Using LINQ how can I transform only the first letter of s.Password to lowercase
if (s.Password == password){}
i want that the first char of s.Password will be in lower case,
i tried :
if( s.Password[0].toString().toLower() + s.Password(1) ) == password ){}
If you would like to make a decision based on an item's position in LINQ, you can use Select that takes a Func with two parameters - the item and its index:
var pwd = "BadPassword";
var res = new string(
pwd.Select((c, i) => i==0 ? char.ToLower(c) : c).ToArray()
); // produces badPassword
The functor above converts the initial character at i==0 to lower case, while leaving all other characters in place.
Demo 1.
Note: LINQ is not necessary for this conversion. You can do the same thing in one line by using Substring:
var res = char.ToLower(pwd[0]) + pwd.Substring(1);
Demo 2.
Related
I want extract from a table all rows where in a column (string) there is at least one word that starts with a specified character.
Example:
Row 1: 'this is the first row'
Row 2: 'this is th second row'
Row 3: 'this is the third row'
If the specified character is T -> I would extract all 3 rows
If the specified character is S -> I would extract only the second column
...
Please help me
Assuming you mean "space delimited sequence of characters, or begin to space or space to end" by "word", then you can split on the delimiter and test them for matches:
var src = new[] {
"this is the first row",
"this is th second row",
"this is the third row"
};
var findChar = 'S';
var lowerFindChar = findChar.ToLower();
var matches = src.Where(s => s.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Any(w => w.ToLower()[0] == lowerFindChar));
The LINQ Enumerable.Any method tests a sequence to see if any element matches, so you can split each string into a sequence of words and see if any word begins with the desired letter, compensating for case.
Try this:
rows.Where(r => Regex.IsMatch(r, " [Tt]"))
You can replace the Tt with Ss (both assuming you want either upper case or lower case).
The problem of course is, what is a "word"?
Is the character sequence 'word' in the sentence above a word according to your definition? It doesn't start with a space, not even a white-space.
A definition of a word could be:
Define wordCharacter: something like A-Z, a-z.
Define word:
- the non-empty sequence of wordCharacters at the beginning of a string followed by a non-wordcharacter
- or the non-empty sequence of wordCharacters at the end of a string preceded by a non-wordcharacter
- any non-empty sequence of wordCharacters in the string preceded and followed by a non-wordcharacter
Define start of word: the first character of a word.
String: "Some strange characters: 'A', 9, äll, B9 C$ X?
- Words: Some, strange characters, A
- Not Words: 9, äll, B9, C$ X?
So you first have to specify precisely what you mean by word, then you can define functions.
I'll write it as an extension method of IEnumerable<string>. Usage will look similar to LINQ. See Extension Methods Demystified
bool IsWordCharacter(char c) {... TODO: implement your definition of word character}
static IEnumerable<string> SplitIntoWords(this string text)
{
// TODO: exception if text null
if (text.Length == 0) return
int startIndex = 0;
while (startIndex != text.Length)
{ // not at end of string. Find the beginning of the next word:
while (startIndex < text.Length && !IsWordCharacter(text[startIndex]))
{
++startIndex;
}
// now startIndex points to the first character of the next word
// or to the end of the text
if (startIndex != text.Length)
{ // found the beginning of a word.
// the first character after the word is either the first non-word character,
// or the end of the string
int indexAfterWord = startWordIndex + 1;
while (indexAfterWord < text.Length && IsWordCharacter(text[indexAfterWord]))
{
++indexAfterWord;
}
// all characters from startIndex to indexAfterWord-1 are word characters
// so all characters between startIndexWord and indexAfterWord-1 are a word
int wordLength = indexAfterWord - startIndexWord;
yield return text.SubString(startIndexWord, wordLength);
}
}
}
Now that you've got a procedure to split any string into your definition of words, your query will be simple:
IEnumerabl<string> texts = ...
char specifiedChar = 'T';
// keep only those texts that have at least one word that starts with specifiedChar:
var textsWithWordThatStartsWithSpecifiedChar = texts
// split the text into words
// keep only the words that start with specifiedChar
// if there is such a word: keep the text
.Where(text => text.SplitIntoWords()
.Where(word => word.Length > 0 && word[0] == specifiedChar)
.Any());
var yourChar = "s";
var texts = new List<string> {
"this is the first row",
"this is th second row",
"this is the third row"
};
var result = texts.Where(p => p.StartsWith(yourChar) || p.Contains(" " + yourChar));
EDITED:
Alternative way (I'm not sure it works in linq query)
var result = texts.Where(p => (" " + p).Contains(" " + yourChar));
you can use .ToLower() if you want Case-insensitive check.
