Develop Reference

ReplaceAll not working

I'm using google volley to retrieve source code from website. Some looping was done to capture the value in the code. I've successfully captured the data I wanted, but error was shown: NumberFormatException: Invalid float: "2,459.00"
My intention was to store the value after the class=ListPrice>
Sample:
RM 2,899.00
The example value of the source code I wanted to save is "RM2,459.00 "
Below is the code I've written:
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_lazada_result);
lelongResult = (TextView) findViewById(R.id.lelong_result);
RequestQueue lelong = MyVolley.getRequestQueue(this);
StringRequest myLel = new StringRequest(
Method.GET,
"http://list.lelong.com.my/Auc/List/List.asp?DA=A&TheKeyword=iphone&x=0&y=0&CategoryID=&PriceLBound=&PriceUBound=",
RetrieveLelong(), createMyReqErrorListener());
lelong.add(myLel);
}
private Response.Listener<String> RetrieveLelong() {
return new Response.Listener<String>() {
#Override
public void onResponse(String response) {
ArrayList<Float> integers = new ArrayList<>();
String to = "class=ListPrice>";
String remainingText = response;
String showP = "";
while (remainingText.indexOf(to) >= 0) {
String tokenString = remainingText.substring(remainingText
.indexOf(to) + to.length());
String priceString = tokenString.substring(0,
tokenString.indexOf("<"));
float price = Float.parseFloat(priceString.replaceAll("[^\\d,]+", "").trim());
integers.add((price / 100));
remainingText = tokenString;
}
for (int i = 0; i < integers.size(); i++) {
String test1 = Float.toString(integers.get(i));
showP += test1 + "\n";
}
lelongResult.setText(showP);
}
};
}
The problem was as below:
I've tried all sort of replaceAll(),
1)replaceAll("[^\d,]+","") result:2,89900
replace all character except digits and comma works.
2)replaceAll("[^\d]+","") result:Invalid int""
replace all character include comma and dot ,not working
3)replaceAll("[^\d.]+,"") result:Invalid int""
replace all character exclude digits and dot, not working
From the experiment 2&3 coding above,I've noticed that if the comma were removed,i cant parseFloat as the value received by it is: "".NumberFormatException:Invalid Float:"" shown.
From the experiment 1,NumberFormatException:Invalid Float "2,45900" is showned.
The problem was replacing comma ,the code will not work but with the presence of comma ,the value cannot be stored into string

try this:
float price = Float.parseFloat(priceString.replaceAll("RM", "").trim());

use `replaceAll(Pattern.quote(","), "");
EDIT
if you want only numbers then use this
String s1= s.replaceAll("\D+","");

Try to parse the number by specifying the Locale.
NumberFormat format = NumberFormat.getInstance(Locale.KOREAN);
Number number = format.parse(priceString.replaceAll("RM", ""));
double d = number.doubleValue();
I'm just guessing the locale, don't know what you should use, depends on country

You need to do it one by one
priceString=priceString.replaceAll("\\D", "");
priceString=priceString.replaceAll("\\s", "");
now
priceString=priceString.trim();
float price = Float.parseFloat(priceString);

the problem is that in your code:
priceString.replaceAll(Pattern.quote(","), "");
float price = Float.parseFloat(priceString.replaceAll("\\D+\\s+", "").trim());
You are replacing coma but not storing the value!
you have to do:
priceString = priceString.replaceAll(",", "");
float price = Float.parseFloat(priceString.replaceAll("\\D+\\s+", "").trim());
I'm not sure of the pattern "\D+\s" because if you remove the coma you don't need to replace anything else (except "RM" that i assume you already removed)
Update: set locale and parse a number:
NumberFormat format = NumberFormat.getInstance(Locale.KOREAN);
Number number = format.parse(priceString.replaceAll("RM", ""));
double d = number.doubleValue();

Android - Check content of a String

I have a string (length 3-8) assigned to a variable (text). I want to check whether the 2nd and 3rd characters are NOT numeric (a letter or symbol or space..or anything other than numbers).

