I am working on an Angular app and having a bit of a problem.
I am trying to test my API by appending a string into a URL.
It works fine when I hardcode the string into the URL but when I append it won't work.
this is a function that will get the string that I want to append.
getString(str: string){
this.strAppend = str
}
this is the URL,
url: string = http://localhost:3000/document/id/${this.strAppend}/transaction?from=1610742245&to=1623439932
notice how I use this.strAppend. Well, this is not working. Is this even the right approach?
You can use Template Literals to solve your problem.
var base = 'url'
getString(strToAdd: string) {
return `${base}/${strToAdd}`;
}
var newStr = getString('test');
First declare the variable in string
for time being refer this
$scope.str1 = 'This is ';
$scope.str2 = 'Sticked Toghether';
$scope.res = '';
$scope.join = function() {
$scope.res = $scope.str1 + $scope.str2;
};
Related
I have a code like this :
// Language = Dart
var someVariable = 'Hello';
var someOtherVariable = 'World';
var str = 'somedomain?x=${someVariable}&y=${someOtherVariable}';
return str;
// Expected:
// somedomain?x=Hello&y=World;
// Actual
// somedomain?x=Hello
If I replace the & character with any alphabets, it is able to successfully concatenate. What am I doing wrong.
This is the actual code which I used in FlutterFlow, and am having issues with:
Future<String> getEventUrlFromReference(BuildContext context, DocumentReference? eventReference) async {
var userId = currentUser?.uid as String;
return "https://somedomain.com/event?eventReference=${eventReference?.id}" + "&invitedBy="+userId;
}
// result: https://somedomain.com/event?eventReference=referencevalue
This was a string encoding issue. I was using the result of my function/code as body text in sms://<number>?&body=<string_containigng_&_character>; The text which is appended to the sms text truncates at the & character, and I made a mistake assuming it's a string concatenation issue.
const test ="[{contactId=2525, additionDetail=samle}]";
I need to convert this string to a JSON object. It will dynamically load like this string. I need to particular string to convert to a JSON object.
JSON.parse(test) command not working for this. I attached the error here.
For that specific string, you'd have to parse it yourself.
const test = '[{contactId=2525, additionDetail=samle}]';
const obj = {};
test.split(/[{}]/)[1].split(/, /).forEach((elm) => {
const entry = elm.split('=');
obj[entry[0]] = entry[1];
});
What I am doing is splitting the string on the braces and selecting the second element (utilising regex) then splitting that on comma and space (again regex) then loop over the result and assign to an object.
You can then JSON.stringify(obj) for the result.
:edit:
For the second string you've asked for there is another, potentially more refined, answer. You'll need to first replace the = with : (I've again used a regex), then you use a regex to match the words and sentence and use a function to add the quotes.
const test = '[{contactId=2525, additionDetail=samle}]';
const test2 = "[{contactId=2525, additionDetail=rrr additional Detail, medicationType={medicationTypeId=3333, medicationType=Tablet}, endDate=2022-12-30}]";
const replaced = test.replace(/=/g,':')
const replaced2 = test2.replace(/=/g, ':');
const replacer = function(match){
return '"' + match + '"';
}
const replacedQuote = replaced.replace(/(?!\s)[-?\w ?]+/g,replacer);
const replaced2Quote = replaced2.replace(/(?!\s)[-?\w ?]+/g,replacer);
const obj = JSON.parse(replacedQuote);
const obj2 = JSON.parse(replaced2Quote);
You should note that Json means javascript object notation, so you need to create a JavaScript object to get started:
const test ="[{contactId=2525, additionDetail=samle}]";
let obj = Object.create(null)
You can now define your variable as one of the object properties :
obj.test = test
Now we have a JavaScript object and we can convert it to json:
let convertedToJson = JSON.stringify(test);
[{contactId=2525, additionDetail=samle}]
this is not a valid JSON-string, and it cannot be parsed by JSON.parse()
the correct JSON-string would be:
const test ='[{"contactId":2525, "additionDetail":"samle"}]';
I need to take 'href'(tag link location value) value from following pattern html text. need some expert help to do it using typescript
String Text one
"<html><body>docker_command.txt </body></html>"
String Text Two
"<html><body>https://www.facebook.com/ </body></html>"
Something like this?