Where as str[] will replace a character, str.insert will insert a character at a position. But it requires two lines of code:
str = "COSO17123456"
str.insert 4, "-"
str.insert 7, "-"
=> "COSO-17-123456"
I was thinking how to do this in one line of code. I came up with the following solution:
str = "COSO17123456"
str.each_char.with_index.reduce("") { |acc,(c,i)| acc += c + ( (i == 3 || i == 5) ? "-" : "" ) }
=> "COSO-17-123456
Is there a built-in Ruby helper for this task? If not, should I stick with the insert option rather than combining several iterators?
Use each to iterate over an array of indices:
str = "COSO17123456"
[4, 7].each { |i| str.insert i, '-' }
str #=> "COSO-17-123456"
You can uses slices and .join:
> [str[0..3], str[4..5],str[6..-1]].join("-")
=> "COSO-17-123456"
Note that the index after the first one (between 3 and 4) will be different since you are not inserting earlier insertion first. ie, more natural (to me anyway...)
You will insert at the absolute index of the original string -- not the moving relative index as insertions are made.
If you want to insert at specific absolute index values, you can also use ..each_with_index and control the behavior character by character:
str2 = ""
tgts=[3,5]
str.split("").each_with_index { |c,idx| str2+=c; str2+='-' if tgts.include? idx }
Both of the above create a new string.
String#insert returns the string itself.
This means you can chain the method calls, which can be a prettier and more efficient if you only have to do it a couple of times like in your example:
str = "COSO17123456".insert(4, "-").insert(7, "-")
puts str
COSO-17-123456
Your reduce version can be therefore more concisely written as:
[4,7].reduce(str) { |str, idx| str.insert(idx, '-') }
I'll bring one more variation to the table, String#unpack:
new_str = str.unpack("A4A2A*").join('-')
# or with String#%
new_str = "%s-%s-%s" % str.unpack("A4A2A*")
i want to give validation condition like,
var BlkAccountIdV = $('#txtBlkAccountId').val();
if (BlkAccountIdV == "") {
document.getElementById('lblAccountId').innerText = "Enter Valid AccountID";
errorflag=1;
}
if condition should be Executed only when entered text box value contains Letter.What value can i give inside Quotation marks(if (BlkAccountIdV == "") )?
var re = /[^0-9]/g;
var error = re.test(BlkAccountIdV);
error will be true if value of BlkAccountIdV is not numeric
I.e, this regex will match everything except numbers
So your code should look somehow like this:
var BlkAccountIdV = $('#txtBlkAccountId').val();
var re = /[^0-9]/g;
if ( re.test(BlkAccountIdV) ){
// found a non-numeric value, handling error
document.getElementById('lblAccountId').innerText = "Enter Valid AccountID";
errorflag=1;
}
if (BlkAccountIdV.match(/[0-9]+/) == null )
In if condition you can use isNaN() function. This checks whether string is 'not a number'
So in your case if this condition not valid, Then string is number
if(!isNaN(BlkAccountIdV) && BlkAccountIdV != ''){
//Your code
}
I'm attempting to parse blocks of text and need a way to detect the difference between apostrophes in different contexts. Possession and abbreviation in one group, quotations in the other.
e.g.
"I'm the cars' owner" -> ["I'm", "the", "cars'", "owner"]
but
"He said 'hello there' " -> ["He","said"," 'hello there' "]
Detecting whitespace on either side won't help as things like " 'ello " and " cars' " would parse as one end of a quotation, same with matching pairs of apostrophes. I'm getting the feeling that there's no way of doing it other than an outrageously complicated NLP solution and I'm just going to have to ignore any apostrophes not occurring mid-word, which would be unfortunate.
EDIT:
Since writing I have realised this is impossible. Any regex-ish based parser would have to parse:
'ello there my mates' dogs
in 2 different ways, and could only do that with understanding of the rest of the sentence. Guess I'm for the inelegant solution of ignoring the least likely case and hoping it's rare enough to only cause infrequent anomalies.
