I want to create an url (not a link) for a text field for a share functionality.
Like for example the one stackoverflow uses
I already hava function that produces the path part for the url, like
toUrl : Route -> String
toUrl route = ...
toUrl (Home (Just "hallo")) --> "/?b=hallo"
and using this string for a link as a href attribute works, but I'm wondering how I could create a complete url from this string.
PS: I'm using a single page application so I get an Url at the beginning.
What you say is "the path part for the url" isn't actually just the path part, but the path and query parts of the URL. Ideally you'd separate them so you can create a well-formed URL representation:
{ initialUrl
| path = "/"
, query = Just "b=hallo"
}
But since it's just a record, with no validation, it'll work if you just use it as the path as well. At least if you later just use Url.toString on it. Other operations might cause unexpected results.
{ initialUrl | path = "/?b=hallo" }
Related
I am trying to obtain the name of a file (JSON format but saved without an extension) within the last directory of a given path. Each file is saved with its own unique subpath inside the app's data container.
I also need to get the full path of the file, including the filename.
From what I've read, I believe it is better to use URLs to do this rather than using string paths.
I have tried the following code:
do {
let enumerator = FileManager.default.enumerator(at: filePath, includingPropertiesForKeys: nil)
while let element = enumerator?.nextObject() as? URL {
var nexObject = element.lastPathComponent
print(nextObject)
}
} catch let error {
print(error.localizedDescription)
}
This does seem to iterate through each level of the path until the end. Great, but what is the best way to get the full path, including the filename, other than concatenation of each object from the above?
All advice gratiously received. Thanks!
As element is an URL, if you're interested in the full path name rather than the last component, just go for:
var nextObject = element.absoluteURL // instead of .lastPathComponent
or just
var nextObject = element.path // or even relativePath
Thank you, #Christophe (+1)
I've also since spotted that the documentation for enumerator(at:includingPropertiesForKeys:options:errorHandler:) provides a nice example, which can be modifed for my purposes by using additional resource keys (e.g. name, path, etc.).
I am currently trying to figure out, how to modify the parameter being integrated into the URL Mapping I am using.
static mappings =
{
"/$controller/$action?/$id?/(.$format)?"
{
constraints {
// apply constraints here
}
}
name test1: "/.../$title/..."{
controller = "study"
action = "st_show"
}
name test2: "/.../$title/..."{
controller = "search"
action = "se_show"
}
The parameter $title is pretty much a dataset, which is pulled from a database and which will get transmitted in the following format [ this is a title ]. So there are square brackets in front and behind the string and words are seperated through blanks.
If I am creating a link through g:link now with the params nested in, it gets put into the url as it is pulled from the database. What I am attempting is to create SEO-URLs, which will present a certain title of a publication devided by hyphens instead of url-encoded "%20".
Until now, I was able to generate dynamic urls looking like this:
http://localhost:8080/projectname/show/%5BAllgemeine%20Bevölkerungs[...]/782/...PARAMS...
Furthermore I already implemented it through JQuery, though it should be static and users should be able to copy the link to open up the page themselves - that wouldn't be possible when changing the url client-side while loading up the page.
Is there a way to define a function with something like replaceAll.(' ', '-'), which can be invoked onto the parameter in the mapping to replace blanks with hyphens and f.e. square brackets with an empty character?
That's pretty much, what I wasn't able to come by through the documentation.
Thank you already in advance for your help!
I managed to solve my problem by creating a service with a function containing a regex and executing this function onto the parameter title in my g:link, which I firstly converted to a string, which gets passed to the function.
<g:link controller="study" action="st_show" params="[data: data, ... title: ConversionService.convert(fieldValue(bean: path).toString(), ... data: data)]"></g:link>
And the function in the ConversionService
public static String convert(String title){
title = title.replaceAll("\\s", "-").replaceAll("[^0-9a-zA-Z\\-]", "");
return title;
}
This url:
url = rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id
I would like to add the extension mp3 to all file name.
In this case the file name is cancion
The id of lesson is a dynamic value.
I would like to get this url something like:
url = rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion.mp3?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id
Thanks!
You can parse the URI, edit the path, then return the value
require 'uri/http'
u = URI.parse('rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id')
u.path += ".mp3"
puts u.to_s
or use a simple regexp replace
u = 'rtmp://xxxxxxxxxxxxxx.cloudfront.net/cfx/st/mp3:audios/lesson/2/cancion?Expires=1386332537&Signature=mysignature__&Key-Pair-Id=my-key-par-id'
u.gsub('?', '.mp3?')
The second approach can be used only if you can assume the format of the input is always the same.
You can do simple gsub since this is URL and you can expect one occurrence of ? so simple do.
url.gsub!('?', '.mp3?')
Usually I would go regex here but no need from previously stated reason.
I am using
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = res.getPath();
String path2 = path.substring(1);
because the output of the method getPath() returns sth like this:
/C:/Users/......
and I need this
C:/Users....
I really need the below address because some external library refuses to work with the slash at the beginning or with file:/ at the beginning or anything else.
I tried pretty much all the methods in URL like toString() toExternalPath() etc. and done the same with URI and none of it returns it like I need it. (I totally don't understand, why it keeps the slash at the beginning).
It is okay to do it on my machine with just erasing the first char. But a friend tried to run it on linux and since the addresses are different there, it does not work...
What should with such problem?
Convert the URL to a URI and use that in the File constructor:
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
File file = new File(res.toURI());
String fileName = file.getPath();
As long as UNIX paths are not supposed to contain drive letters, you may try this:
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = res.getPath();
char a_char = text.charAt(2);
if (a_char==':') path = path.substring(1);
Convert to a URI, then use Paths.get().
URL res = this.getClass().getClassLoader().getResource(dictionaryPath);
String path = Paths.get(res.toURI()).toString();
You could probably just format the string once you get it.
something like this:
path2= path2[1:];
I was searching for one-line solution, so the best what i came up with was deleting it manually like this:
String url = this.getClass().getClassLoader().getResource(dictionaryPath).getPath().replaceFirst("/","");
In case if someone also needs to have it on different OS, you can make IF statement with
System.getProperty("os.name");
I am having trouble with building a URL with a query string. I have this code that does what I want it to:
formatoptions: { baseLinkUrl: '#Url.Action("UserInformation", "UserList")', idName: 'Id' }
This makes the proper URL (/UserInformation?Id=4)
This is the section that I am having trouble with, basically trying to replicate what is above, but the syntax is different and I am not sure what's wrong.
results.Add(New SearchResult With {.Link = Url.Action("UserInformation", "UserList", New With {.id = use.Id}), .Text = use.ToString, .Type = "User"})
This make the URL a bit off (/UserInformation/4), it causes problems when redirecting from that page. I'd like to edit this to replicate the proper URL string.
This is a bit of a hack that we figured out to make it work..but I'd like to do it 'properly' if possible
results.Add(New SearchResult With {.Link = Url.Content("~/UserList/UserInformation?Id=" & use.Id), .Text = use.ToString, .Type = "User"})
It's because of your default route having the Id as optional. You can either change the default route which might make all your other Url's look not as clean, or just pick a different parameter name for the Id on the UserInformation action such as userId.