const a = '"<html><body>docker_command.txt </body></html>"';
const b = '"<html><body>https://www.facebook.com/ </body></html>"';
function getHref(html: string): string|null {
if (!html) {
return null;
}
return html.match(/ href=("|')([^'"]*?)('|")/i)[2];
}
console.log(getHref(a));
console.log(getHref(b));
Please can you tell me how to make shorter this code. i wanna do it in one line.
var htmlstring = "{{1}}Hello {{2}}, clic here !";
var firststep = htmlstring.replace('{{1}}', "http://google.fr");
var secondstep = htmlstring.replace('{{1}}', "http://google.fr");
var thirdstep = secondstep.replacee('{{2}}', "Mister");
In short,
I have it:
{{1}}Hello {{2}}, clic here !
I wanna have this at the end:
http://google.frHello Mister, clic here !"
Sure, use String.replace and regexp:
const html = '{{1}}Hello {{2}}, click here !'
const url = 'http://google.fr';
const name = 'Mister';
const output = html.replace(/\{\{1\}\}/g, url).replace(/\{\{2\}\}/g, name);
I have no Idea how you can replace both {{1}} and {{2}} in one go, but you can replace both occurences of {{1}} by using this function:
String.prototype.replaceAll = function(search, replacement) {
var target = this;
return target.replace(new RegExp(search, 'g'), replacement);
};
Use it like this:
var htmlstring = '{{1}}Hello {{2}}, clic here !';
htmlstring.replaceAll("{{1}}", "http://google.fr");
htmlstring.replaceAll("{{2}}", "Mister");
ES6 Update:
You can now use an even better function for the replaceAll() method:
String.prototype.replaceAll = function(search, replacement) {
return this.split(search).join(replacement);
};
This will utilize .split() to first make an array without the search value and then .join() that Array with the replacement:
String.prototype.replaceAll = function(search, replacement) {return this.split(search).join(replacement);};
let htmlstring = `{{1}}Hello {{2}}, click here !`;
console.log(htmlstring.replaceAll("{{1}}","https://google.fr").replaceAll("{{2}}", "Mister"));
I was using this, in Swift 1.2
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
This now gives me a warning asking me to use
stringByAddingPercentEncodingWithAllowedCharacters
I need to use a NSCharacterSet as an argument, but there are so many and I cannot determine what one will give me the same outcome as the previously used method.
An example URL I want to use will be like this
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA
The URL Character Set for encoding seems to contain sets the trim my
URL. i.e,
The path component of a URL is the component immediately following the
host component (if present). It ends wherever the query or fragment
component begins. For example, in the URL
http://www.example.com/index.php?key1=value1, the path component is
/index.php.
However I don't want to trim any aspect of it.
When I used my String, for example myurlstring it would fail.
But when used the following, then there were no issues. It encoded the string with some magic and I could get my URL data.
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
As it
Returns a representation of the String using a given encoding to
determine the percent escapes necessary to convert the String into a
legal URL string
Thanks
For the given URL string the equivalent to
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
is the character set URLQueryAllowedCharacterSet
let urlwithPercentEscapes = myurlstring.stringByAddingPercentEncodingWithAllowedCharacters( NSCharacterSet.URLQueryAllowedCharacterSet())
Swift 3:
let urlwithPercentEscapes = myurlstring.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed)
It encodes everything after the question mark in the URL string.
Since the method stringByAddingPercentEncodingWithAllowedCharacters can return nil, use optional bindings as suggested in the answer of Leo Dabus.
