I want to remove all occurrence of 0 pairs before appearing any digit(1-9) from NSString
if 000001234000 required 01234000
if 0000123400 required 123400
if 012340000 required 012340000
if 00000012 required 12
Can anyone help ? thanks.
Perhaps not the most elegant solution (but only trims leading '00'):
- (NSString *)trimLeadingDoubleZerosFrom:(NSString *)str {
if (str.length > 1 ) {
if ([[str substringWithRange:NSMakeRange(0, 2)] isEqualToString:#"00"]) {
return [self trimLeadingDoubleZerosFrom:[str substringFromIndex:2]];
}
}
return str;
}
Seems to work for your examples:
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"000001234000"]);// returns 01234000
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"0000123400"]); // returns 123400
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"012340000"]); // returns 012340000
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"00000012"]); // returns 12
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"12"]); // returns 12
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"02"]); // returns 02
NSLog(#"%#", [self trimLeadingDoubleZerosFrom:#"2"]); // returns 2
NSString *MyString = #"00000123050060007";
NSString *NewString = [self RemovePairOfZero:MyString];
NSLog(#"OUTPUT:: %#", NewString);
- (NSString *)RemovePairOfZero:(NSString *)Param
{
if ([Param length] > 1 )
{
if ([[Param substringWithRange:NSMakeRange(0, 2)] isEqualToString:#"00"])
{
return [self RemovePairOfZero:[Param substringFromIndex:2]];
}
}
return Param;
}
//2013-09-24 13:15:58.527 Test[1584:907] OUTPUT:: 0123050060007
NSString* a = #"00000123";
while ([a hasPrefix:#"00"]) {
a = [a stringByReplacingCharactersInRange:NSMakeRange(0, 2) withString:#""];
}
You can try using stringByreplacingOccurancesOfString:#"00" withString:#""
NSString *s= #"00000123";
NSLog(#"%#",s);
s = [s stringByReplacingOccurrencesOfString:#"00" withString:#"."];
NSLog(#"%#",s);
Try this one..
That can be done by simply replacing #"00" with #""
NSString *string = #"00000123";
string = [string stringByReplacingOccurrencesOfString:#"00" withString:#""];
Hope that helps!
Related
Code Snippet:
NSString *tempStr = self.consumerNumber.text;
if ([tempStr hasPrefix:#"0"] && [tempStr length] > 1) {
tempStr = [tempStr substringFromIndex:1];
[self.consumerNumbers addObject:tempStr];>
}
I tried those things and removing only one zero. how to remove more then one zero
Output :001600240321
Expected result :1600240321
Any help really appreciated
Thanks in advance !!!!!
Try to use this one
NSString *stringWithZeroes = #"001600240321";
NSString *cleanedString = [stringWithZeroes stringByReplacingOccurrencesOfString:#"^0+" withString:#"" options:NSRegularExpressionSearch range:NSMakeRange(0, stringWithZeroes.length)];
NSLog(#"Clean String %#",cleanedString);
Clean String 1600240321
convert string to int value and re-assign that value to string,
NSString *cleanString = [NSString stringWithFormat:#"%d", [string intValue]];
o/p:-1600240321
You can add a recursive function that is called until the string begin by something else than a 0 :
-(NSString*)removeZerosFromString:(NSString *)anyString
{
if ([anyString hasPrefix:#"0"] && [anyString length] > 1)
{
return [self removeZerosFromString:[anyString substringFromIndex:1]];
}
else
return anyString;
}
so you just call in your case :
NSString *tempStr = [self removeZerosFromString:#"000903123981000"];
NSString *str = #"001600240321";
NSString *newStr = [#([str integerValue]) stringValue];
If the NSString contains numbers only.
Other wise use this:
-(NSString *)stringByRemovingStartingZeros:(NSString *)string
{
NSString *newString = string;
NSInteger count = 0;
for(int i=0; i<[string length]; i++)
{
if([[NSString stringWithFormat:#"%c",[string characterAtIndex:i]] isEqualToString:#"0"])
{
newString = [newString stringByReplacingCharactersInRange:NSMakeRange(i-count, 1) withString:#""];
count++;
}
else
{
break;
}
}
return newString;
}
Simply call this method:-
NSString *stringWithZeroes = #"0000000016909tthghfghf";
NSLog(#"%#", [self stringByRemovingStartingZeros:stringWithZeroes]);
OutPut: 16909tthghfghf
Try the `stringByReplacingOccurrencesOfString´ methode like this:
NSString *new = [old stringByReplacingOccurrencesOfString: #"0" withString:#""];
SORRY: This doesn't help you due to more "0" in the middle part of your string!
i have this code:
NSString * firstdDigitTest = [numberString substringToIndex:1];
if (! [firstdDigitTest isEqualToString:#"0"])
{
numberString = [numberString substringFromIndex:1];
self.firstImageName = [namefile stringByAppendingString:firstdDigitTest];
}
self.number = [numberString integerValue];
i check if the first number is not 0 and if it's not 0 i need to insert it to the firstImageName. But when i do it (using the substringFromIndex) and i try to use integerValue it's dosent work!
