How to pass variable in SQL from Powershell script? [duplicate] - sql

I have a string that I want to insert dynamically a variable. Ex;
$tag = '{"number" = "5", "application" = "test","color" = "blue", "class" = "Java"}'
I want to accomplish:
$mynumber= 2
$tag = '{"number" = "$($mynumber)", "application" = "test","color" = "blue", "class" = "Java"}'
What I want is to have the variable inserted on the string, But it is not going through. I guess the '' sets all as a string. Any recomendations on how should I approach this?
thanks!
powershell test and trial and error. Also Google.

The reason your current attempt doesn't work is that single-quoted (') string literals in PowerShell are verbatim strings - no attempt will be made at expanding subexpression pipelines or variable expressions.
If you want an expandable string literal without having to escape all the double-quotes (") contained in the string itself, use a here-string:
$mynumber = 2
$tag = #"
{"number" = "$($mynumber)", "application" = "test","color" = "blue", "class" = "Java"}
"#

To add to Mathias' helpful answer:
Mistakenly expecting string interpolation inside '...' strings (as opposed to inside "...") has come up many times before, and questions such as yours are often closed as a duplicate of this post.
However, your question is worth answering separately, because:
Your use case introduces a follow-up problem, namely that embedded " characters cannot be used as-is inside "...".
More generally, the linked post is in the context of argument-passing, where additional rules apply.
Note: Some links below are to the relevant sections of the conceptual about_Quoting_Rules help topic.
In PowerShell:
only "..." strings (double-quoted, called expandable strings) perform string interpolation, i.e. expansion of variable values (e.g. "... $var" and subexpressions (e.g., "... $($var.Prop)")
not '...' strings (single-quoted, called verbatim strings), whose values are used verbatim (literally).
With "...", if the string value itself contains " chars.:
either escape them as `" or ""
E.g., with `"; note that while use of $(...), the subexpression operator never hurts (e.g. $($mynumber)), it isn't necessary with stand-alone variable references such as $mynumber:
$mynumber= 2
$tag = "{`"number`" = `"$mynumber`", `"application`" = `"test`",`"color`" = `"blue`", `"class`" = `"Java`"}"
Similarly, if you want to selectively suppress string interpolation, escape $ as `$
# Note the ` before the first $mynumber.
# -> '$mynumber = 2'
$mynumber = 2; "`$mynumber` = $mynumber"
See the conceptual about_Special_Characters help topic for info on escaping and escape sequences.
If you need to embed ' inside '...', use '', or use a (single-quoted) here-string (see next).
or use a double-quoted here-string instead (#"<newline>...<newline>"#):
See Mathias' answer, but generally note the strict, multiline syntax of here-strings:
Nothing (except whitespace) must follow the opening delimiter on the same line (#" / #')
The closing delimiter ("# / '#) must be at the very start of the line - not even whitespace may come before it.
Related answers:
Overview of PowerShell's expandable strings
Overview of all forms of string literals in PowerShell
When passing strings as command arguments, they are situationally implicitly treated like expandable strings (i.e. as if they were "..."-enclosed); e.g.
Write-Output $HOME\projects - see this answer.
Alternatives to string interpolation:
Situationally, other approaches to constructing a string dynamically can be useful:
Use a (verbatim) template string with placeholders, with -f, the format operator:
$mynumber= 2
# {0} is the placeholder for the first RHS operand ({1} for the 2nd, ...)
'"number" = "{0}", ...' -f $mynumber # -> "number" = "2", ...
Use simple string concatenation with the + operator:
$mynumber= 2
'"number" = "' + $mynumber + '", ...' # -> "number" = "2", ...

Related

Printing Unnecessary escape character [duplicate]

I tried many ways to get a single backslash from an executed (I don't mean an input from html).
I can get special characters as tab, new line and many others then escape them to \\t or \\n or \\(someother character) but I cannot get a single backslash when a non-special character is next to it.
I don't want something like:
str = "\apple"; // I want this, to return:
console.log(str); // \apple
and if I try to get character at 0 then I get a instead of \.
(See ES2015 update at the end of the answer.)
You've tagged your question both string and regex.
In JavaScript, the backslash has special meaning both in string literals and in regular expressions. If you want an actual backslash in the string or regex, you have to write two: \\.
The following string starts with one backslash, the first one you see in the literal is an escape character starting an escape sequence. The \\ escape sequence tells the parser to put a single backslash in the string:
var str = "\\I have one backslash";
The following regular expression will match a single backslash (not two); again, the first one you see in the literal is an escape character starting an escape sequence. The \\ escape sequence tells the parser to put a single backslash character in the regular expression pattern:
var rex = /\\/;
If you're using a string to create a regular expression (rather than using a regular expression literal as I did above), note that you're dealing with two levels: The string level, and the regular expression level. So to create a regular expression using a string that matches a single backslash, you end up using four:
// Matches *one* backslash
var rex = new RegExp("\\\\");
That's because first, you're writing a string literal, but you want to actually put backslashes in the resulting string, so you do that with \\ for each one backslash you want. But your regex also requires two \\ for every one real backslash you want, and so it needs to see two backslashes in the string. Hence, a total of four. This is one of the reasons I avoid using new RegExp(string) whenver I can; I get confused easily. :-)
ES2015 and ES2018 update
Fast-forward to 2015, and as Dolphin_Wood points out the new ES2015 standard gives us template literals, tag functions, and the String.raw function:
// Yes, this unlikely-looking syntax is actually valid ES2015
let str = String.raw`\apple`;
str ends up having the characters \, a, p, p, l, and e in it. Just be careful there are no ${ in your template literal, since ${ starts a substitution in a template literal. E.g.:
let foo = "bar";
let str = String.raw`\apple${foo}`;
...ends up being \applebar.
Try String.raw method:
str = String.raw`\apple` // "\apple"
Reference here: String.raw()
\ is an escape character, when followed by a non-special character it doesn't become a literal \. Instead, you have to double it \\.
console.log("\apple"); //-> "apple"
console.log("\\apple"); //-> "\apple"
There is no way to get the original, raw string definition or create a literal string without escape characters.
please try the below one it works for me and I'm getting the output with backslash
String sss="dfsdf\\dfds";
System.out.println(sss);

