Linkify Android Transform Question - android

Trying to do something fairly simple.
Taking text like this
User Name: This is a comment I am making
It is in a single TextView. I want to make the User Name a link. I decided that the easiest thing would be to surround the User Name with something like "$#" so it becomes
"$#User Name:$# This is a comment I am making
That way I can use the following regular expression
Pattern userName = Pattern.compile(".*\\$#(.+)\\$#.*");
with Linkify and make it a link. However, clearly I need to remove the delimiters, so the following is the code
title.setText(titleText);
Linkify.TransformFilter transformer = new Linkify.TransformFilter() {
#Override
public String transformUrl(Matcher match, String url) {
return match.group(1);
}
};
Linkify.addLinks(title, userName, "content://user=", null, transformer);
For some reason however, the whole text becomes one giant link, and the text isn't being transformed at all.

It actually did turned out to be pretty easy. I ended up not using the crazy "$#" to delimit the username, instead sticking with just
User Name: This is a comment I am making
so I ended up using the following pattern
Pattern userName = Pattern.compile("(.+:)");
Very simple, and the code becomes just
title.setText(titleText);
Linkify.addLinks(title, GlobalUtil.userName, "user://" + userId + "/");
Thank you to nil for the original suggestion. I was indeed matching the whole string instead of just the userName which is the link.

My best guess is the regex you're using is the problem, where you're telling it to basically pick out the entire string if it matches, including everything before and after what you're looking for. So, the TransformFilter is probably being passed the entire matched string. transformUrl as far as I can tell expects you to return the URL, so the entire string is linked to the first match group.
So, with that in mind, it's probably in your best interest to change the regex to something along the lines of "\\$#(.+?)\\$#" (with an added ? in the group to make the match non-greedy) so as to avoid matching the entire string and just picking out the bit you want to URL-ize (for lack of a better term, plus adding -ize to words sounds cool).

Why not put the delimiters inside the pattern to change ?
Pattern userName = Pattern.compile(".*(\\$#.+\\$#).*");
Then change the transform filter to remove the start and end patterns when changing into the URL...

Related

String to text field in android app

I have the following data scanned from a pdf417 and need to extract certain text to certain text fields (already created), not sure how to go about this... Data scanned with manatee works plugin and android app using android studio.
All help will be appreciated.
Data that was returned from scan -
%MVL1CC18%0154%4025M003%4025012RP01C%DC62XBGP%NISSAN%SILVER/SILWER%
Each part between the %'s need to go to a text field. I know that I need to make use of String substr=mysourcestring.substring(startIndex,endIndex); but this will work up to the first 2 % signs. How do I continue to the next few?
Thanks.
If you want to split string based on a delimiter, use the following
String delimitter="%";
String[] parts = inputString.split(delimitter);
Why not use String.split()?
In your case it would look something like this:
String[] extractedStrings = mysourcestring.split("%");
You can work on your string by using split method:
String yourString = "%MVL1CC18%0154%4025M003%4025012RP01C%DC62XBGP%NISSAN%SILVER/SILWER%";
String[] split = yourString.split("%");
In this way you will get an array where each item is a substring between two % chars.

output text line by line

I write app for Android such gets data from server in JSON format. Now I get this value in string, but in my application it must look like:
Route:
1)first point
2)secon point
3).....
n) n point
I read that in Android in textView I can do it if string will be with html tags but I think it is not the best variant. After Android I must do it in iPhone now I don't know how to do that there. Send Routes as Array is not good variant too. Can you say what is the best way to decide this problem?
Have a look here you will have to find the good pattern .
Hence you have separated strings just use a list View with an ArrayAdapter.
I am not so good with regex but i think it should like : [1-9][0-9]) [[a-f][0-9]]+
I couldn't comment b/c of rep, sorry. Could you provide an example of returned JSON string. I think JSON format can be parsed with ease.
If this the case you can parse it in a loop (or another way. I'm not that good at it)
String[] parseIt (String JSON){
String[] list=JSON.split("\\d\\)");
String[] rlist=new String[list.length-1];
for(int i=0;i<list.length-1;i++){
rlist[i]=list[i+1].trim();
}
return rlist;
}
This might do trick. But you should edit result. I didn't test yet
Edit: I edited code. It simply return the address now with leading whitespace. You can get rid off them using. String trim() method like;
list[1].trim();
Do it in loop and don't care about first element (index 0).
Edit 2: Now it should work

Android - Make String SQL Safe

My CO-worker has created the below code for iOS to make a user defined string SQL search safe.
i.e to remove the possibility of SQL-injections and that kind of thing
char * sqlSafeQuery = sqlite3_mprintf("%q",[searchTerm UTF8String]);
searchTerm = [NSString stringWithFormat:#"%s", sqlSafeQuery];
sqlite3_free(sqlSafeQuery);
Is there a way in android to so something similar? I can't see much on the Google's :)
Thanks.
You may use prepared statements, see an Android question on how to use these in sqlite. As the statement is parsed without the user input in it, the parameters cannot change the statement itself; thus the injection cannot happen.

