I am using react 18 and react-location. I save some filter in the URL, and they are type string but effectively a number like "38" so url looks like id=32
Now, when I use
const search = useSearch<MakeGenerics<{ Search: { id: string } }>>();
what I actually get in code is a number 38 and not my string "38".
But I never asked react-location to actually convert my search parameters to a number, what I want is to get the string as my type is suggesting.
is there a way to avoid this behavior ?
How do I get the text from the li tag? I want to find the text "Password is required." only, not the text inside strong tag.
<li><strong>Error:</strong> Password is required.</li>
You need to show your code for somebody to give a complete answer. I guess that you already know how to do something like the following
WebElement something = driver.FindElement(By.CssSelector(?))
string s = something.Text;
The next bit seems to be where you are stuck. There you need to parse the string s. That is nothing to do with Selenium-Webdriver. You could do something like
string[] s2 = s.split(new string[] {">","<"});
were the last element in s2 would be your answer here. This would be totally non generic though. Is this a situation in which you always want to purge html?
Here is the method developed in python.
def get_text_exclude_children(element):
return driver.execute_script(
"""
var parent = arguments[0];
var child = parent.firstChild;
var textValue = "";
while(child) {
if (child.nodeType === Node.TEXT_NODE)
textValue += child.textContent;
child = child.nextSibling;
}
return textValue;""",
element).strip()
How to use in this:
liElement = driver.find_element_by_xpath("//li")
liOnlyText = get_text_exclude_children(liElement)
print(liOnlyText)
Please use your possible strategy to get the element, this method need an element from which you need the text (without children text).
The setup: A UITextField and a Tableview with suggested users
I try to have the following result:
I want users to be able to link other users.
Its working fine as long as I search with my last word in the array
let caption = captionTextView.text
let words = caption?.components(separatedBy: .whitespacesAndNewlines)
guard let searchingWord = words?.last else {return}
if searchingWord.hasPrefix("#") {
self.indicator.startAnimating()
let search = searchingWord.trimmingCharacters(in: CharacterSet.punctuationCharacters).lowercased()
}
But in case a user wants to adjust a username in the middle or at least not at the end of the array, the searching functions doesn't work properly as it still searches with the last word in the array
Example:
"Hey how are you #Lisa #Marcel #Thomas"
In case a user wants to change "#Lisa" to "#Lisbeth" the search function will search with Thomas as its the last word in the array
I wasn't able to get the word I am working at, only last and first words in the array, however I am able to get the current cursor location with
let cursor = captionTextView.cursorOffset!
which is an extension.
So how do I get the word I am working at up until the next "#" to the left und the next blank space to the right? Thanks in advance!
Maybe try something like this:
if let selectedRange = textview.selectedTextRange {
let cursorOffset = textview.offset(from: textview.beginningOfDocument, to: selectedRange.start)
let text = textview.text
let substring = text?.prefix(cursorOffset)
let editedWord = substring?.split(separator: "#")
}
(written on a phone, and untested)
One solution is Regular Expression
let string = "Hey how are you #Lisa #Marcel #Thomas"
let searchingWord = "Lisa"
let replacingWord = "Lisbeth"
let pattern = "#\(searchingWord)\\s"
string.replacingOccurrences(of: pattern, with: "#\(replacingWord) ", options: .regularExpression)
The pattern searches for # followed by the searching word followed by a whitespace character.
Since you say things are working the way you want if the last word is the one that has a username in it you just need to loop over all the words. Depending on your needs you may need to keep track of the usernames that were in the text before to save you from searching for the same user multiple times, but an array of used usernames should sort that for you.
Also, unless you want to prevent users from having underscores and the such in their names you should tweak the way in which you remove the # symbol as well.
guard let words = captionTextView.text?.components(separatedBy: .whitespacesAndNewlines) else { return }
for word in words where word.hasPrefix("#") {
self.indicator.startAnimating()
let search = word.replacingOccurrences(of: "#", with: "").lowercased()
}
Sticking the above code into a playground that uses the sample string you supplied in place of captionTextView.text? and printing search each time yielded…
lisa
marcel
thomas
Can someone tell me what "PathSegment model = info.getPathSegments().get(1);" do, specifically, what does he getPathSegments().get(1) mean? Please provide a sample URL for demonstration. The book didn't give an example URL for this one.
Also, is there such a thing as get(0); ?
#Path("/cars/{make}")
public class CarResource
{
#GET
#Path("/{model}/{year}")
#Produces("image/jpeg")
public Jpeg getPicture(#Context UriInfo info)
{
String make = info.getPathParameters().getFirst("make");
PathSegment model = info.getPathSegments().get(1);
String color = model.getMatrixParameters().getFirst("color");
...
}
}
Thanks again,
If you split the path of a URL by a '/' you'll get a list of path-segments. So e.g. the path /cars/ford/mustang/1976 contains the four segments [cars, ford, mustang, 1976]. info.getPathSegments().get(1) should return the segment ford.
The PathSegment holds also the associated MatrixParameters of the current segment. MatrixParameters can be used if you want to filter the resources with a parameter that affects only one segment like here:
/cars/ford/mustang;generation=two/1976
I wanna create an array in javascript which looks like this:
[0,0,0,0,0,1,1,0,0,1,1,1,1,0,0],[0,0,0,0,1,1,1,1,0,1,0,1,1,1,0],[0,0,0,0,1,1,1,1,0,1,0,0,1,1,0],[0,0,0,0,0,1,1,0,1,0,1,0,1,0,0]
My problem is that I don't know how to add the opening and closing square brackets to the start and the end of the output string.
here's my code:
game = new Array();
for(row=0;row<matrix.length;++row){
game[row]=matrix[row].join(',');
}
document.getElementById('jsvalue').value=game.join('],[');
document.getElementById('name2').value = name;
I tried a few things, but they didn't seem to work and all I got were errors or this output:
0,0,0,0,0,1,1,0,0,1,1,1,1,0,0],[0,0,0,0,1,1,1,1,0,1,0,1,1,1,0],[0,0,0,0,1,1,1,1,0,1,0,0,1,1,0],[0,0,0,0,0,1,1,0,1,0,1,0,1,0,0
How could I add them? Is there a simple array method that I missed and would solve my problem?
Thanks in advance!
It looks like you are trying to set the value of an HTML element to the format you described in your question. However, you are not setting the value of that HTML element to an Array - you are setting it to a string. the .join function outputs a string. If indeed you want the value to be set to a string formatted in the way you described, then you could take advantage of .join, but have to do a little bit in addition to what you are doing:
game = new Array();
for(row=0;row<matrix.length;++row){
game[row]= "[" + matrix[row].join(',') + "]";
}
document.getElementById('jsvalue').value=game.join(',');
document.getElementById('name2').value = name;
If you are using join to create the string, then why not just manually add the brackets?
For example:
document.getElementById('jsvalue').value= '[' + game.join('],[') + ']';