Hm, I'm afraid this won't be easy. Here's a regex that kinda works, alas only for stuff like "I'm" and "I've":
>> s1 =~ /[\w\s]*((?<!I)'(?:[^']+)')[\w\s]*/
=> nil
>> s2 =~ /[\w\s]*((?<!I)'(?:[^']+)')[\w\s]*/
=> 0
>> $1
=> "'hello there'"
If you play around with it a bit more, you may be able to eliminate some other common contractions, which might still be better than nothing.
Some rules to think about:
Quotes will start with an apostrophe with a whitespace character or nothing before it.
Quotes will end with an apostrophe with punctuation or a whitespace character after it.
Some words may look like the end of quotes, e.g., peoples'.
Quote delimiting apostrophes will never have letters directly before and after them.
Use a very simple two-phase process.
In pass 1 of 2, start with this regular expression to break the text down into alternating segments of word and non-word characters.
/(\w+)|(\W+)/gi
Store the matches in a list like this (I'm using AS3-style pseudo-code, since I don't work with ruby):
class MatchedWord
{
var text:String;
var charIndex:int;
var isWord:Boolean;
var isContraction:Boolean = false;
function MatchedWord( text:String, charIndex:int, isWord:Boolean )
{
this.text = text; this.charIndex = charIndex; this.isWord = isWord;
}
}
var match:Object;
var matched_word:MatchedWord;
var matched_words:Vector.<MatchedWord> = new Vector.<MatchedWord>();
var words_regex:RegExp = /(\w+)|(\W+)/gi
words_regex.lastIndex = 0; //this is where to start looking for matches, and is updated to the end of the last match each time exec is called
while ((match = words_regex.exec( original_text )) != null)
matched_words.push( new MatchedWord( match[0], match.index, match[1] != null ) ); //match[0] is the entire match and match[1] is the first parenthetical group (if it's null, then it's not a word and match[2] would be non-null)
In pass 2 of 2, iterate over the list of matches to find contractions by checking to see if each (trimmed, non-word) match ENDS with an apostrophe. If it does, then check the next adjacent (word) match to see if it matches one of only 8 common contraction endings. Despite all the two-part contractions I could think of, there are only 8 common endings.
d
l
ll
m
re
s
t
ve
Once you've identified such a pair of matches (non-word)="'" and (word)="d", then you just include the preceding adjacent (word) match and concatenate the three matches to get your contraction.
Understanding the process just described, one modification you must make is expand that list of contraction endings to include contractions that start with apostrophe, such as "'twas" and "'tis". For those, you simply don't concatenate the preceding adjacent (word) match, and you look at the apostrophe match a little more closely to see if it included other non-word character before it (that's why it's important it ends with an apostrophe). If the trimmed string EQUALS an apostrophe, then merge it with the next match, and if it only ENDS with an apostrophe, then strip off the apostrophe and merge it with the following match. Likewise, conditions that will include the prior match should first check to ensure the (trimmed non-word) match ending with an apostrophe EQUALS an apostrophe, so there are no extra non-word characters included accidentally.
Another modification you may need to make is expand that list of 8 endings to include endings that are whole words such as "g'day" and "g'night". Again, it's a simple modification involving a conditional check of the preceding (word) match. If it's "g", then you include it.
That process should capture the majority of contractions, and is flexible enough to include new ones you can think of.
The data structure would look like this.
Condition(Ending, PreCondition)
where PreCondition is
"*", "!", or "<exact string>"
The final list of conditions would look like this:
new Condition("d","*") //if apostrophe d is found, include the preceding word string and count as successful contraction match
new Condition("l","*");
new Condition("ll","*");
new Condition("m","*");
new Condition("re","*");
new Condition("s","*");
new Condition("t","*");
new Condition("ve","*");
new Condition("twas","!"); //if apostrophe twas is found, exclude the preceding word string and count as successful contraction match
new Condition("tis","!");
new Condition("day","g"); //if apostrophe day is found and preceding word string is g, then include preceding word string and count as successful contraction match
new Condition("night","g");
If you just process those conditions as I explained, that should cover all of these 86 contractions (and more):
'tis 'twas ain't aren't can't could've couldn't didn't doesn't don't
everybody's g'day g'night hadn't hasn't haven't he'd he'll he's how'd
how'll how's I'd I'll I'm I've isn't it'd it'll it's let's li'l
might've mightn't mustn't needn't nobody's nothing's shan't she'd
she'll she's should've shouldn't that'd that'll that's there's they'd
they'll they're they've wasn't we'd we'll we're we've weren't what'll
what're what'd what's what've when'd when'll when's where'd where'll
where's who's who'll who're who'd who'll who's who've why'd why'll
why's won't would've wouldn't you'd you'll you're you've
On a side note, don't forget about slang contractions that don't use apostrophes such as "gotta" > "got to" and "gonna" > "going to".