without the substringFromIndex it's works! :(
NSString *str0 = #"XXX10098";
NSString *str1 = [str0 substringWithRange:NSMakeRange(4, str0.length-4)];
NSLog(#"%#", str1);
NSLog(#"%d", [str1 intValue]);
2012-10-28 10:52:07.309 iFoto[5652:907] 0098
2012-10-28 10:52:07.312 iFoto[5652:907] 98
check [numberString intValue];
how can I display 5 raise to 1/3 in iphone i.e I want 1/3 written above 5 can anyone help please
I Found this solution, hope so it would be helpful for you.
x to the power of y in a UILabel could be easy. Just replace your indices with unicode superscript characters... I use the following method to turn an integer into a string with superscript characters.
+(NSString *)convertIntToSuperscript:(int)i
{
NSArray *array = [[NSArray alloc] initWithObjects:#"⁰", #"¹", #"²", #"³", #"⁴", #"⁵", #"⁶", #"⁷", #"⁸", #"⁹", nil];
if (i >= 0 && i <= 9) {
NSString *myString = [NSString stringWithFormat:#"%#", [array objectAtIndex:i]];
[array release];
return myString;
}
else {
NSString *base = [NSString stringWithFormat:#"%i", i];
NSMutableString *newString = [[NSMutableString alloc] init];
for (int b = 0; b<[base length]; b++) {
int temp = [[base substringWithRange:NSMakeRange(b, 1)] intValue];
[newString appendString:[array objectAtIndex:temp]];
}
[array release];
NSString *returnString = [NSString stringWithString:newString];
[newString release];
return returnString;
}
}
Try this NSString *cmsquare=#"cm\u00B2";
It will display cm².
Yes you can do that but you need custom UILabel, either Make it by yourself or Get it Open Source..
I need to convert a hex string to binary form in objective-c, Could someone please guide me?
For example if i have a hex string 7fefff78, i want to convert it to 1111111111011111111111101111000?
BR,
Suppi
Nice recursive solution...
NSString *hex = #"49cf3e";
NSUInteger hexAsInt;
[[NSScanner scannerWithString:hex] scanHexInt:&hexAsInt];
NSString *binary = [NSString stringWithFormat:#"%#", [self toBinary:hexAsInt]];
-(NSString *)toBinary:(NSUInteger)input
{
if (input == 1 || input == 0)
return [NSString stringWithFormat:#"%u", input];
return [NSString stringWithFormat:#"%#%u", [self toBinary:input / 2], input % 2];
}
Simply convert each digit one by one: 0 -> 0000, 7 -> 0111, F -> 1111, etc. A little lookup table could make this very concise.
The beauty of number bases that are powers of another base :-)
In case you need leading zeros, for example 18 returns 00011000 instead of 11000
-(NSString *)toBinary:(NSUInteger)input strLength:(int)length{
if (input == 1 || input == 0){
NSString *str=[NSString stringWithFormat:#"%u", input];
return str;
}
else {
NSString *str=[NSString stringWithFormat:#"%#%u", [self toBinary:input / 2 strLength:0], input % 2];
if(length>0){
int reqInt = length * 4;
for(int i= [str length];i < reqInt;i++){
str=[NSString stringWithFormat:#"%#%#",#"0",str];
}
}
return str;
}
}
NSString *hex = #"58";
NSUInteger hexAsInt;
[[NSScanner scannerWithString:hex] scanHexInt:&hexAsInt];
NSString *binary = [NSString stringWithFormat:#"%#", [self toBinary:hexAsInt strLength:[hex length]]];
NSLog(#"binario %#",binary);
I agree with kerrek SB's answer and tried this.
Its work for me.
+(NSString *)convertBinaryToHex:(NSString *) strBinary
{
NSString *strResult = #"";
NSDictionary *dictBinToHax = [[NSDictionary alloc] initWithObjectsAndKeys:
#"0",#"0000",
#"1",#"0001",
#"2",#"0010",
#"3",#"0011",
#"4",#"0100",
#"5",#"0101",
#"6",#"0110",
#"7",#"0111",
#"8",#"1000",
#"9",#"1001",
#"A",#"1010",
#"B",#"1011",
#"C",#"1100",
#"D",#"1101",
#"E",#"1110",
#"F",#"1111", nil];
for (int i = 0;i < [strBinary length]; i+=4)
{
NSString *strBinaryKey = [strBinary substringWithRange: NSMakeRange(i, 4)];
strResult = [NSString stringWithFormat:#"%#%#",strResult,[dictBinToHax valueForKey:strBinaryKey]];
}
return strResult;
}
sorry but i'm still struggling to make this code working. It works if with a 2 digits number but it crashes with a single digit number. How can I verify if the NSString *secDigit has a value or is null. I hope my question is clear. Thanks in advance.