Escape String interpolation in a string literal

In a normal String I can escape the ${variable} with a backslash:
"You can use \${variable} syntax in Kotlin."
Is it possible to do the same in a String literal? The backslash is no longer an escape character:
// Undesired: Produces "This \something will be substituted.
"""This \${variable} will be substituted."""
So far, the only solutions I see are String concatenation, which is terribly ugly, and nesting the interpolation, which starts to get a bit ridiculous:
// Desired: Produces "This ${variable} will not be substituted."
"""This ${"\${variable}"} will not be substituted."""
From kotlinlang.org:
If you need to represent a literal $ character in a raw string (which doesn't
support backslash escaping), you can use the following syntax:
val price = """
${'$'}9.99
"""
So, in your case:
"""This ${'$'}{variable} will not be substituted."""
As per String templates docs you can represent the $ directly in a raw string:
Templates are supported both inside raw strings and inside escaped strings. If you need to represent a literal $ character in a raw string (which doesn't support backslash escaping), you can use the following syntax:
val text = """This ${'$'}{variable} will be substituted."""
println(text) // This ${variable} will be substituted.

How do I replace duplicate whitespaces in a String in Kotlin?

Say I have a string: "Test me".
how do I convert it to: "Test me"?
I've tried using:
string?.replace("\\s+", " ")
but it appears that \\s is an illegal escape in Kotlin.
replace function in Kotlin has overloads for either raw string and regex patterns.
"Test me".replace("\\s+", " ")
This replaces raw string \s+, which is the problem.
"Test me".replace("\\s+".toRegex(), " ")
This line replaces multiple whitespaces with a single space.
Note the explicit toRegex() call, which makes a Regex from a String, thus specifying the overload with Regex as pattern.
There's also an overload which allows you to produce the replacement from the matches. For example, to replace them with the first whitespace encountered, use this:
"Test\n\n me".replace("\\s+".toRegex()) { it.value[0].toString() }
By the way, if the operation is repeated, consider moving the pattern construction out of the repeated code for better efficiency:
val pattern = "\\s+".toRegex()
for (s in strings)
result.add(s.replace(pattern, " "))

