swift - replacing characters from keyboard [duplicate] - swift

I have tried to print it but it just by passes because it's an escaped character.
e.g output should be as follows.
\correct

For that and also future reference:
\0 – Null character (that is a zero after the slash)
\\ – Backslash itself. Since the backslash is used to escape other characters, it needs a special escape to actually print itself.
\t – Horizontal tab
\n – Line Feed
\r – Carriage Return
\” – Double quote. Since the quotes denote a String literal, this is necessary if you actually want to print one.
\’ – Single Quote. Similar reason to above.

Use the following code for Swift 5, Xcode 10.2
let myText = #"This is a Backslash: \"#
print(myText)
Output:
This is a Backslash: \
Now not required to add a double slash to use a single slash in swift 5, even now required slash before some character, for example, single quote, double quote etc.
See this post for latest update about swift 5
https://www.hackingwithswift.com/articles/126/whats-new-in-swift-5-0

var s1: String = "I love my "
let s2: String = "country"
s1 += "\"\(s2)\""
print(s1)
It will print I love my "country"

The backslash character \ acts as an escape character when used in a string. This means you can use, for example, double quotes, in a string by pre-pending them with \. The same also applies for the backslash character itself, which is to say that println("\\") will result in just \ being printed.

Related

Output is not generating while running the bash script [duplicate]

