i have a string,
mystr = 'public\uploads\file-1490095922739.jpg';
i want to replace
public\uploads
with " ", so that i just want to extract only file name ie
file-1490095922739.jpg
or like,
\uploads\file-1490095922739.jpg
how can i do this, is there any methods for this in js or can we do it by replace method.
i am performing the following steps,
var imagepath1;
var imagepath = 'public\uploads\file-1490095922739.jpg';
unwantedChar = 'public|uploads';
regExp = new RegExp(unwantedChar , 'gi');
imagepath = imagepath.replace(regExp , '');
imagepath1 = imagepath;
$scope.model.imagepath = imagepath1.replace(/\\/g, "");
please suggest me optimized method.
var input = "public\\uploads\\file-1490095922739.jpg";
var result = input.replace("public\\uploads\\", "");
This is what you're looking for, no need for fancy regexs :). More information about replace can be found here.
Maybe I don't understand the issue - but wouldn't this work?
var mystr = 'public\uploads\file-1490095922739.jpg';
var filename = mystr.replace('public\uploads', '');
If you want to get the part of the string after the last backslash character, you can use this:
var filename = mystr.substr(mystr.lastIndexOf('\\') + 1);
Also note that you need to escape the backslash characters in your test string:
var mystr = 'public\\uploads\\file-1490095922739.jpg';
What about just doing:
var imagepath = 'public\\uploads\\file-1490095922739.jpg';
$scope.model.imagepath = imagepath.replace('public\\uploads\\', '');
instead of using a bunch of unnecessary variables?
This way you're getting the file path, removing public\uploads\ and then setting the file path to $scope.model.imagepath
Note that this will only work if the image file path always matches 'public\uploads\*FILENAME*'.
var url = '/anysource/anypath/myfilename.gif';
var filename = url.slice(url.lastIndexOf('/')+1,url.length);
Search for the last forward slash, and slice the string (+1 because you don't want the slash), with the length of the string to get the filename. This way, you don't have to worry about the path is at all times.
Related
lets say I have this image address like
https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b
how is it possible to replace FILE_NAME.jpg with THUMB_FILE_NAME.jpg
Note: FILE_NAME and THUMB_FILE_NAME are not static and fix.
the FILE_NAME is not fixed and I can't use string.replace method.
eventually I don't know the File_Name
Use replace
.replace(/(?<=\/)[^\/]*(?=(.jpg))/g, "THUMB_FILE_NAME")
or if you want to support multiple formats
.replace(/(?<=\/)[^\/]*(?=(.(jpg|png|jpeg)))/g, "THUMB_FILE_NAME")
Demo
var output = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b".replace(/(?<=\/)[^\/]*(?=(.jpg))/g, "THUMB_FILE_NAME");
console.log( output );
Explanation
(?<=\/) matches / but doesn't remember the match
[^\/]* matches till you find next /
(?=(.jpg) ensures that match ends with .jpg
To match the FILE_NAME, use
.match(/(?<=\/)[^\/]*(?=(.(jpg|png|jpeg)))/g)
var pattern = /[\w-]+\.(jpg|png|txt)/
var c = 'https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b
'
c.replace(pattern, 'YOUR_FILE_NAME.jpg')
you can add any format in the pipe operator
You can use the String's replace method.
var a = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b";
a = a.replace('FILE_NAME', 'THUMB_FILE_NAME');
If you know the format, you can use the split and join to replace the FILE_NAME.
let str = "https://firebasestorage.googleapis.com/v0/b/myproj-d.appspot.com/o/FILE_NAME.jpg?alt=media&token=124bb2bf-c6ef-432b-92c7-7032563ba31b";
let str_pieces = str.split('/');
let str_last = str_pieces[str_pieces.length - 1];
let str_last_pieces = str_last.split('?');
str_last_pieces[0] = 'THUMB_' + str_last_pieces[0];
str_last = str_last_pieces.join('?');
str_pieces[str_pieces.length - 1] = str_last;
str = str_pieces.join('/');
I have a url that looks like this:
http://mysite/#/12345
How do I retrieve the text using regex after the /#/ which is essentially a token that I would like to use elsewhere in my javascript application?
