Anyone have a regex to strip lat/long from a string? such as:
ID: 39.825 -86.88333
To match one value
-?\d+\.\d+
For both values:
(-?\d+\.\d+)\ (-?\d+\.\d+)
And if the string always has this form:
"ID: 39.825 -86.88333".match(/^ID:\ (-?\d+\.\d+)\ (-?\d+\.\d+)$/)
var latlong = 'ID: 39.825 -86.88333';
var point = latlong.match( /-?\d+\.\d+/g );
//result: point = ['39.825', '-86.88333'];
function parseLatLong(str) {
var exp = /ID:\s([-+]?\d+\.\d+)\s+([-+]?\d+\.\d+)/;
return { lat: str.replace(exp, "$1"), long: str.replace(exp, "$2") };
}
function doSomething() {
var res = parseLatLong("ID: 39.825 -86.88333");
alert('Latitude is ' + res.lat + ", Longitude is " + res.long);
}
Related
ViewModel
fun changeQty(textField: TextFieldValue) {
val temp1 = textField.text
Timber.d("textField: $temp1")
val temp2 = temp1.replace("[^\\d]".toRegex(), "")
Timber.d("temp2: $temp2")
_qty.value = textField.copy(temp2)
}
TextField
OutlinedTextField(
modifier = Modifier
.focusRequester(focusRequester = focusRequester)
.onFocusChanged {
if (it.isFocused) {
keyboardController?.show()
}
},
value = qty.copy(
text = qty.text.trim()
),
onValueChange = changeQty,
label = { Text(text = qtyHint) },
singleLine = true,
keyboardOptions = KeyboardOptions(
keyboardType = KeyboardType.Number,
imeAction = ImeAction.Done
),
keyboardActions = KeyboardActions(
onDone = {
save()
onDismiss()
}
)
)
Set KeyboardType.Number, it display 1,2,3,4,5,6,7,8,9 and , . - space.
I just want to get integer like -10 or 10 or 0.
But I type the , or . or -(not the front sign), it show as it is.
ex)
typing = -10---------
hope = -10
display = -10---------
I put regular expression in
val temp2 = temp1.replace("[^\\d]".toRegex(), "")
But, it doesn't seem to work.
How I can get only integer(also negative integer)?
Use this regex (?<=(\d|-))(\D+) to replace all non digit characters, except first -.
fun getIntegersFromString(input: String): String {
val pattern = Regex("(?<=(\\d|-))(\\D+)")
val formatted = pattern.replace(input, "")
return formatted
}
Check it here
I have a function that takes in text and checks for the # symbol. The output is the same text, but any words following the # symbol will be coloured, similar to that of a social media mention. The problem is that it adds an extra empty space to the front of the original text. How do I modify the output to remove the empty space it adds to the front of the new text?
func textWithHashtags(_ text: String, color: Color) -> Text {
let words = text.split(separator: " ")
var output: Text = Text("")
for word in words {
if word.hasPrefix("#") { // Pick out hash in words
output = output + Text(" ") + Text(String(word))
.foregroundColor(color) // Add custom styling here
} else {
output = output + Text(" ") + Text(String(word))
}
}
return output
}
Just call the function in a view like
textWithHashtags("Hello #stackoverflow how is it going?", color: .red)
try something like this:
func textWithHashtags(_ text: String, color: Color) -> Text {
let words = text.split(separator: " ")
var output: Text = Text("")
var firstWord = true // <-- here
for word in words {
let spacer = Text(firstWord ? "" : " ") // <-- here
if word.hasPrefix("#") { // Pick out hash in words
output = output + spacer + Text(String(word))
.foregroundColor(color) // Add custom styling here
} else {
output = output + spacer + Text(String(word))
}
firstWord = false
}
return output
}
The problem is that \b doesn't work with Russian and Ukrainian letters.
