Extract just the path from CSS and exclude anything after the path
2
Using the regex below, I can extract the path from the CSS, but the result also includes the bit after the "#" and "?". Is there any way I can just extract the path
Regex
url([s]?["|']?(.*?)["|']?[s]?)
String
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix') format('embedded-opentype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff2') format('woff2')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff') format('woff')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.ttf') format('truetype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.svg#OpenSans') format('svg')
Expected
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot
Actual
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix
php css regex
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
add a comment |
2
Using the regex below, I can extract the path from the CSS, but the result also includes the bit after the "#" and "?". Is there any way I can just extract the path
Regex
url([s]?["|']?(.*?)["|']?[s]?)
String
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix') format('embedded-opentype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff2') format('woff2')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff') format('woff')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.ttf') format('truetype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.svg#OpenSans') format('svg')
Expected
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot
Actual
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix
php css regex
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
Tryurl([s'"]*K[^'"()?#]+. See live demo here
– revo
Nov 26 '18 at 10:54
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58
add a comment |
2
2
2
Using the regex below, I can extract the path from the CSS, but the result also includes the bit after the "#" and "?". Is there any way I can just extract the path
Regex
url([s]?["|']?(.*?)["|']?[s]?)
String
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix') format('embedded-opentype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff2') format('woff2')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff') format('woff')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.ttf') format('truetype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.svg#OpenSans') format('svg')
Expected
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot
Actual
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix
php css regex
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
Using the regex below, I can extract the path from the CSS, but the result also includes the bit after the "#" and "?". Is there any way I can just extract the path
Regex
url([s]?["|']?(.*?)["|']?[s]?)
String
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix') format('embedded-opentype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff2') format('woff2')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.woff') format('woff')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.ttf') format('truetype')
url('../fonts/Google_OpenSans/open-sans-v15-latin-italic.svg#OpenSans') format('svg')
Expected
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot
Actual
../fonts/Google_OpenSans/open-sans-v15-latin-italic.eot?#iefix
php css regex
php css regex
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
edited Nov 26 '18 at 9:35
Micha Wiedenmann
10.4k1364104
10.4k1364104
asked Nov 26 '18 at 9:34
ReadoReado
59641442
asked Nov 26 '18 at 9:34
ReadoReado
59641442
asked Nov 26 '18 at 9:34
ReadoReado
59641442
59641442
Tryurl([s'"]*K[^'"()?#]+. See live demo here
– revo
Nov 26 '18 at 10:54
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58
add a comment |
Tryurl([s'"]*K[^'"()?#]+. See live demo here
– revo
Nov 26 '18 at 10:54
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58
Try
url([s'"]*K[^'"()?#]+. See live demo here– revo
Nov 26 '18 at 10:54
Try
url([s'"]*K[^'"()?#]+. See live demo here– revo
Nov 26 '18 at 10:54
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58
add a comment |
2 Answers
2
active
oldest
votes
1
You may use
url(s*(["']?)([^()?#]*)(?:[#?].*?)?1s*)
See the regex demo. The result will be in Group 2.
Details
url(-url(substring
s*- 0+ whitespaces
(["']?)- Group 1: an optional'or"
([^()?#]*)- Group 2: any 0+ chars other than?,#,)and(
(?:[#?].*?)?- an optional substring starting with?or#and then having any 0+ chars other than line break chars, as few as possible (to make it even more efficient, replace.*?here with[^()]*, cf. with this demo)
1- same value as captured into Group 1
s*- 0+ whitespaces
)- a)char.
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
add a comment |
0
If you're looking for a non-regex based option, I wrote this little function that can parse a URL or path and gives an object of the correct pieces:
function parseURL (url) {
url = decodeURI(url);
let split = {}
split["?"] = url.split("?")
split["path"] = split["?"][0]
split["tmp"] = split["?"][1].split("#")
split["#"] = split["tmp"][1]
split["?"] = split["tmp"][0]
let params = {};
split["?"].split("&").forEach(i => {
let tmp = i.split("=");
params[tmp[0]] = tmp[1];
})
return {
path: encodeURI(split["path"]),
params,
fragments: split["#"]
}
}
All you need to do is get the path property from the returned object.
$ node
> parseURL("https://example.com/path/to/resource?name=param&purpose=none#extrainfo")
> {
path: "https://example.com/path/to/resource",
params: {
name: "param",
purpose: "none"
},
fragments: "extrainfo"
}
>
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You may use
url(s*(["']?)([^()?#]*)(?:[#?].*?)?1s*)
See the regex demo. The result will be in Group 2.
Details
url(-url(substring
s*- 0+ whitespaces
(["']?)- Group 1: an optional'or"
([^()?#]*)- Group 2: any 0+ chars other than?,#,)and(
(?:[#?].*?)?- an optional substring starting with?or#and then having any 0+ chars other than line break chars, as few as possible (to make it even more efficient, replace.*?here with[^()]*, cf. with this demo)
1- same value as captured into Group 1
s*- 0+ whitespaces
)- a)char.
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
add a comment |
1
You may use
url(s*(["']?)([^()?#]*)(?:[#?].*?)?1s*)
See the regex demo. The result will be in Group 2.
Details
url(-url(substring
s*- 0+ whitespaces
(["']?)- Group 1: an optional'or"
([^()?#]*)- Group 2: any 0+ chars other than?,#,)and(
(?:[#?].*?)?- an optional substring starting with?or#and then having any 0+ chars other than line break chars, as few as possible (to make it even more efficient, replace.*?here with[^()]*, cf. with this demo)
1- same value as captured into Group 1
s*- 0+ whitespaces
)- a)char.
