I have this string: https://2352353252142dsbxcs35#github.com/happy.git
I want to get result: https://github.com/happy.git (without random string after second / and after # but without #).
Now I have something like this:
var s = 'https://2352353252142dsbxcs35#github.com/happy.git';
var d = s.substring(s.indexOf('/')+2, s.indexOf('#')+1;
s = s.replace(d, "");
it works, but I know it's an ugly solution.
What is the most efficient and more universal solution?
Try this:
const indexOfAtSign: number = receivedMessage.indexOf('#')+1
const httpsString: string = 'https://'
const trimmedString: string = s.slice(indexOfAtSign)
const requiredURL: string = httpsString.concat(trimmedString)
// Print this value of requiredURL wherever you want.
So here what my code does is, it gets position of # and removes everything before it along with the sign itself. Then using the slice() function, we are left with the remaining part which I named as trimmedString. Now I have pre-defined the `https string, anf we just need to merge them now. Done :-)
I had tried this out in my telegram bot and here's how it works:
Goal: To write the file content in json format using Node js. Upon opening the file manually, content should be displayed in json format
I tried both fs-extra module functions - outputJsonSync or writeFileSync to write json content to a file. They write the content inline as below
{"a":"1", "b":"2"}
However, I would like to see the content as below when I open the file manually:
{
"a" : "1",
"b" : "2"
}
I tried jsome and pretty-data on the data as follows:
fs.outputJsonSync(jsome(data))
fs.outputJsonSync(pd.json(data))
They also write data inline only with extra \ or \n and tabs added to the data but doesn't open in formatted style.
Any inputs are highly appreciated. Thanks!
[Update]
Other scenario:
const obj = {"a":"1", "b":"2"}
var string = "abc" + "splitIt" + obj
doSomething(string)
And inside the function implementation:
doSomething(string){
var arr = string.split("splitIt")
var stringToWrite = JSON.stringify(arr[1], null, ' ').replace(/: "(?:[^"]+|\\")*"$/, ' $&')
fs.writeFileSync(filePath, stringToWrite)
}
Following output is displayed when I open the file:
"[object Object]"
Once you have the object, you can specify a replacer function to separate each key-value pair by a newline, then use a regular expression to trim the leading spaces, then use another regular expression to insert a space before the : of a key-value pair. Then, just write the formatted string to a file:
const obj = {"a":"1", "b":"2"};
const stringToWrite = JSON.stringify(obj, null, ' ')
// Trim leading spaces:
.replace(/^ +/gm, '')
// Add a space after every key, before the `:`:
.replace(/: "(?:[^"]+|\\")*",?$/gm, ' $&');
console.log(stringToWrite);
Though, you may find the leading spaces more readable:
const obj = {"a":"1", "b":"2"};
const stringToWrite = JSON.stringify(obj, null, ' ')
// Add a space after every key, before the `:`:
.replace(/: "(?:[^"]+|\\")*",?$/gm, ' $&');
console.log(stringToWrite);
JSON.stringify, has two optional parameters, the first one being a replacer function, the second one(What you want) is for spacing.
const obj = {"a":"1", "b":"2"}
console.log(JSON.stringify(obj, null, 2))
This will give you:
{
"a": "1",
"b": "2"
}
In sourceConfigPath variable, it has a path like "conf/test.json", or it might have another layer, like "test/conf/test.json". I want to get the "test.json" part only.
I tried the indexOf function to get the position, then use either slice or substr function to get the 'test.json' part. But it always return 0 when do indexOf.
Can anyone please help here? many thanks!
var position = sourceConfigPath.indexOf('conf');
var newsourceConfigPath = sourceConfigPath.slice(position+4);
Or is there any better way to do this? Many thanks!
The best way is to use path.basename
The path.basename() methods returns the last portion of a path,
similar to the Unix basename
const path = require('path');
const newSource = path.basename('conf/test.json'); // test.json
You can use lastIndexOf instead of indexOf, but path.basename is recommended.
const filepath = '/path/to/file.json';
const position = filepath.lastIndexOf('/') + 1; // +1 is to remove '/'
console.log(filepath.substr(position));
Say I have a URL
http://example.com/query?q=
and I have a query entered by the user such as:
random word £500 bank $
I want the result to be a properly encoded URL:
http://example.com/query?q=random%20word%20%A3500%20bank%20%24
What's the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right.
URLEncoder is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);
When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString() as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8".
Note that spaces in query parameters are represented by +, not %20, which is legitimately valid. The %20 is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?), not in query string (the part after ?).
Also note that there are three encode() methods. One without Charset as second argument and another with String as second argument which throws a checked exception. The one without Charset argument is deprecated. Never use it and always specify the Charset argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.
All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
See also:
What every web developer must know about URL encoding
I would not use URLEncoder. Besides being incorrectly named (URLEncoder has nothing to do with URLs), inefficient (it uses a StringBuffer instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.
Instead I would use URIBuilder or Spring's org.springframework.web.util.UriUtils.encodeQuery or Commons Apache HttpClient.
The reason being you have to escape the query parameters name (ie BalusC's answer q) differently than the parameter value.
The only downside to the above (that I found out painfully) is that URL's are not a true subset of URI's.