Here I try to find all matches of a word 'февраля' it the text, change them to tempword, then make it a link and change it back to 'февраля'.
function addLinks(word, siteurl) {
var id = 'doc\'s ID';
var doc = DocumentApp.openById(id);
var body = doc.getBody();
var tempword = 'ASDFDSGDDKDSL2';
var searchText = "\\b"+word+"\\b";
var element = body.findText(searchText);
console.log(element);
while (element) {
var start = element.getStartOffset();
var text = element.getElement().asText();
text.replaceText(searchText, tempword);
text.setLinkUrl(start, start + tempword.length - 1, siteurl);
element = body.findText(searchText);
}
body.replaceText(tempword, word);
}
addLinks('февраля', 'example.com');
It works as it should, if I change Russian word 'февраля' to English 'february'.
addLinks('february', 'example.com');
I need regular expression, because if I just look for 'февраля' script will apply it to other words like 'февралям', 'февралями' etc.
So, it is a question, how to make it work.
Mistake "Exception: Invalid regular expression pattern" occurs with this code:
var searchText = "(?<=[\\s,.:;\"']|^)"+word+"(?=[\\s,.:;\"']|$)";
or this:
var searchText = "(^|\s)"+word+"(?=\s|$)";
and some other.
Here is my solution:
function main() {
addLinks('февраля', 'example.com');
}
function addLinks(word, url) {
var doc = DocumentApp.getActiveDocument();
var pgfs = doc.getParagraphs();
var bound = '[^А-яЁё]'; // any letter except Russian one
var patterns = [
{regex: bound + word + bound, start: 1, end: 1}, // word inside of line
{regex: '^' + word + bound, start: 0, end: 1}, // word at the start
{regex: bound + word + '$', start: 1, end: 0}, // word at the end
{regex: '^' + word + '$', start: 0, end: 0} // word = line
];
for (var pgf of pgfs) for (var pattern of patterns) {
var location = pgf.findText(pattern.regex);
while (location) {
var start = location.getStartOffset() + pattern.start;
var end = location.getEndOffsetInclusive() - pattern.end;
pgf.editAsText().setLinkUrl(start, end, url);
location = pgf.findText(pattern.regex, location);
}
}
}
Test output:
It handles well the word placed at the start or at the end of the line (or both). And it gives no the weird error message.
im looking for the regexp that make able to do this tasks
message Body Input: Test1 (Test2) (test3) (ti,ab(text(text here(possible text)text(possible text(more text))))) end (text)
the result that i want Result: (text(text here(possible text)text(possible text(more text))))
I want to collect everything that is inside ti,ab(................)
var messageBody = message.getPlainBody()
var ssFile = DriveApp.getFileById(id);
DriveApp.getFolderById(folder.getId()).addFile(ssFile);
var ss = SpreadsheetApp.open(ssFile);
var sheet = ss.getSheets()[0];
sheet.insertColumnAfter(sheet.getLastColumn());
SpreadsheetApp.flush();
var sheet = ss.getSheets()[0];
var range = sheet.getRange(1, 1, sheet.getLastRow(), sheet.getLastColumn() + 1)
var values = range.getValues();
values[0][sheet.getLastColumn()] = "Search Strategy";
for (var i = 1; i < values.length; i++) {
//here my Regexp
var y = messageBody.match(/\((ti,ab.*)\)/ig);
if (y);
values[i][values[i].length - 1] = y.toString();
range.setValues(values);
The only solution you may use here is to extract all substrings inside parentheses and then filter them to get all those that start with ti,ab:
var a = [], r = [], result;
var txt = "Test1 (Test2) (test3) (ti,ab(text(text here(possible text)text(possible text(more text))))) end (text)";
for(var i=0; i < txt.length; i++){
if(txt.charAt(i) == '(') {
a.push(i);
}
if(txt.charAt(i) == ')') {
r.push(txt.substring(a.pop()+1,i));
}
}
result = r.filter(function(x) { return /^ti,ab\(/.test(x); })
.map(function(y) {return y.substring(6,y.length-1);})
console.log(result);
The nested parentheses function is borrowed from Nested parentheses get string one by one. The /^ti,ab\(/ regex matches ti,ab( at the start of the string.