Here is the final AS3 code. Overall, you're looking at less than 50 lines of code to parse the text into alternating word and non-word groups, and identify and merge contractions. Simple. You could even add a Boolean "isContraction" variable to the MatchedWord class and set the flag in the code below when a contraction is identified.
//Automatically merge known contractions
var conditions:Array = [
["d","*"], //if apostrophe d is found, include the preceding word string and count as successful contraction match
["l","*"],
["ll","*"],
["m","*"],
["re","*"],
["s","*"],
["t","*"],
["ve","*"],
["twas","!"], //if apostrophe twas is found, exclude the preceding word string and count as successful contraction match
["tis","!"],
["day","g"], //if apostrophe day is found and preceding word string is g, then include preceding word string and count as successful contraction match
["night","g"]
];
for (i = 0; i < matched_words.length - 1; i++) //not a type-o, intentionally stopping at next to last index to avoid a condition check in the loop
{
var m:MatchedWord = matched_words[i];
var apostrophe_text:String = StringUtils.trim( m.text ); //check if this ends with an apostrophe first, then deal more closely with it
if (!m.isWord && StringUtils.endsWith( apostrophe_text, "'" ))
{
var m_next:MatchedWord = matched_words[i + 1]; //no bounds check necessary, since loop intentionally stopped at next to last index
var m_prev:MatchedWord = ((i - 1) >= 0) ? matched_words[i - 1] : null; //bounds check necessary for previous match, since we're starting at beginning, since we may or may not need to look at the prior match depending on the precondition
for each (var condition:Array in conditions)
{
if (StringUtils.trim( m_next.text ) == condition[0])
{
var pre_condition:String = condition[1];
switch (pre_condition)
{
case "*": //success after one final check, include prior match, merge current and next match into prior match and delete current and next match
if (m_prev != null && apostrophe_text == "'") //EQUAL apostrophe, not just ENDS with apostrophe
{
m_prev.text += m.text + m_next.text;
m_prev.isContraction = true;
matched_words.splice( i, 2 );
}
break;
case "!": //success after one final check, do not include prior match, merge current and next match, and delete next match
if (apostrophe_text == "'")
{
m.text += m_next.text;
m.isWord = true; //match now includes word text so flip it to a "word" block for logical consistency
m.isContraction = true;
matched_words.splice( i + 1, 1 );
}
else
{ //strip apostrophe off end and merge with next item, nothing needs deleted
//preserve spaces and match start indexes by manipulating untrimmed strings
var apostrophe_end:int = m.text.lastIndexOf( "'" );
var apostrophe_ending:String = m.text.substring( apostrophe_end, m.text.length );
m.text = m.text.substring( 0, m.text.length - apostrophe_ending.length); //strip apostrophe and any trailing spaces
m_next.text = apostrophe_ending + m_next.text;
m_next.charIndex = m.charIndex + apostrophe_end;
m_next.isContraction = true;
}
break;
default: //conditional success, check prior match meets condition
if (m_prev != null && m_prev.text == pre_condition)
{
m_prev.text += m.text + m_next.text;
m_prev.isContraction = true;
matched_words.splice( i, 2 );
}
break;
}
}
}
}
}
I have a file with two columns the first one with a name and the second one with a number.
The size of the number column is 20 chars, the numbers use to be less than 2 chars size the rest of the chars are complite with 0.
I need to take out all the zeros before the comma. I should use a tMap, How?
The solution:
Using a tMap, put a Var in the midle of both files (Input and output).
In the var use:
"0"+row1.numberField.split(",")[0].replace("0", "") + "." + row1.numberField.split(",")[1]
Example:
000000001,58
Result:
01.58
Solution 2:
Define your own routine:
public static String calcImp(String theNumber) {
Float theFNumber = new Float(theNumber.replace(",", "."));
return Float.toString(theFNumber).replace(".", ",");
}
Example:
000000001,587
Result:
1,587