Thanks.
You don't need regex here, just String#substr using String#indexOf:
var s = 'http://mysite/#/12345';
var p ='/#/'; // search needle
var r= s.substr(s.indexOf(p) + p.length);
//=> 12345
Let the browser do it for you
var parser = document.createElement('a');
parser.href = "http://mysite/#/12345";
alert(parser.hash.substring(2)); //This is just to remove the #/ at the start of the string
JSFiddle: http://jsfiddle.net/gibble/uvhqa4yv/
Try with JavaScript String methods.
var str='http://mysite/#/12345';
alert(str.substring(str.lastIndexOf("/#/")+3));
You can try with String'smatch() method as well that uses regex expression.
Just get the matched group from index 1 that is captured by enclosing inside the parenthesis (...)
var str='http://mysite/#/12345';
alert(str.match(/\/#\/(.*)$/)[1]);
Using the browser to parse the URL and getting the hash would probably be most reliable and would work with any valid URL
var url = 'http://mysite/#/12345';
var ele = document.createElement('a');
ele.href = url;
var result = ele.hash.slice(2);
FIDDLE
or you can just split and pop it
var result = url.split('#/').pop();
I'm trying to find a specific character, for example '?' and then remove all text behind the char until I hit a whitespace.
So that:
var string = '?What is going on here?';
Then the new string would be: 'is going on here';
I have been using this:
var mod_content = content.substring(content.indexOf(' ') + 1);
But this is not valid anymore, since the specific string also can be in the middle of a string also.
I haven't really tried anything but this. I have no idea at all how to do it.
use:
string = string.replace(/\?\S*\s+/g, '');
Update:
If want to remove the last ? too, then use
string = string.replace(/\?\S*\s*/g, '');
var firstBit = str.split("?");
var bityouWant = firstBit.substring(firstBit.indexOf(' ') + 1);
The pattern in this code does not replace the parenthesis. I've also tried "/(|)/g".
var re = "/[^a-z]/g",
txt = navsel.options[i].text.split(" ")[0], // here I get the text from a select and I split it.
// What I expect is strings like "(en)" , "(el)" etc
txt = txt.replace(re," ")
Thanks in advance
Your regex is a string, this will try to replace that exact string. Regex objects don't have quotes around them, just the delimiters. Try it like this:
var re = /[^a-z]/g,
txt = navsel.options[i].text.split(" ")[0], // here I get the text from a select and I split it.
txt = txt.replace(re," ");
Or if you prefer strings (and a more explicit type):
var re = new RegExp("[^a-z]", "g")
I have a url :
http://www.xyz.com/a/test.jsp?a=b&c=d
How do I get test.jsp of it ?
This should do it:
var path = document.location.pathname,
file = path.substr(path.lastIndexOf('/'));
Reference: document.location, substr, lastIndexOf
I wont just show you the answer, but I'll give you direction to it. First... strip out everything after the "?" by using a string utility and location.href.status (that will give you the querystring). Then what you will be left with will be the URL; get everything after the last "/" (hint: lastindexof).
Use a regular expression.
var urlVal = 'http://www.xyz.com/a/test.jsp?a=b&c=d';
var result = /a\/(.*)\?/.exec(urlVal)[1]
the regex returns an array, use [1] to get the test.jsp
This method does not depend on pathname:
<script>
var url = 'http://www.xyz.com/a/test.jsp?a=b&c=d';
var file_with_parameters = url.substr(url.lastIndexOf('/') + 1);
var file = file_with_parameters.substr(0, file_with_parameters.lastIndexOf('?'));
// file now contains "test.jsp"
</script>
var your_link = "http://www.xyz.com/a/test.jsp?a=b&c=d";
// strip the query from the link
your_link = your_link.split("?");
your_link = your_link[0];
// get the the test.jsp or whatever is there
var the_part_you_want = your_link.substring(your_link.lastIndexOf("/")+1);
Try this:
/\/([^/]+)$/.exec(window.location.pathname)[1]