NSString *depositOverTotalRwy = [NSString stringWithFormat:#"%#", [deposit text]];
NSArray *components = [depositOverTotalRwy
componentsSeparatedByString:#"/"];
NSString *firstThird = [components objectAtIndex:0];
char firstChar = [firstThird characterAtIndex:0];
char secChar = [firstThird characterAtIndex:1];
NSString *firstDigit = [NSString stringWithFormat:#"%c",firstChar];
NSString *secDigit = [NSString stringWithFormat:#"%c", secChar];
NSLog(#" i'm %#", firstDigit);
NSLog(#" i'm %#", secDigit);
if ([firstDigit isEqualToString: #"1"]) {
firstDigit=#"wet";
NSLog(#"wet");
}
if ([firstDigit isEqualToString: #"2"]) {
firstDigit=#"wet";
NSLog(#"snow");
}
if ([firstDigit isEqualToString: #"3"]) {
firstDigit=#"wet";
NSLog(#"ice");
}
if ([secDigit isEqualToString: #"1"]) {
secDigit=#"wet";
NSLog(#"wet");
}
if ([secDigit isEqualToString: #"2"]) {
secDigit=#"snow";
NSLog(#"snow");
}
if ([secDigit isEqualToString: #"3"]) {
secDigit=#"ice";
NSLog(#"ice");
}
thanks to all of you..... here my code (working now):
NSString *depositOverTotalRwy = [NSString stringWithFormat:#"%#", [deposit text]];
NSArray *components = [depositOverTotalRwy
componentsSeparatedByString:#"/"];
NSString *firstThird = [components objectAtIndex:0];
char firstChar = [firstThird characterAtIndex:0];
NSString *firstDigit = [NSString stringWithFormat:#"%c",firstChar];
NSLog(#" i'm %#", firstDigit);
if ([firstDigit isEqualToString: #"1"]) {
firstDigit=#"wet";
NSLog(#"wet");
}
if ([firstDigit isEqualToString: #"2"]) {
firstDigit=#"wet";
NSLog(#"snow");
}
if ([firstDigit isEqualToString: #"3"]) {
firstDigit=#"wet";
NSLog(#"ice");
}
if ([firstThird length] >1) {
char secChar = [firstThird characterAtIndex:1];
NSString *secDigit = [NSString stringWithFormat:#"%c", secChar];
if ([secDigit isEqualToString: #"1"]) {
secDigit=#"wet";
NSLog(#"wet");
}
if ([secDigit isEqualToString: #"2"]) {
secDigit=#"snow";
NSLog(#"snow");
}
if ([secDigit isEqualToString: #"3"]) {
secDigit=#"ice";
NSLog(#"ice");
}
}
I guess you code crashes in this line:
char secChar = [firstThird characterAtIndex:1];
This is because you try to access a character outside of the string bounds. You need to guard against this by checking the length of the string first:
if ([firstThird count] > 1) {
// String has 2 or more characters, do all the stuff that involves
// a second character.
char secChar = [firstThird characterAtIndex:1];
NSString *secDigit = [NSString stringWithFormat:#"%c", secChar];
if ([secDigit isEqualToString: #"1"]) {
secDigit=#"wet";
NSLog(#"wet");
}
}
But I'd also like to recommend to not use an NSString here, as you already have a char. Just do something like this:
if ([firstThird count] > 1) {
// String has 2 or more characters, do all the stuff that involves
// a second character.
char secChar = [firstThird characterAtIndex:1];
if (secChar == '1') {
secDigit=#"wet";
NSLog(#"wet");
}
}
I think this is what you are looking for:
char secChar;
if(firstThird.length > 1)
{
secChar = [firstThird characterAtIndex:1];
}
According to this
http://developer.apple.com/library/ios/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html
NSString "Raises an NSRangeException if index lies beyond the end of the receiver"
So, your code:
char secChar = [firstThird characterAtIndex:1];
Is the problem (you should see that in the debugger Console)
Check the length first with
if ([firstThird length] < 2) {
// handle the case where it is one digit
}
You can check number of characters in a string using NSString length. and modify your code
as
NSString *depositOverTotalRwy = [NSString stringWithFormat:#"%#", #"23"];
NSArray *components = [depositOverTotalRwy
componentsSeparatedByString:#"/"];
NSString *firstThird = [components objectAtIndex:0];
unichar firstChar;
unichar secChar;
if([firstThird length]>1){
firstChar = [firstThird characterAtIndex:0];
secChar = [firstThird characterAtIndex:1];
} else {
firstChar = [firstThird characterAtIndex:0];
secChar = 0;
}
switch (firstChar) {
case '1': /* Do your stuff*/break;
case '2': /* Do your stuff*/break;
case '3': /* Do your stuff*/break;
default:
break;
}
switch (secChar) {
case '1': /* Do your stuff*/break;
case '2': /* Do your stuff*/break;
case '3': /* Do your stuff*/break;
default:
break;
}
you can use unichar instead of char. And can perform check in switch statements.
If you use char, a casting is done from unichar to char and for some characters you may lose actual value. So it is safe to use unichar...
If you want to convert unichar to string simply code
NSString * stringChar = [NSString StringWithFormat:#"%C",unicharVariable];
Thats it ...