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
add a comment |
1
1
1
You may use
url(s*(["']?)([^()?#]*)(?:[#?].*?)?1s*)
See the regex demo. The result will be in Group 2.
Details
url(-url(substring
s*- 0+ whitespaces
(["']?)- Group 1: an optional'or"
([^()?#]*)- Group 2: any 0+ chars other than?,#,)and(
(?:[#?].*?)?- an optional substring starting with?or#and then having any 0+ chars other than line break chars, as few as possible (to make it even more efficient, replace.*?here with[^()]*, cf. with this demo)
1- same value as captured into Group 1
s*- 0+ whitespaces
)- a)char.
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
You may use
url(s*(["']?)([^()?#]*)(?:[#?].*?)?1s*)
See the regex demo. The result will be in Group 2.
Details
url(-url(substring
s*- 0+ whitespaces
(["']?)- Group 1: an optional'or"
([^()?#]*)- Group 2: any 0+ chars other than?,#,)and(
(?:[#?].*?)?- an optional substring starting with?or#and then having any 0+ chars other than line break chars, as few as possible (to make it even more efficient, replace.*?here with[^()]*, cf. with this demo)
1- same value as captured into Group 1
s*- 0+ whitespaces
)- a)char.
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
answered Nov 26 '18 at 9:38
Wiktor StribiżewWiktor Stribiżew
315k16133213
315k16133213
add a comment |
add a comment |
0
If you're looking for a non-regex based option, I wrote this little function that can parse a URL or path and gives an object of the correct pieces:
function parseURL (url) {
url = decodeURI(url);
let split = {}
split["?"] = url.split("?")
split["path"] = split["?"][0]
split["tmp"] = split["?"][1].split("#")
split["#"] = split["tmp"][1]
split["?"] = split["tmp"][0]
let params = {};
split["?"].split("&").forEach(i => {
let tmp = i.split("=");
params[tmp[0]] = tmp[1];
})
return {
path: encodeURI(split["path"]),
params,
fragments: split["#"]
}
}
All you need to do is get the path property from the returned object.
$ node
> parseURL("https://example.com/path/to/resource?name=param&purpose=none#extrainfo")
> {
path: "https://example.com/path/to/resource",
params: {
name: "param",
purpose: "none"
},
fragments: "extrainfo"
}
>
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
add a comment |
0
If you're looking for a non-regex based option, I wrote this little function that can parse a URL or path and gives an object of the correct pieces:
function parseURL (url) {
url = decodeURI(url);
let split = {}
split["?"] = url.split("?")
split["path"] = split["?"][0]
split["tmp"] = split["?"][1].split("#")
split["#"] = split["tmp"][1]
split["?"] = split["tmp"][0]
let params = {};
split["?"].split("&").forEach(i => {
let tmp = i.split("=");
params[tmp[0]] = tmp[1];
})
return {
path: encodeURI(split["path"]),
params,
fragments: split["#"]
}
}
All you need to do is get the path property from the returned object.
$ node
> parseURL("https://example.com/path/to/resource?name=param&purpose=none#extrainfo")
> {
path: "https://example.com/path/to/resource",
params: {
name: "param",
purpose: "none"
},
fragments: "extrainfo"
}
>
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
add a comment |
0
0
0
If you're looking for a non-regex based option, I wrote this little function that can parse a URL or path and gives an object of the correct pieces:
function parseURL (url) {
url = decodeURI(url);
let split = {}
split["?"] = url.split("?")
split["path"] = split["?"][0]
split["tmp"] = split["?"][1].split("#")
split["#"] = split["tmp"][1]
split["?"] = split["tmp"][0]
let params = {};
split["?"].split("&").forEach(i => {
let tmp = i.split("=");
params[tmp[0]] = tmp[1];
})
return {
path: encodeURI(split["path"]),
params,
fragments: split["#"]
}
}
All you need to do is get the path property from the returned object.
$ node
> parseURL("https://example.com/path/to/resource?name=param&purpose=none#extrainfo")
> {
path: "https://example.com/path/to/resource",
params: {
name: "param",
purpose: "none"
},
fragments: "extrainfo"
}
>
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
If you're looking for a non-regex based option, I wrote this little function that can parse a URL or path and gives an object of the correct pieces:
function parseURL (url) {
url = decodeURI(url);
let split = {}
split["?"] = url.split("?")
split["path"] = split["?"][0]
split["tmp"] = split["?"][1].split("#")
split["#"] = split["tmp"][1]
split["?"] = split["tmp"][0]
let params = {};
split["?"].split("&").forEach(i => {
let tmp = i.split("=");
params[tmp[0]] = tmp[1];
})
return {
path: encodeURI(split["path"]),
params,
fragments: split["#"]
}
}
All you need to do is get the path property from the returned object.
$ node
> parseURL("https://example.com/path/to/resource?name=param&purpose=none#extrainfo")
> {
path: "https://example.com/path/to/resource",
params: {
name: "param",
purpose: "none"
},
fragments: "extrainfo"
}
>
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
answered Nov 26 '18 at 10:37
Jacob SchneiderJacob Schneider
456316
456316
add a comment |
add a comment |
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Try
url([s'"]*K[^'"()?#]+. See live demo here– revo
Nov 26 '18 at 10:54
Reado, did you have time to check my approach?
– Wiktor Stribiżew
Nov 27 '18 at 7:58