In Bash, what are the differences between single quotes ('') and double quotes ("")?
Single quotes won't interpolate anything, but double quotes will. For example: variables, backticks, certain \ escapes, etc.
Example:
$ echo "$(echo "upg")"
upg
$ echo '$(echo "upg")'
$(echo "upg")
The Bash manual has this to say:
3.1.2.2 Single Quotes
Enclosing characters in single quotes (') preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.
3.1.2.3 Double Quotes
Enclosing characters in double quotes (") preserves the literal value of all characters within the quotes, with the exception of $, `, \, and, when history expansion is enabled, !. The characters $ and ` retain their special meaning within double quotes (see Shell Expansions). The backslash retains its special meaning only when followed by one of the following characters: $, `, ", \, or newline. Within double quotes, backslashes that are followed by one of these characters are removed. Backslashes preceding characters without a special meaning are left unmodified. A double quote may be quoted within double quotes by preceding it with a backslash. If enabled, history expansion will be performed unless an ! appearing in double quotes is escaped using a backslash. The backslash preceding the ! is not removed.
The special parameters * and # have special meaning when in double quotes (see Shell Parameter Expansion).
The accepted answer is great. I am making a table that helps in quick comprehension of the topic. The explanation involves a simple variable a as well as an indexed array arr.
If we set
a=apple # a simple variable
arr=(apple) # an indexed array with a single element
and then echo the expression in the second column, we would get the result / behavior shown in the third column. The fourth column explains the behavior.
#
Expression
Result
Comments
1
"$a"
apple
variables are expanded inside ""
2
'$a'
$a
variables are not expanded inside ''
3
"'$a'"
'apple'
'' has no special meaning inside ""
4
'"$a"'
"$a"
"" is treated literally inside ''
5
'\''
invalid
can not escape a ' within ''; use "'" or $'\'' (ANSI-C quoting)
6
"red$arocks"
red
$arocks does not expand $a; use ${a}rocks to preserve $a
7
"redapple$"
redapple$
$ followed by no variable name evaluates to $
8
'\"'
\"
\ has no special meaning inside ''
9
"\'"
\'
\' is interpreted inside "" but has no significance for '
10
"\""
"
\" is interpreted inside ""
11
"*"
*
glob does not work inside "" or ''
12
"\t\n"
\t\n
\t and \n have no special meaning inside "" or ''; use ANSI-C quoting
13
"`echo hi`"
hi
`` and $() are evaluated inside "" (backquotes are retained in actual output)
14
'`echo hi`'
`echo hi`
`` and $() are not evaluated inside '' (backquotes are retained in actual output)
15
'${arr[0]}'
${arr[0]}
array access not possible inside ''
16
"${arr[0]}"
apple
array access works inside ""
17
$'$a\''
$a'
single quotes can be escaped inside ANSI-C quoting
18
"$'\t'"
$'\t'
ANSI-C quoting is not interpreted inside ""
19
'!cmd'
!cmd
history expansion character '!' is ignored inside ''
20
"!cmd"
cmd args
expands to the most recent command matching "cmd"
21
$'!cmd'
!cmd
history expansion character '!' is ignored inside ANSI-C quotes
See also:
ANSI-C quoting with $'' - GNU Bash Manual
Locale translation with $"" - GNU Bash Manual
A three-point formula for quotes
If you're referring to what happens when you echo something, the single quotes will literally echo what you have between them, while the double quotes will evaluate variables between them and output the value of the variable.
For example, this
#!/bin/sh
MYVAR=sometext
echo "double quotes gives you $MYVAR"
echo 'single quotes gives you $MYVAR'
will give this:
double quotes gives you sometext
single quotes gives you $MYVAR
Others explained it very well, and I just want to give something with simple examples.
Single quotes can be used around text to prevent the shell from interpreting any special characters. Dollar signs, spaces, ampersands, asterisks and other special characters are all ignored when enclosed within single quotes.
echo 'All sorts of things are ignored in single quotes, like $ & * ; |.'
It will give this:
All sorts of things are ignored in single quotes, like $ & * ; |.
The only thing that cannot be put within single quotes is a single quote.
Double quotes act similarly to single quotes, except double quotes still allow the shell to interpret dollar signs, back quotes and backslashes. It is already known that backslashes prevent a single special character from being interpreted. This can be useful within double quotes if a dollar sign needs to be used as text instead of for a variable. It also allows double quotes to be escaped so they are not interpreted as the end of a quoted string.
echo "Here's how we can use single ' and double \" quotes within double quotes"
It will give this:
Here's how we can use single ' and double " quotes within double quotes
It may also be noticed that the apostrophe, which would otherwise be interpreted as the beginning of a quoted string, is ignored within double quotes. Variables, however, are interpreted and substituted with their values within double quotes.
echo "The current Oracle SID is $ORACLE_SID"
It will give this:
The current Oracle SID is test
Back quotes are wholly unlike single or double quotes. Instead of being used to prevent the interpretation of special characters, back quotes actually force the execution of the commands they enclose. After the enclosed commands are executed, their output is substituted in place of the back quotes in the original line. This will be clearer with an example.
today=`date '+%A, %B %d, %Y'`
echo $today
It will give this:
Monday, September 28, 2015
Since this is the de facto answer when dealing with quotes in Bash, I'll add upon one more point missed in the answers above, when dealing with the arithmetic operators in the shell.
The Bash shell supports two ways to do arithmetic operation, one defined by the built-in let command and the other the $((..)) operator. The former evaluates an arithmetic expression while the latter is more of a compound statement.
It is important to understand that the arithmetic expression used with let undergoes word-splitting, pathname expansion just like any other shell commands. So proper quoting and escaping need to be done.
See this example when using let:
let 'foo = 2 + 1'
echo $foo
3
Using single quotes here is absolutely fine here, as there isn't any need for variable expansions here. Consider a case of
bar=1
let 'foo = $bar + 1'
It would fail miserably, as the $bar under single quotes would not expand and needs to be double-quoted as
let 'foo = '"$bar"' + 1'
This should be one of the reasons, the $((..)) should always be considered over using let. Because inside it, the contents aren't subject to word-splitting. The previous example using let can be simply written as
(( bar=1, foo = bar + 1 ))
Always remember to use $((..)) without single quotes
Though the $((..)) can be used with double quotes, there isn't any purpose to it as the result of it cannot contain content that would need the double quote. Just ensure it is not single quoted.
printf '%d\n' '$((1+1))'
-bash: printf: $((1+1)): invalid number
printf '%d\n' $((1+1))
2
printf '%d\n' "$((1+1))"
2
Maybe in some special cases of using the $((..)) operator inside a single quoted string, you need to interpolate quotes in a way that the operator either is left unquoted or under double quotes. E.g., consider a case, when you are tying to use the operator inside a curl statement to pass a counter every time a request is made, do
curl http://myurl.com --data-binary '{"requestCounter":'"$((reqcnt++))"'}'
Notice the use of nested double quotes inside, without which the literal string $((reqcnt++)) is passed to the requestCounter field.
There is a clear distinction between the usage of ' ' and " ".
When ' ' is used around anything, there is no "transformation or translation" done. It is printed as it is.
With " ", whatever it surrounds, is "translated or transformed" into its value.
By translation/ transformation I mean the following:
Anything within the single quotes will not be "translated" to their values. They will be taken as they are inside quotes. Example: a=23, then echo '$a' will produce $a on standard output. Whereas echo "$a" will produce 23 on standard output.
A minimal answer is needed for people to get going without spending a lot of time as I had to.
The following is, surprisingly (to those looking for an answer), a complete command:
$ echo '\'
whose output is:
\
Backslashes, surprisingly to even long-time users of bash, do not have any meaning inside single quotes. Nor does anything else.