Elementary way to do this could be:
if(((text.charAt(1)-'0')>=0)&&(text.charAt(1)-'0')<10))||((text.charAt(2)-'0')>=0)&&(text.charAt(2)-'0')<10)))
{
//do nothing, since this means 2nd and/or 3rd characters in the string are numeric
}
else
{
// Your condition is met
}
You could also use REGEX's , if your checking is still more complicated.

Here is Another way to achieve this:
boolean isNumeric = true;
String test = "testing";
char second = test.charAt(1);
char third = test.charAt(2);
try {
Integer.parseInt(String.valueOf(second));
Integer.parseInt(String.valueOf(third));
} catch(NumberFormatException e) {
isNumeric = false;
}
System.out.println("Contains Number in 2nd and 3rd or both position: " + isNumeric);

You might make use of the String.IndexOf(String) method, like:
String digits = "0123456789";
String s2 = text.substring(2,3);
String s3 = text.substring(3,4);
boolean valid = (digits.indexOf(s2) > -1) && (digits.indexOf(s3) > -1);

Getting a utf-8 encoded string from a database, then displaying in a webview

I am attempting to scoop some utf-8 encoded (Spanish text) data from a sqlite database and then display it in a webview. The essential parts of the code are as follows:
String web_data_string = cursor.getString(0);
webview.loadData(web_data_string, "text/html", "utf-8");
The string is a word containing all normal characters except one which should be a lower-case e+acute. In the webview all the normal characters appear correctly, but the e+acute appears as two characters, an A+tilde followed by a copyright symbol.
Where did I go wrong?
EDIT: +1 to dd619 for a solution that can be made to work, but I am hoping for a more general solution where there may be many more special characters, in other languages too.

This does the trick...
String web_data_string = cursor.getString(0);
webview.loadData(utf_eight_to_web(web_data_string), "text/html", "utf-8");
String utf_eight_to_web(String str)
{
String ans = "";
Character c;
for(int i = 0;i<str.length();i++)
{
c = str.charAt(i);
if (c <= 127)
{
ans += str.charAt(i);
}
else
{
ans += String.format("&#%d;",(int)c);
}
}
return ans;
}

If you have any special character/letter then you need to replace it with respective escape character sequence.
Have a look at this useful link.
I had the same problem with some Spanish characters e.g.á and i solved it by replacing á with \u00e1
e.g. if you want to print "Parámetros" then simply do.
String str="Parámetros";
str=str.replace("á","\u00e1");
Log.i("MyClass",str);
Now you can str in database or java or in C++ or any web platform!

The Mick's answer could be improved using StringBuilder:
private String utfEightToWeb(String str)
{
StringBuilder stringBuilder = new StringBuilder();
Character c;
for (int i = 0; i<str.length(); i++) {
c = str.charAt(i);
if (c <= 127) {
stringBuilder.append(str.charAt(i));
} else {
stringBuilder.append(String.format("&#%d;",(int)c));
}
}
return stringBuilder.toString();
}

\n (Newline) deleting Android

There was some XML parsed text that looked like this:
06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href="..."> ... <br>15:45 Something..
and there was a lot of it..
Well, I have done this:
String mim =ses.replaceAll("(?s)\\<.*?\\>", " \n");
there was no other way to show text nicely.
Now, after few showings, and some time, I need that same text separated into alone strings like this:
06:00 Vesti
... or
07:15 Something Else
I've tried something like this, but it does not work:
char[] rast = description.toCharArray();
int brojac = 0;
for(int q=0; q<description.length(); q++){
if(rast[q]=='\\' && rast[q+1]=='n' ) brojac++;
}
String[] niz = new String[brojac];
int bf1=0;
int bf2=0;
int bf3=0;
int oo=0;
for(int q=0; q<description.length(); q++){
if(rast[q]=='\\'&& rast[q+1]=='n'){
bf3=bf1;
bf1=q;
String lol = description.substring(bf3, bf1);
niz[oo]=lol;
oo++;
}
}
I know that in description.substring(bf3,bf1) are not set as they should be but I think that this:
if(rast[q]=='\\' && rast[q+1]=='n)
does not work that way.. is there any other solution?
Note. there is no other way to get that resource. , It must be through this.