Sample code:
import org.apache.http.client.utils.URIBuilder;
URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();
// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24
You need to first create a URI like:
String urlStr = "http://www.example.com/CEREC® Materials & Accessories/IPS Empress® CAD.pdf"
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
Then convert that URI to an ASCII string:
urlStr = uri.toASCIIString();
Now your URL string is completely encoded. First we did simple URL encoding and then we converted it to an ASCII string to make sure no character outside US-ASCII remained in the string. This is exactly how browsers do it.
Guava 15 has now added a set of straightforward URL escapers.
The code
URL url = new URL("http://example.com/query?q=random word £500 bank $");
URI uri = new URI(url.getProtocol(), url.getUserInfo(), IDN.toASCII(url.getHost()), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String correctEncodedURL = uri.toASCIIString();
System.out.println(correctEncodedURL);
Prints
http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$
What is happening here?
1. Split URL into structural parts. Use java.net.URL for it.
2. Encode each structural part properly!
3. Use IDN.toASCII(putDomainNameHere) to Punycode encode the hostname!
4. Use java.net.URI.toASCIIString() to percent-encode, NFC encoded Unicode - (better would be NFKC!). For more information, see: How to encode properly this URL
In some cases it is advisable to check if the URL is already encoded. Also replace '+' encoded spaces with '%20' encoded spaces.
Here are some examples that will also work properly
{
"in" : "http://نامهای.com/",
"out" : "http://xn--mgba3gch31f.com/"
},{
"in" : "http://www.example.com/‥/foo",
"out" : "http://www.example.com/%E2%80%A5/foo"
},{
"in" : "http://search.barnesandnoble.com/booksearch/first book.pdf",
"out" : "http://search.barnesandnoble.com/booksearch/first%20book.pdf"
}, {
"in" : "http://example.com/query?q=random word £500 bank $",
"out" : "http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$"
}
The solution passes around 100 of the test cases provided by Web Platform Tests.
Using Spring's UriComponentsBuilder:
UriComponentsBuilder
.fromUriString(url)
.build()
.encode()
.toUri()
The Apache HttpComponents library provides a neat option for building and encoding query parameters.
With HttpComponents 4.x use:
URLEncodedUtils
For HttpClient 3.x use:
EncodingUtil
Here's a method you can use in your code to convert a URL string and map of parameters to a valid encoded URL string containing the query parameters.
String addQueryStringToUrlString(String url, final Map<Object, Object> parameters) throws UnsupportedEncodingException {
if (parameters == null) {
return url;
}
for (Map.Entry<Object, Object> parameter : parameters.entrySet()) {
final String encodedKey = URLEncoder.encode(parameter.getKey().toString(), "UTF-8");
final String encodedValue = URLEncoder.encode(parameter.getValue().toString(), "UTF-8");
if (!url.contains("?")) {
url += "?" + encodedKey + "=" + encodedValue;
} else {
url += "&" + encodedKey + "=" + encodedValue;
}
}
return url;
}
In Android, I would use this code:
Uri myUI = Uri.parse("http://example.com/query").buildUpon().appendQueryParameter("q", "random word A3500 bank 24").build();
Where Uri is a android.net.Uri
In my case I just needed to pass the whole URL and encode only the value of each parameters.
I didn't find common code to do that, so (!!) so I created this small method to do the job:
public static String encodeUrl(String url) throws Exception {
if (url == null || !url.contains("?")) {
return url;
}
List<String> list = new ArrayList<>();
String rootUrl = url.split("\\?")[0] + "?";
String paramsUrl = url.replace(rootUrl, "");
List<String> paramsUrlList = Arrays.asList(paramsUrl.split("&"));
for (String param : paramsUrlList) {
if (param.contains("=")) {
String key = param.split("=")[0];
String value = param.replace(key + "=", "");
list.add(key + "=" + URLEncoder.encode(value, "UTF-8"));
}
else {
list.add(param);
}
}
return rootUrl + StringUtils.join(list, "&");
}
public static String decodeUrl(String url) throws Exception {
return URLDecoder.decode(url, "UTF-8");
}
It uses Apache Commons' org.apache.commons.lang3.StringUtils.
Use this:
URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());
or this:
URLEncoder.encode(query, "UTF-8");
You can use the following code.
String encodedUrl1 = UriUtils.encodeQuery(query, "UTF-8"); // No change
String encodedUrl2 = URLEncoder.encode(query, "UTF-8"); // Changed
String encodedUrl3 = URLEncoder.encode(query, StandardCharsets.UTF_8.displayName()); // Changed
System.out.println("url1 " + encodedUrl1 + "\n" + "url2=" + encodedUrl2 + "\n" + "url3=" + encodedUrl3);
I am trying to split a string into individual characters.
The string I want to split: let lastName = "Kocsis" so that is returns something like: ["K","o","c","s","i","s"]
So far I have tried:
var name = lastName.componentsSeparatedByString("")
This returns the original string
name = lastName.characters.split{$0 == ""}.map(String.init)
This gives me an error: Missing argument for parameter #1 in call. So basically it does't accept "" as an argument.
name = Array(lastName)
This does't work in Swift2
name = Array(arrayLiteral: lastName)
This doesn't do anything.
How should I do this? Is There a simple solution?
Yes, there is a simple solution
let lastName = "Kocsis"
let name = Array(lastName.characters)
The creation of a new array is necessary because characters returns String.CharacterView, not [String]