If I do var str1 = "a string" then I get a string literal (for want of a better term) when I look at the string in the debugger. i.e. it just displays "a string"
However if i do var str2 = new Ember.String("another string") then it appears as a String object with the chars listed as array items rather than just as "another string"
This is a problem because if i have an array like var myarray = ["str1", "str2","str3"] and do myarray.contains(myvar) it won't work if myvar is of type String.
Now I came across this because of an action handler where I had passed "this" and found it to have become type String. What on earth is going on!
How can I turn my String back into a raw js string so it can be used in lookups in my array of constants?
The object you're getting is not an 'Ember string', it's just a Javascript string object created via new String(''). I don't know why you're coming across one, but if you do, just convert it to a string primitive. You can either do that with the toString() method, or (the safer way) by just concatenating an empty string: str2 + ''.
Related
I'm doing a small javascript method, which receive a list of point, and I've to read those points to create a Polygon in a google map.
I receive those point on the form:
(lat, long), (lat, long),(lat, long)
So I've done the following regex:
\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)
I've tested it with RegexPal and the exact data I receive:
(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)
and it works, so why when I've this code in my javascript, I receive null in the result?
var polygons="(25.774252, -80.190262),(18.466465, -66.118292),(32.321384, -64.75737),(25.774252, -80.190262)";
var reg = new RegExp("/\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g");
var result = polygons.match(reg);
I've no javascript error when executing(with debug mode of google chrome). This code is hosted in a javascript function which is in a included JS file. This method is called in the OnLoad method.
I've searched a lot, but I can't find why this isn't working. Thank you very much!
Use a regex literal [MDN]:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
You are making two errors when you use RegExp [MDN]:
The "delimiters" / are should not be part of the expression
If you define an expression as string, you have to escape the backslash, because it is the escape character in strings
Furthermore, modifiers are passed as second argument to the function.
So if you wanted to use RegExp (which you don't have to in this case), the equivalent would be:
var reg = new RegExp("\\(\\s*([0-9.-]+)\\s*,\\s([0-9.-]+)\\s*\\)", "g");
(and I think now you see why regex literals are more convenient)
I always find it helpful to copy and past a RegExp expression in the console and see its output. Taking your original expression, we get:
/(s*([0-9.-]+)s*,s([0-9.-]+)s*)/g
which means that the expressions tries to match /, s and g literally and the parens () are still treated as special characters.
Update: .match() returns an array:
["(25.774252, -80.190262)", "(18.466465, -66.118292)", ... ]
which does not seem to be very useful.
You have to use .exec() [MDN] to extract the numbers:
["(25.774252, -80.190262)", "25.774252", "-80.190262"]
This has to be called repeatedly until the whole strings was processed.
Example:
var reg = /\(\s*([0-9.-]+)\s*,\s([0-9.-]+)\s*\)/g;
var result, points = [];
while((result = reg.exec(polygons)) !== null) {
points.push([+result[1], +result[2]]);
}
This creates an array of arrays and the unary plus (+) will convert the strings into numbers:
[
[25.774252, -80.190262],
[18.466465, -66.118292],
...
]
Of course if you want the values as strings and not as numbers, you can just omit the +.
I recently discovered thatString#split method returns a one-element array with the original string in it if no matches found:
"azerty keyboard".split(/\n/)
=> ["azerty keyboard"]
"azerty keyboard".split(',')
=> ["azerty keyboard"]
What would be a solution to not split if there are no matches found? Just use:
if "azerty keyboard".match(/\n/)
...# split here
end
or there is a better way?
Since we're playing with Ruby, you could open up the String class and create a new function that works the way you expect:
class String
def split_if(matches)
match(matches) ? split(matches) : self
end
end
Which would then return:
"azerty keyboard".split_if(/\n/)
=> "azerty keyboard"
"azerty keyboard".split_if(' ')
=> ["azerty", "keyboard"]
Although keep in mind that just because we can, doesn't always mean we should. Some folks think opening up classes is a big no-no. So should you do this? That's up to you, your team and the needs of your application.
Clearly you wouldn't run this if you know what the string already contains but assume you have a string object:
string = "azerty keyboard"
split_string = string.split(/\n/) if string.include?("\n")
=>nil
You can also give match a block instead of using the if statement:
str = 'azerty keyboard'
str.match(/\n/) {|m| str.split(m[0])} || str
i'm trying to capture a value between two strings using VB.NET
Each line from the file i'm reading in from can contain many different parameters, in any order, and I'd like to store the values of these parameters in their own variables. Two sample lines would be:
identifier="121" messagecount="112358" timestamp="11:31:41.622" column="5" row="98" colour="ORANGE" value="Hello"
or it could be:
identifier="1121" messagecount="1123488" timestamp="19:14:41.568" valid="true" state="running"
Also, this may not be the sole text in the string, there may be other values before and after (and in between) the parameters i would like to capture.
So essentially i'd need to store everything between 'identifier="' and it's closing '"' into an identifier variable, and so on... As the order of these parameters within each line can change, i can't simply stick the first value in one variable each time, I have to refer to them specifically by what their name is (identifier, messagecount) etc.
Can anyone help? Thanks. I guess it would be via a regular expression, but i'm not too hot on those. I'd prefer to have each expression for each paramater within it's own statement, rather than being all in one, thanks.
Here is a sample how you can go about that. It converts one line into a dictionary.
This will capture any string consisting of a-z-characters (case-insensitive) as the attribute name, and then catch any character other than " in the value string. (If " can occur in the string as "" you need to add some treatment for that.)
Imports System.Text.RegularExpressions
[...]
