It should be something like "Hello, my name is ??????????." in Open Office Base.
I am receiving a string from a website which is always build up like this:
"Hello, my name is " +name+ "."
The variable name can be ANY name with a length of up to 10 characters and it changes everyday.
I am trying to check if the string, that is provided by the website, contains this sentence (Example: "Hello, my name is John.").
BUT I don't know the name that the website provides, so I have to ask:
if(string1.contains("Hello, my name is " + ANY 10 CHARACTERS + "."))
{
return true;
}
First need to parse out the name from the String by removing all characters besides the name.
String name = stringFromWebsite.replace("Hello, my name is ", ""); //remove "Hello, my name is "
name = name.substring(0, name.length() - 1) //remove "." at the end
name = name.trim(); //trim any whitespaces
Next concatenate the base String + your variable.
String newString = "Hello, my name is " + name;
Now run the contains function against the new String
if (string1.contains(newString)){
return true;
}
Related
I am fetching number from contact book and sending it to server. i get number like this (+91)942 80-60 135 but i want result like this +9428060135.+ must be first character of string number.
Given your example you want to replace the prefix with a single + character. You also want to remove other non-numeric characters from the number string. Here's how you can do that:
String number = "(+91)942 80-60 135";
number = "+" + number.replaceAll("\\(\\+\\d+\\)|[^\\d]", "");
The regex matches any prefix (left paren followed by a + followed by one or more digits, followed by a right paren) or any non digit character, and removes them. This is concatenated to a leading + as required. This code will also handle + characters within the number string, e.g. +9428060135+++ and +(+91)9428060135+++.
If you simply wanted to remove any character that is not a digit nor a +, the code would be:
String number = "(+91)942 80-60 135";
number = number.replaceAll("[^\\d+]", "");
but be aware that this will retain the digits in the prefix, which is not the same as your example.
You can use String.replace(oldChar, newChar). Use the code below
String phone = "(+91)942 80-60 135"; // fetched string
String trimmedPhone = phone.replace("(","").replace(")","").replace("-","").trim();
I hope it will work for you.
check this. Pass your string to this function or use as per code goes
String inputString = "(+91)942 80-60 135";
public void removeSpecialCharacter(String inputString) {
String replaced = inputString.replaceAll("[(\\-)]", "");
String finalString = replaced.replaceAll(" ", "");
Log.e("String Output", " " + replaced + " " + second);
}
I want use register in my application and i should send password and verifyCode with SMS to users phones.
But i should read verifyCode from message and set automatically number into verifyCode EditText.
My message format :
Hi, welcome to our service.
your password 12345
your verifyCode 54321
How can i do it? Please help me <3
Assuming that the number of digits are fixed in password and verify codes (Generally they are same as default values), We can extract digits from the string and then find substring which has verify code. This assumption is for simplicity.
String numberOnly= str.replaceAll("[^0-9]", "");
String verifyCode = numberOnly.substring(6);
Here String verifyCode = numberOnly.substring(6); is getting last 5 digits of the string which is your verification code. You can also write numberOnly.substring(6,10); to avoid confusions.
But this is prone to errors like StringIndexOutOfBoundsException, So whenever you want to get substring which is starting from index i till the end of the string, always write numberOnly.substring(i).
There are a lot ways to do this. You can use some complicated regex or use a simple spilt method.
Try this,
String str = "Hi, welcome to our service.\n"
+ "\n"
+ "your password \n"
+ "12345\n"
+ "\n"
+ "your verifyCode \n"
+ "54321";
// Solution #1
String[] parts = str.split("\n");
System.out.println(parts[3]);
System.out.println(parts[6]);
// Solution #2
String PAT = "(password|verifyCode)\\s+(\\d+)";
Pattern pats = Pattern.compile(PAT);
Matcher m = pats.matcher(str);
while (m.find()) {
String grp = m.group(2);
System.out.println(grp);
}
sumTextView.setText(Integer.toString(a) + " + " + Integer.toString(b));
This Line show warning you see in pic..
Use String.format();
sumTextView.setText(String.format("%1$d + %2$d", a, b));
With this you can format a string correctly with multiple variables, no matter whether they are strings or integers. This example takes the value of variable a and replaces the placeholder %1$d with it. Same goes for the other variable.
take an string copy whole line in it, then show string in setText
String str = (Integer.toString(a) + " + " + Integer.toString(a));
sumTextView.setText(str);
1. The First String Says that do not concate string with setText property.
String txt = String.valueOf(a) + " + " + String.valueOf(b);
sumTextView.setText(str);
2. Second warning says that your program have possibility to crash or genearte an exception in case if value of a or b is null or not an integer.
So check condition if(a!=null and b!=null) then display text in if condition.
I have a string
String retail = c.getString(c.getColumnIndex("retail"));
The date is being passed as "99999", I need it to print out as "999.99", how can I do this?
If you always have to add the "." 2 character before the end, this should work:
retail = retail.substring(0, retail.length()-2) + "." + retail.substring(retail.length()-2,retail.length());
This will add a dot two character before the end of the String, as you need.
It seems that if you call
String text = "String<br>String";
Log.d(TAG, text);
it automatically parses the String to take two lines. The same goes for new line (\n) characters. That makes debugging more complicated. Is there a way to tell the logger to give me the exact String?
The arguments to the methods in Log class are Java strings, so escaping special characters is just like in Java. For example,
String text = "String\nString";
Log.d("TEST!!", text);
Will give you:
D/TEST!!﹕ String
String
while:
String text = "String\\nString";
Log.d("TEST!!", text);
will give you:
D/TEST!!﹕ String\nString
in the logcat.
As far as <BR>, I'm not seeing the same effect as you. Specifically,
String text = "String<br>String";
Log.d("TEST!!", text);
Produces:
D/TEST!!﹕ String<br>String
So I am unable to reproduce your actual problem. However, in general special characters in the Log strings are escaped just like any other Java strings. The logger is dumb and there's no settings to automatically escape special characters; you'll have to do this yourself for arbitrary strings. The Log methods just turn around and call println_native.
I use System.getProperty("line.separator")
ArrayList<String> txts = new ArrayList<String>();
txts.add("aoeuaeou");
txts.add("snhsnthsnth");
String msg = TextUtils.join(System.getProperty("line.separator"),txts);
Log.d(TAG, "Bla bla bla: "+ msg );
show in the log like
Bla bla bla: aoeuaeou
snhsnthsnth
At the end of the message, a trailing space seems to be needed.
Log.i("tag", "My message with a blank line following.\n ");
or
Log.i("tag", "Variable 1: " + v1 + " Variable 2: " + v2 + "\n ");