how to avoid WHERE to be seen as attribute in parsing sql using haskell

I am parsing SQL in Haskell using Parsec. How can I ensure that a statement with a where clause will not treat the WHERE as a table name? Find below some part of my coding. The p_Combination works but it sees the WHERE as part of the list of attributes instead of the where clause.
--- from clause
data Table_clause = Table {table_name :: String, alias :: Maybe String} deriving Show
p_Table_clause:: Parser Table_clause
p_Table_clause = do
t <- word
skipMany (space <?> "require space at the Table clause")
a <- optionMaybe (many1 (alphaNum)) <?> "aliase for table or nothing"
return $ Table t a
newtype From_clause = From [Table_clause] deriving Show
p_From_clause :: Parser From_clause
p_From_clause = do
string "FROM" <?> "From";
skipMany1 (space <?> "space in the from clause ")
x <- sepBy p_Table_clause (many1(char ',' <|> space))
return $ From x
-- where clause conditions elements
data WhereClause = WhereFCondi String deriving Show
p_WhereClause :: Parser WhereClause
p_WhereClause = do
string "WHERE"
skipMany1 space
x <- word
return $ WhereFCondi x
data Combination = FromWhere From_clause (Maybe WhereClause) deriving Show
p_Combination :: Parser Combination
p_Combination = do
x <- p_From_clause
skipMany1 space
y <- optionMaybe p_WhereClause
return $ FromWhere x y
Normal SQL parsers have a number of reserved words, and they’re often not context-sensitive. That is, even where a where might be unambiguous, it is not allowed simply because it is reserved. I’d guess most implementations do this by first lexing the source in a conceptually separate stage from parsing the lexed tokens, but we do not need to do that with Parsec.
Usually the way we do this with Parsec is by using Text.Parsec.Token. To use it, you first create a LanguageDef defining some basic characteristics about the language you intend to parse: how comments work, the reserved words, whether it’s case sensitive or not, etc. Then you use makeTokenParser to get a record full of functions tailored to that language. For example, identifier will not match any reserved word, they are all careful to require whitespace where necessary, and when they are skipping whitespace, comments are also skipped.
If you want to stay with your current approach, using only Parsec primitives, you’ll probably want to look into notFollowedBy. This doesn’t do exactly what your code does, but it should provide some inspiration about how to use it:
string "FROM" >> many1 space
tableName <- many1 alphaNum <* many1 space
aliasName <- optionMaybe $ notFollowedBy (string "WHERE" >> many1 space)
>> many1 alphaNum <* many1 space
Essentially:
Parse a FROM, then whitespace.
Parse a table name, then whitespace.
If WHERE followed by whitespace is not next, parse an alias name then whitespace.
I guess the problem is that p_Table_clause accepts "WHERE". To fix this, check for "WHERE" and fail the parser:
p_Table_clause = do
t <- try (do w <- word
if w == "WHERE"
then unexpected "keyword WHERE"
else return w)
...
I guess there might be a try missing in sepBy p_Table_clause (many1 (char ',' <|> space)). I would try sepBy p_Table_clause (try (many1 (char ',' <|> space))).
(Or actually, I would follow the advice from the Parsec documentation and define a lexeme combinator to handle whitespace).
I don't really see the combinator you need right away, if there is one. Basically, you need p_Combination to try (string "WHERE" >> skipMany1 space) and if it succeeds, parse a WHERE "body" and be done. If it fails, try p_Table_clause, if it fails be done. If p_Table_clause succeeds, read the separator and loop. After the loop is done prepend the Table_clause to the results.
There's some other problems with your parser, too. many1 (char ',' <|> space) matches " ,,, , ,, " which is not a valid separator between tables in a from clause, for example. Also, SQL keywords are case-insensitive, IIRC.
In general, you want to exclude keywords from matching identifiers, with something like:
keyword :: Parser Keyword
keyword = string "WHERE" >> return KW_Where
<|> string "FROM" >> return KW_From
<|> string "SELECT" >> return KW_Select
identifier :: Parser String
identifier = try (keyword >> \kw -> fail $ "Expected identifier; got:" ++ show kw)
<|> (liftA2 (:) identiferStart (many identifierPart))
If two (or more) or your keywords have common prefixes, you'll want to combine them for more efficiency (less backtracking) like:
keyword :: Parser Keyword
keyword = char 'D' >> ( string "ROP" >> KW_Drop
<|> string "ELETE" >> KW_Delete
)
<|> string "INSERT" >> return KW_Insert

why does using "\" shows error in jython

I am trying to use a copy command for Windows and we have directories such as c:\oracle.
While trying to execute one such, we get the following error:
source_file=folder+"\"
^
SyntaxError: Lexical error at line 17, column 23. Encountered: "\r" (13), after : ""
Here folder is my path of c:\oracle and while trying to add file to it like:
source=folder+"\"+src_file
I am not able to do so. Any suggestion on how to solve this issue?
I tried with / but my copy windows calling source in os.command is getting "the syntax is incorrect" and the only way to solve it is to use \ but I am getting the above error in doing so.
Please suggest. Thanks for your help
Thanks.
Short answer:
You need:
source_file = folder + "\\" + src_file
Long answer:
The problem with
source_file = folder + "\" + src_file
is that \ is the escape character. What it's doing in this particular case is escaping the " so that it's treated as a character of the string rather than the string terminator, similar to:
source_file = folder + "X + src_file
which would have the same problem.
In other words, you're trying to construct a string consisting of ", some other text and the end of line (\r, the carriage return character). That's where your error is coming from:
Encountered: "\r" (13)
Paxdiablo is absolutely correct about why \ isn't working for you. However, you could also solve your problem by using os.path.normpath instead of trying to construct the proper platform-specific path characters yourself.
In all programming languages I know of, you can't put a quote inside a string like this: "this is a quote: "." The reason for this is that the first quote opens the string, the second then closes it (!), and then the third one opens another string - with the following two problems:
whatever is between the quotes #2 and #3 is probably not valid code;
the quote #3 is probably not being closed.
There are two common mechanisms of solving this: doubling and escaping. Escaping is far more common, and what it means is you put a special character (usually \) in front of characters that you don't want to be interpreted in their usual value. Thus, "no, *this* is a quote: \"." is a proper string, where the quote #2 is not closing the string - and the character \ does not appear.
However, now you have another problem - how do you actually make the escape character appear in a string? Simple: escape it! "This is an escape: \\!" is how you do it: the backslash #1 is the escape character, and the backslash #2 is the escapee: it will not be interpreted with its usual escape semantics, but as a simple backslash character.
Thus, your line should say this:
source=folder+"\\"+src_file
BTW: upvote for both #paxdiablo (who got in before my diatribe) and #Nick (who has a proper Pythonic way to do what you want to do)