Scan string for characters and return bounded text

I am writing a dictionary-type app. I have a list of hash-mapped terms and definitions. The basic premise is that there is a list of words that you tap on to see the definitions.
I have this functionality up and running - I am now trying to put dynamic links between the definitions.
Example: say the user taps on an item in the list, "dog". The definition might pop up, saying "A small furry [animal], commonly kept as a pet. See also [cat].". The intention is that the user can click on the word [animal] or [cat] and go to the appropriate definition. I've already gone to the trouble of making sure that any links in definitions are bounded by square brackets, so it's just a case of scanning the pop-up string for text [surrounded by brackets] and providing a link to that definition.
Note that definitions can contain multiple links, whilst some don't contain any links.
I have access to the string before it is displayed, so I guess the best way to do this is to do the scanning and ready the links before the dialog box is displayed.
The question is, how would I go about scanning for text surrounded by square brackets, and returning the text contained within those brackets?
Ideally the actual dialog box that is displayed would be devoid of the square brackets, and I need to also figure out a way of putting hyperlinks into a dialog box's text, but I'll cross that bridge when I come to it.
I'm new to Java - I've come from MATLAB and am just about staying afloat, but this is a less common task than I've had to deal with so far!
You could probably do this with a regular expression; something like this:
([^[]*)(\[[^]]+\])
which describes two "match groups"; the first of which means any string of zero or more characters that aren't "[" and the second of which means any string starting with "[", containing one or more characters that aren't "]", and ending with "]".
Then you could scan through your input for matches to this pattern. The first match group is passed through unchanged, and the second match group gets converted to a link. When the pattern stops matching your input, take whatever's left over and transmit that unchanged as well.
You'll have to experiment a little; regular expressions typically take some debugging. If your link text can only contain alphanumerics and spaces, your pattern would look more like this:
([^[]*)(\[[\s\w]+\])
Also, you may find that regular expression matching under Android is too slow to be practical, in which case you'll have to use wasyl's suggestion.
Quite simple, I think... As the text is in brackets, you need to scan every letter. So the basic recipe would be :
in a while loop scan every character (let's say, while i < len(text))
If scanned character is [:
i++;
Add letter at index i to some temporary variable
while (character # i) != ']' append it to the temporary variable
store this temporary variable in a list of results.
Some tips:
If you use solution above, use StringBuilder to append text (as regular string is immutable)
You might also want (and it's better, I think) to store starting and ending positions of all square brackets first, and then use string.substring() on each pair to get the text inside. This way you'd first iterate definition to find brackets (maybe catch unmatched ones, for early error handling), then iterate pairs of indices...
As for links, maybe this will be of use: How can I get clickable hyperlinks in AlertDialog from a string resource?

Only get part of string in android?

Okay I wasn't really sure how to word this question, but basically what I want to do is, I got a url from a webView in android, and I need to put part of that url into a string, the url will look something like this: http://localhost/?code=4/3pakksajdfASDFwek.4nsKfAYN7XQVshQV0ieZDAp-PrgEcAI and I only want the part after code=, is that possible? Thanks
int start = my_string.indexOf("=");
String suffix = my_string.substring(start + 1);
If other parameters can be on the URL, or code is not always first parameter:
String url = "http://localhost/?code=4/3pakksajdfASDFwek.4nsKfAYN7XQVshQV0ieZDAp-PrgEcAI";
String code = url.replaceAll(".*(?:[?]|[&])code=([^&]+)","\1");
tests here: http://fiddle.re/nxfv
If you are totally sure the URL never has other form, you can use indexOf and substring.
Otherwise, it is better if you use URI class to extract out the query part of the URL (use getRawQuery just to be safe), then tokenize it with split along & character and find the correct key-value pair to obtain the correct value. This method is not as brittle as indexOf method above.

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