Dim s As String =
"identifier=""121"" messagecount=""112358"" " &
"timestamp=""11:31:41.622"" column=""5"" row=""98"" " &
"colour=""ORANGE"" value=""Hello"""
Dim d As New Dictionary(Of String, String)
Dim rx As New Regex("([a-z]+)=""(.*?)""", RegexOptions.IgnoreCase)
Dim rxM As MatchCollection = rx.Matches(s)
For Each M As Match In rxM
d.Add(M.Groups(1).Value, M.Groups(2).Value)
Next
' Dictionary is ready
' test output
For Each k As String In d.Keys
MsgBox(String.Format("{0} => {1}", k, d(k)))
Next
You just need to split the data into manageable clumps, and then go through it. Something like this to start you off.
Private Sub ProcessMyData(LineOfData As String)
' NOTE! This assumes all your 'names' have no spaces in!
Dim vElements = LineOfData.Split({" "c}, StringSplitOptions.RemoveEmptyEntries)
For Each vElement In vElements
Dim vPair = vElement.Split({"="c})
Dim vResult = vPair(1).Trim(Convert.ToChar(34))
Select Case vPair(0).ToLower
Case "identifier"
MyIDVariable = CInt(vResult)
Case "colour"
MyColourVariable = vResult
' etc., etc.
End Select
Next
End Sub
You can define the variables you want locally in the sub [function], and then return a list/dictionary/custom class of the things you're interested in.
I'm currently in an Intro C++ class, and I'm learning about Strings and Member Functions of them.
I have questions that are like this:
Assume that name is a variable of type string that has been assigned a value. Write an expression whose value is a string containing the first character of the value of name . So if the value of name were "Smith" the expression's value would be "S".
or
Assume that name is a variable of type string that has been assigned a value. Write an expression whose value is a string containing the last character of the value of name . So if the value of name were "Smith" the expression's value would be "h".
or
Assume that word is a variable of type string that has been assigned a value. Write an expression whose value is a string consisting of the last three characters of the value of word . So if the value of word were "biggest" the expression's value would be "est".
I know things like name[0] and name[name.length() - 1], but I don't know how to turn those into a string in one expression. I've been looking for a table or list of member functions that can help me do this, but I'm stuck. Any directions or aid would be great. :D
Take a look at the substr method.
http://www.cplusplus.com/reference/string/string/substr/
I would use substring:
//assume that the string in question is held in a variable with the name str
string s1 = str.substr(0,1);
string s2 = str.substr(str.length() - 2, 1);
string s3 = str.substr(str.length() = 4, 3);
Hopefully that helps!
There are quite a few number of ways to do it. For instance, an std::string is also a standard container of char. So you could begin with an empty string, and append one char to it.
Using Actionscript 3.0 (Within Flash CS5)
A standard regex to match any digit is:
var myRegexPattern:Regex = /\d/g;
What would the regex look like to incorporate a string variable to match?
(this example is an 'IDEAL' not a 'WORKING' snippet) ie:
var myString:String = "MatchThisText"
var myRegexPatter_WithString:Regex = /\d[myString]/g;
I've seen some workarounds which involve creating multiple regex instances, then combine them by source, with the variable in question, which seems wrong. OR using the flash string to regex creator, but it's just plain sloppy with all the double and triple escape sequences required.
There must be some pain free way that I can't find in the live docs or on google. Does AS3 hold this functionality even? If not, it really should.
Or I am missing a much easier means of simply avoiding this task that I'm simply naive too due to my newness to regex?
I've actually blogged about this, so I'll just point you there: http://tyleregeto.com/using-vars-in-regular-expressions-as3 It talks about the possible solutions, but there is no ideal one like you mention.
EDIT
Here is a copy of the important parts of that blog entry:
Here is a regex to strip the tags from a block of text.
/<("[^"]*"|'[^']*'|[^'">])*>/ig
This nifty expression works like a charm. But I wanted to update it so the developer could limit which tags it stripped to those specified in a array. Pretty straight forward stuff, to use a variable value in a regex you first need to build it as a string and then convert it. Something like the following:
var exp:String = 'start-exp' + someVar + 'more-exp';
var regex:Regexp = new RegExp(exp);
Pretty straight forward. So when approaching this small upgrade, that's what I did. Of course one big problem was pretty clear.
var exp:String = '/<' + tag + '("[^"]*"|'[^']*'|[^'">])*>/';
Guess what, invalid string! Better escape those quotes in the string. Whoops, that will break the regex! I was stumped. So I opened up the language reference to see what I could find. The "source" parameter, (which I've never used before,) caught my eye. It returns a String described as "the pattern portion of the regular expression." It did the trick perfectly. Here is the solution:
var start:Regexp = /])*>/ig;
var complete:RegExp = new RegExp(start.source + tag + end.source);
You can reduce it down to this for convenience:
var complete:RegExp = new RegExp(/])*>/.source + tag, 'ig');
As Tyler correctly points out (and his answer works just fine), you can assemble your regex as a string end then pass this string to the RegExp constructor with the new RegExp("pattern", "flags") syntax.
function assembleRegex(myString) {
var re = new RegExp('\\d' + myString, "i");
return re;
}
Note that when using a string to store a regex pattern, you do need to add some extra backslashes to get it to work right (e.g. to get a \d in the regex, you need to specify \\d in the string). Note also that the string pattern does not use the forward slash delimiters. In other words, the following two statements are equivalent:
var re1 = /\d/ig;
var re2 = new Regexp("\\d", "ig");
Additional note: You may need to process the myString variable to escape any backslashes it might contain (if they are to be interpreted as literal). If this is the case the function becomes:
function assembleRegex(myString) {
myString = myString.replace(/\\/, '\\\\');
var re = new RegExp('\\d' + myString);
return re;
}