It will depend on your url. If your url is a path you can use the character set
urlPathAllowed
let myFileString = "My File.txt"
if let urlwithPercentEscapes = myFileString.addingPercentEncoding(withAllowedCharacters: .urlPathAllowed) {
print(urlwithPercentEscapes) // "My%20File.txt"
}
Creating a Character Set for URL Encoding
urlFragmentAllowed
urlHostAllowed
urlPasswordAllowed
urlQueryAllowed
urlUserAllowed
You can create also your own url character set:
let myUrlString = "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red Lion&location=19036&location=1090 N Charlotte St, Lancaster, PA"
let urlSet = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
extension CharacterSet {
static let urlAllowed = CharacterSet.urlFragmentAllowed
.union(.urlHostAllowed)
.union(.urlPasswordAllowed)
.union(.urlQueryAllowed)
.union(.urlUserAllowed)
}
if let urlwithPercentEscapes = myUrlString.addingPercentEncoding(withAllowedCharacters: .urlAllowed) {
print(urlwithPercentEscapes) // "http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA"
}
Another option is to use URLComponents to properly create your url
Swift 3.0 (From grokswift)
Creating URLs from strings is a minefield for bugs. Just miss a single / or accidentally URL encode the ? in a query and your API call will fail and your app won’t have any data to display (or even crash if you didn’t anticipate that possibility). Since iOS 8 there’s a better way to build URLs using NSURLComponents and NSURLQueryItems.
func createURLWithComponents() -> URL? {
var urlComponents = URLComponents()
urlComponents.scheme = "http"
urlComponents.host = "www.mapquestapi.com"
urlComponents.path = "/geocoding/v1/batch"
let key = URLQueryItem(name: "key", value: "YOUR_KEY_HERE")
let callback = URLQueryItem(name: "callback", value: "renderBatch")
let locationA = URLQueryItem(name: "location", value: "Pottsville,PA")
let locationB = URLQueryItem(name: "location", value: "Red Lion")
let locationC = URLQueryItem(name: "location", value: "19036")
let locationD = URLQueryItem(name: "location", value: "1090 N Charlotte St, Lancaster, PA")
urlComponents.queryItems = [key, callback, locationA, locationB, locationC, locationD]
return urlComponents.url
}
Below is the code to access url using guard statement.
guard let url = createURLWithComponents() else {
print("invalid URL")
return nil
}
print(url)
Output:
http://www.mapquestapi.com/geocoding/v1/batch?key=YOUR_KEY_HERE&callback=renderBatch&location=Pottsville,PA&location=Red%20Lion&location=19036&location=1090%20N%20Charlotte%20St,%20Lancaster,%20PA
In Swift 3.1, I am using something like the following:
let query = "param1=value1¶m2=" + valueToEncode.addingPercentEncoding(withAllowedCharacters: .alphanumeric)
It's safer than .urlQueryAllowed and the others, because it this will encode every characters other than A-Z, a-z and 0-9. This works better when the value you are encoding may use special characters like ?, &, =, + and spaces.
In my case where the last component was non latin characters I did the following in Swift 2.2:
extension String {
func encodeUTF8() -> String? {
//If I can create an NSURL out of the string nothing is wrong with it
if let _ = NSURL(string: self) {
return self
}
//Get the last component from the string this will return subSequence
let optionalLastComponent = self.characters.split { $0 == "/" }.last
if let lastComponent = optionalLastComponent {
//Get the string from the sub sequence by mapping the characters to [String] then reduce the array to String
let lastComponentAsString = lastComponent.map { String($0) }.reduce("", combine: +)
//Get the range of the last component
if let rangeOfLastComponent = self.rangeOfString(lastComponentAsString) {
//Get the string without its last component
let stringWithoutLastComponent = self.substringToIndex(rangeOfLastComponent.startIndex)
//Encode the last component
if let lastComponentEncoded = lastComponentAsString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.alphanumericCharacterSet()) {
//Finally append the original string (without its last component) to the encoded part (encoded last component)
let encodedString = stringWithoutLastComponent + lastComponentEncoded
//Return the string (original string/encoded string)
return encodedString
}
}
}
return nil;
}
}
Swift 4.0
let encodedData = myUrlString.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlHostAllowed)