]\[)([^\[\]]+)(\]\[0\])$/, '$130333$3')
Test here
^ - match beginning of string
first group - match "asd[", any chars except [ and ], "]["
second group - match any chars except [ and ]
third group - match exactly: "][0]"
$ - match end of string
There are many ways to do this. One possible pattern is
str.replace(/^(.+)(\[.+\])(\[.+\])(\[.+\])$/, `$1$2[${strToReplace}]$4`)
You can see that $<number> is referred to captured string from regex (string groups in parentheses). We can refer to those and rearrange it however we want.
You can use Regular Expression like this /\[[0-9]+\]/ as below.
var str = "asd[595442/A][30327][0]";
var strToReplace = "30333";
var strDesiredResult = str.replace(/\[[0-9]+\]/, '[' + strToReplace + ']');
console.log(strDesiredResult); //"asd[595442/A][30333][0]";
I want regular expressions to match string that starts with either ${ , "${ , '${ and ends with } , }' , }". My string may be ${anythinghere}, "${anythinghere}", '${anythinghere}'. I tried with the following
var str = "${a}";
var patt = /^("|'|${)*(}|'|")$/;
var res = patt.test(str);
alert(res);
But my above code always returns false.
You have to escape the inner $ with a backslash, since it's a special character. But that alone won't fix the problem. As it is above, patt matches any string that begins with " or ' or ${, 0 or more times, and ends with } or ' or ". The regular expression that you want would be something like this:
var str = "${a}";
var patt = /^(['"]?)\${.*}\1$/;
var res = patt.test(str);
alert(res);
Here is what each part of patt is doing:
^(['"]?): The string must begin with 0 or 1 single quote, or with 0 or 1 double quote. This is in parentheses so that we can reference it at the end of the regexp
\${: Next must be a dollar sign followed by an open curly bracket
.*: Next must be 0 or more of any character (other than a newline)
}: Next must be a closed curly bracket
\1$: Finally, the string must end with whatever pattern was matched at the beginning of the string. \1 is a "back-reference" to the first capturing group (in the parentheses), so if the string began with a single quote, it will only match if it also ends with a single quote. Same goes for double quotes, and no quotes at all
$ has a special meaning in regex, so it must be escaped using a backslash \. If you want 1 or more wildcards, that's represented with a .+, not a *. The following code will match the test pattern:
/^'?"?\${.+}'?"?$/
You can use
var a=new RegExp("\^(\'\\$\{)[a-z]+(\}\')|(\"\\$\{)[a-z]+(\}\")|(\\$\{)[a-z]+(\})\$");
a.test("${vijay}")//true
a.test("\'${vijay}\'")//true
a.test("\"${vijay}\"")//true
a.test("\'${vijay}\"")//false
If you use only \$ it takes as end og expresion.So use\\$ means $ as a special character
This works:
var str = "\"${a}\"";
var patt = /^\"\$\{((?!\").)+\}\"$/;
var res = patt.test(str);
alert(res);
Note that the value assigned to str has been changed to \"${a}\" in order to reflect what you stated targets. You must escape (put a backslash before) any character you wish to be read literally and not as a metacharacter. In order to have this match any one of your targets at once, simply repeat the pattern in patt three times, separated by pipes as usual, replacing the \" with \' or nothing at all (in which case, you should change the \" inside the inner most parenthesis to }). These changes are shown below:
var patt = /^\"\$\{((?!\").)+\}\"$|^\'\$\{((?!\').)+\}\'$|^\$\{((?!\}).)+\}$/;
I want to match some keywords in the url
var parentURL = document.referrer;
var greenPictures = /redwoods-are-big.jpg|greendwoods-are-small.jpg/;
var existsGreen = greenPictures.test(parentURL);
var existsGreen turns true when it finds greendwoods-are-small.jpg but also when it finds small.jpg
What can i do that it only turns true if there is exactly greendwoods-are-small.jpg?
You can use ^ to match the beginning of a string and $ to match the end:
var greenPictures = /^(redwoods-are-big.jpg|greendwoods-are-small.jpg)$/;
var existsGreen = greenPictures.test(parentURL);
But of cause the document.referrer is not equal ether redwoods-are-big.jpg or greendwoods-are-small.jpg so i would match /something.png[END]:
var greenPictures = /\/(redwoods-are-big\.jpg|greendwoods-are-small\.jpg)$/; // <-- See how I escaped the / and the . there? (\/ and \.)
var existsGreen = greenPictures.test(parentURL);
Try this regex:
/(redwoods-are-big|greendwoods-are-small)\.jpg/i
I used the i flag for ignoring the character cases in parentURL variable.
Description
Demo
http://regex101.com/r/aI4yJ6
Dashes does not have any special meaning outside character sets, e.g.:
[a-f], [^x-z] etc.
The characters with special meaning in your regexp is | and .
/redwoods-are-big.jpg|greendwoods-are-small.jpg/
| denotes either or.
. matches any character except the newline characters \n \r \u2028 or \u2029.
In other words: There is something else iffy going on in your code.
More on RegExp.
Pages like these can be rather helpful if you struggle with writing regexp's:
regex101 (with sample)
RegexPlanet
RegExr
Debuggex
etc.