'getting image src in fabricjs
hi i am using fabric js for image resize and and drag and drop. now i want active object src . so i try like this way
canvas.on('object:selected', function () {
currentObject = canvas.getActiveObject();
var imagesrc =currentObject._originalElement.currentSrc;
but the result getting like
i need only the fullbar_1.svg from this .how i seperate it from this
Solution 1:[1]
First save your path in one variable and then split & get your file name
var url="http://localhost:2200/Images/design/fullbar_1.svg";
var check = url.split("/");
alert(check[check.length-1]);
Solution 2:[2]
A solution : http://locutus.io/php/url/parse_url/
module.exports = function parse_url (str, component) { // eslint-disable-line camelcase
// discuss at: http://locutus.io/php/parse_url/
// original by: Steven Levithan (http://blog.stevenlevithan.com)
// reimplemented by: Brett Zamir (http://brett-zamir.me)
// input by: Lorenzo Pisani
// input by: Tony
// improved by: Brett Zamir (http://brett-zamir.me)
// note 1: original by http://stevenlevithan.com/demo/parseuri/js/assets/parseuri.js
// note 1: blog post at http://blog.stevenlevithan.com/archives/parseuri
// note 1: demo at http://stevenlevithan.com/demo/parseuri/js/assets/parseuri.js
// note 1: Does not replace invalid characters with '_' as in PHP,
// note 1: nor does it return false with
// note 1: a seriously malformed URL.
// note 1: Besides function name, is essentially the same as parseUri as
// note 1: well as our allowing
// note 1: an extra slash after the scheme/protocol (to allow file:/// as in PHP)
// example 1: parse_url('http://user:pass@host/path?a=v#a')
// returns 1: {scheme: 'http', host: 'host', user: 'user', pass: 'pass', path: '/path', query: 'a=v', fragment: 'a'}
// example 2: parse_url('http://en.wikipedia.org/wiki/%22@%22_%28album%29')
// returns 2: {scheme: 'http', host: 'en.wikipedia.org', path: '/wiki/%22@%22_%28album%29'}
// example 3: parse_url('https://host.domain.tld/[email protected]/folder')
// returns 3: {scheme: 'https', host: 'host.domain.tld', path: '/[email protected]/folder'}
// example 4: parse_url('https://gooduser:[email protected]/[email protected]/folder?foo=bar')
// returns 4: { scheme: 'https', host: 'www.example.com', path: '/[email protected]/folder', query: 'foo=bar', user: 'gooduser', pass: 'secretpassword' }
var query
var mode = (typeof require !== 'undefined' ? require('../info/ini_get')('locutus.parse_url.mode') : undefined) || 'php'
var key = [
'source',
'scheme',
'authority',
'userInfo',
'user',
'pass',
'host',
'port',
'relative',
'path',
'directory',
'file',
'query',
'fragment'
]
// For loose we added one optional slash to post-scheme to catch file:/// (should restrict this)
var parser = {
php: new RegExp([
'(?:([^:\\/?#]+):)?',
'(?:\\/\\/()(?:(?:()(?:([^:@\\/]*):?([^:@\\/]*))?@)?([^:\\/?#]*)(?::(\\d*))?))?',
'()',
'(?:(()(?:(?:[^?#\\/]*\\/)*)()(?:[^?#]*))(?:\\?([^#]*))?(?:#(.*))?)'
].join('')),
strict: new RegExp([
'(?:([^:\\/?#]+):)?',
'(?:\\/\\/((?:(([^:@\\/]*):?([^:@\\/]*))?@)?([^:\\/?#]*)(?::(\\d*))?))?',
'((((?:[^?#\\/]*\\/)*)([^?#]*))(?:\\?([^#]*))?(?:#(.*))?)'
].join('')),
loose: new RegExp([
'(?:(?![^:@]+:[^:@\\/]*@)([^:\\/?#.]+):)?',
'(?:\\/\\/\\/?)?',
'((?:(([^:@\\/]*):?([^:@\\/]*))?@)?([^:\\/?#]*)(?::(\\d*))?)',
'(((\\/(?:[^?#](?![^?#\\/]*\\.[^?#\\/.]+(?:[?#]|$)))*\\/?)?([^?#\\/]*))',
'(?:\\?([^#]*))?(?:#(.*))?)'
].join(''))
}
var m = parser[mode].exec(str)
var uri = {}
var i = 14
while (i--) {
if (m[i]) {
uri[key[i]] = m[i]
}
}
if (component) {
return uri[component.replace('PHP_URL_', '').toLowerCase()]
}
if (mode !== 'php') {
var name = (typeof require !== 'undefined' ? require('../info/ini_get')('locutus.parse_url.queryKey') : undefined) || 'queryKey'
parser = /(?:^|&)([^&=]*)=?([^&]*)/g
uri[name] = {}
query = uri[key[12]] || ''
query.replace(parser, function ($0, $1, $2) {
if ($1) {
uri[name][$1] = $2
}
})
}
delete uri.source
return uri
}
Solution 3:[3]
I Know I am late here but you can get image URL using
activeObject.getSrc()
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Navneeth |
Solution 2 | neopheus |
Solution 3 | Umashankar |