The above solution allows extracting nested parentheses inside nested parentheses. If you do not need it, use
var txt = "Test1 (Test2) ((ti,ab(text(text here))) AND ab(test3) Near Ti(test4) NOT ti,ab,su(test5) NOT su(Test6))";
var start=0, r = [], level=0;
for (var j = 0; j < txt.length; j++) {
if (txt.charAt(j) == '(') {
if (level === 0) start=j;
++level;
}
if (txt.charAt(j) == ')') {
if (level > 0) {
--level;
}
if (level === 0) {
r.push(txt.substring(start, j+1));
}
}
}
console.log("r: ", r);
var rx = "\\b(?:ti|ab|su)(?:,(ti|ab|su))*\\(";
var result = r.filter(function(y) { return new RegExp(rx, "i").test(y); })
.map(function(x) {
return x.replace(new RegExp(rx, "ig"), '(')
});
console.log("Result:",result);
The pattern used to filter and remove the unnecessary words
\b(?:ti|ab|su)(?:,(ti|ab|su))*\(
Details
\b - a word boundary
(?:ti|ab|su) - 1 of the alternatives,
(?:,(ti|ab|su))* - 0 or more repetitions of , followed with 1 of the 3 alternatives
\( - a (.
The match is replaced with ( to restore it in the match.
I have a regex statement with multiple capture groups which are separated by | operator. How can I find out which capture group is matched? Only way I can think of -for this example- is counting the number of characters if something is matched.
var string = "1234567897"
var pattern = "(^\\d{9}$)|(^\\d{10}$)|(^\\d{13}$)|(^[a-zA-Z]{2}\\d{9}[a-zA-Z]{2}$)"
var myRegex = NSRegularExpression(pattern: pattern, options: nil, error: nil)!
if let myMatch = myRegex.firstMatchInString(string, options: nil,
range: NSRange(location: 0, length: string.utf16Count)) {
println((string as NSString).substringWithRange(myMatch.rangeAtIndex(0)))
}
I wrote a code which worked for my example. I am sure it can be written better way but it works for now.
Swift 2.3
var string = "123456789"
var pattern = "(^\\d{9}$)|(^\\d{10}$)|(^\\d{13}$)|(^[a-zA-Z]{2}\\d{9}[wW]{2}$)"
var myRegex = try! NSRegularExpression(pattern: pattern, options: [])
if let myMatch = myRegex.firstMatchInString(string, options: NSMatchingOptions.init(rawValue: 0), range: NSRange(location: 0, length: string.utf16.count)) {
var matchedGroup = 0
for var i in 1..<myMatch.numberOfRanges {
if myMatch.rangeAtIndex(i).length != 0 {
matchedGroup = i
break
}
}
print(matchedGroup)
print((string as NSString).substringWithRange(myMatch.rangeAtIndex(0))) //whatever the range you want to print
}
Swift 3
var string = "123456789"
var pattern = "(^\\d{9}$)|(^\\d{10}$)|(^\\d{13}$)|(^[a-zA-Z]{2}\\d{9}[wW]{2}$)"
var myRegex = try! NSRegularExpression(pattern: pattern, options: [])
if let myMatch = myRegex.firstMatch(in: string, options: NSRegularExpression.MatchingOptions.init(rawValue: 0), range: NSRange(location: 0, length: string.utf16.count)) {
var matchedGroup = 0
for var i in 1..<myMatch.numberOfRanges {
if myMatch.rangeAt(i).length != 0 {
matchedGroup = i
break
}
}
print(matchedGroup)
print((string as NSString).substring(with: myMatch.rangeAt(0))) //whatever the range you want to print
}