Calling Html.fromHtml(String) will properly translate the <br> into \n.
String html = "06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href=\"...\"> ... <br>15:45 Something..";
String str = Html.fromHtml(html).toString();
String[] arr = str.split("\n");
Then, just split it on a line basis - no need for regexps (which you shouldn't be using to parse HTML in the first case).
Edit: Turning everything into a bunch of Dates
// Used to find the HH:mm, in case the input is wonky
Pattern p = Pattern.compile("([0-2][0-9]:[0-5][0-9])");
SimpleDateFormat fmt = new SimpleDateFormat("HH:mm");
SortedMap<Date, String> programs = new TreeMap<Date, String>();
for (String row : arr) {
Matcher m = p.matcher(row);
if (m.find()) {
// We found a time in this row
ParsePosition pp = new ParsePosition(m.start(0));
Date when = fmt.parse(row, pp);
String title = row.substring(pp.getIndex()).trim();
programs.put(when, title);
}
}
// Now programs contain the sorted list of programs. Unfortunately, since
// SimpleDateFormat is stupid, they're all placed back in 1970 :-D.
// This would give you an ordered printout of all programs *AFTER* 08:00
Date filter = fmt.parse("08:00");
SortedMap<Date, String> after0800 = programs.tailMap(filter);
// Since this is a SortedMap, after0800.values() will return the program names in order.
// You can also iterate over each entry like so:
for (Map.Entry<Date,String> program : after0800.entrySet()) {
// You can use the SimpleDateFormat to pretty-print the HH:mm again.
System.out.println("When:" + fmt.format(program.getKey()));
System.out.println("Title:" + program.getValue());
}

Use regex:
List<String> results = new ArrayList<String>();
Pattern pattern = Pattern.compile("(\d+:\d+ \w+)<?");
Matcher matcher = pattern.matcher("06:00 Vesti<br>07:15 Something Else<br>09:10 Movie<a href="..."> ... <br>15:45 Something..");
while(matcher.find()) {
results.add(matcher.group(0));
}
results will end up as a list of strings:
results = List[
"06:00 Vesti",
"07:15 Something Else",
"09:10 Movie",
"15:45 Something.."]
See Rexgex Java Tutorial for an idea of how javas regex library works.

How to split a string and get a specific string in android?

I want to split a string and get a word finally. My data in database is as follows.
Mohandas Karamchand Gandhi (1869-1948), also known as Mahatma Gandhi, was born in Porbandar in the present day state of Gujarat in India on October 2, 1869.
He was raised in a very conservative family that had affiliations with the ruling family of Kathiawad. He was educated in law at University College, London.
src="/Leaders/gandhi.png"
From the above paragraph I want get the image name "gandhi". I am getting the index of "src=". But now how can I get the image name i.e "gandhi" finally.
My Code:
int index1;
public static String htmldata = "src=";
if(paragraph.contains("src="))
{
index1 = paragraph.indexOf(htmldata);
System.out.println("index1 val"+index1);
}
else
System.out.println("not found");

You can use the StringTokenizer class (from java.util package ):
StringTokenizer tokens = new StringTokenizer(CurrentString, ":");
String first = tokens.nextToken();// this will contain one word
String second = tokens.nextToken();// this will contain rhe other words
// in the case above I assumed the string has always that syntax (foo: bar)
// but you may want to check if there are tokens or not using the hasMoreTokens method

Try this code. Check if it working for you..
public String getString(String input)
{
Pattern pt = Pattern.compile("src=.*/(.*)\\..*");
Matcher mt = pt.matcher(input);
if(mt.find())
{
return mt.group(1);
}
return null;
}
Update:
Change for multiple item -
public ArrayList<String> getString(String input)
{
ArrayList<String> ret = new ArrayList<String>();
Pattern pt = Pattern.compile("src=.*/(.*)\\..*");
Matcher mt = pt.matcher(input);
while(mt.find())
{
ret.add(mt.group(1));
}
return ret;
}
Now you'll get an arraylist with all the name. If there is no name then you'll get an empty arraylist (size 0). Always make a check for size.