confused about what must be escaped for sed

I want to replace specific strings in php files automatically using sed. Some work, and some do not. I already investigated this is not an issue with the replacement string but with the string that is to be replaced. I already tried to escape [ and ] with no success. It seems to be the whitespace within the () - not whitespaces in general. The first whitespaces (around the = ) do not have any problems. Please can someone point me to the problem:
sed -e "1,\$s/$adm = substr($path . rawurlencode($upload['name']) , 16);/$adm = rawurlencode($upload['name']); # fix 23/g" -i administration/identify.php
I already tried to shorten the string which should be replaced and the result was if I cut it directly behind $path it works, with the following whitespace it does not. Escaping whitespace has no effect...
what must be escaped for sed
The following characters have special meaning in sed and have to be escaped with \ for the regex to be taken literally:
\
[
the character used in separating s command parts, ie. / here
.
*
& only replacement string
Newline character is handled specially as the end of the string, but can be replaced for \n.
So first escape all special characters in input and then pass it to sed:
rgx="$adm = substr($path . rawurlencode($upload['name']) , 16);"
rgx_escaped=$(sed 's/[\\\[\.\*\/&]/\\&/g' <<<"$rgx")
sed "s/$rgx_escaped/ etc."
See Escape a string for a sed replace pattern for a generic escaping solution.
You may use
sed -i 's/\$adm = substr(\$path \. rawurlencode(\$upload\['"'"'name'"'"']) , 16);/$adm = rawurlencode($upload['"'"'name'"'"']); # fix 23/g' administration/identify.php
Note:
the sed command is basically wrapped in single quotes, the variable expansion won't occur inside single quotes
In the POSIX BRE syntax, ( matches a literal (, you do not need to escape ) either, but you need to escape [ and . that must match themselves
The single quotes require additional quoting with concatenation.

How to Write Multiline String With Significant Whitespace at the End

I’m converting a previously concatenated string to the new multi-line string syntax introduced in Swift 4. It contains signifiant whitespace at the end of lines, so I’m using the string newline escaping to explicitly terminate the line with \n\ after the significant whitespace, to protect it from the editor (Xcode) that strips the whitespace from the end of the line.
let asciiArt = """
\ / \n\
V \
"""
Problem is with the last line, where I don’t want the newline to be part of the string, but cannot use just a single \ because it gives me error:
Escaped newline at the last line is not allowed
How do I write multiline string with significant trailing whitespace that doesn’t contain newline at the end?
Turns out the error is due to quirks in Swift parsing grammar and the workaround in Swift 4.0 and 4.1 is to insert extra newline at the end, to make the compiler happy:
let asciiArt = """
\ / \n\
V \
"""
Notice the empty line before the closing """!
In Summary: to write a multiline string that ends with significant whitespace and doesn’t contain a trailing newline, insert one more newline at the end. The compiler removes the newlines after the opening """, before the closing """ and after every line-trailing \, which leaves you with:
a string that ends in whitespace without \n,
a content compiler and
a question why people follow made-up rules ad-absurdum like robots... 🤖🤦‍♂️🤷‍♂️

Why does q/\\a/ equal q/\a/?

The following example prints "SAME":
if (q/\\a/ eq q/\a/) {
print "SAME\n";
}
else {
print "DIFFERENT\n";
}
I understand this is consistent with the documentation. But I think this behavior is undesirable. Is there a need to escape a backlash lilteral in single-quoted string? If I wanted 2 backlashes, I'd need to specify 4; this does not seem convenient.
Shouldn't Perl detect whether a backslash serves as an escape character or not? For instance, when a backslash does not precede a delimiter, it should be treated as a literal; and if that were the case, I wouldn't need 3 backslashes to express two, e.g.,
q<a\\b>
instead of
q<a\\\b>.
Is there a need to escape a backlash in single-quoted string?
Yes, if the backslash is followed by another backslash, or is the last character in the string:
$ perl -e'print q/C:\/'
Can't find string terminator "/" anywhere before EOF at -e line 1.
$ perl -e'print q/C:\\/'
C:\
This makes it possible to include any character in a single-quoted string, including the delimiter and the escape character.
If I wanted 2 backlashes, I'd need to specify 4; this does not seem convenient.
Actually, you only need three (because the second backslash isn't followed by another backslash). But as an alternative, if your string contains a lot of backslashes you can use a single-quoted heredoc, which requires no escaping:
my $path = <<'END';
C:\a\very\long\path
END
chomp $path;
print $path; # C:\a\very\long\path
Note that the heredoc adds a newline to the end, which you can remove with chomp.
In single-quoted string literals,
A backslash represents a backslash unless followed by the delimiter or another backslash, in which case the delimiter or backslash is interpolated.
In other words,
You must escape delimiters.
You must escape \ that are followed by \ or the delimiter.
You may escape \ that aren't followed by \ or the delimiter.
So,
q/\// ⇒ /
q/\\\\a/ ⇒ \\a
q/\\\a/ ⇒ \\a
q/\\a/ ⇒ \a
q/\a/ ⇒ \a
Is there a need to escape a backlash in single-quoted string?
Yes, if it's followed by another backslash or the delimiter.
If I wanted 2 backlashes, I'd need to specify 4
Three would suffice.
this does not seem convenient.
It's more convenient than double-quoted strings, where backslashes must always be escaped. Single-quoted string require the minimum amount of escaping possible without losing the ability to produce the delimiter.

Set String with quotes to label in objective C programming?

trackLbl2.text = #""Nonstop Bollywood Music"";
I try this but not working. How to text with double quotes.
Try this:
trackLbl2.text = #"\"Nonstop Bollywood Music\"";
You have to put \ before any special character to print it out.
The user before me provided a good answer, but I would like to elaborate on your problem a little bit. It's a good thing to know.
In most (all?) programming languages, certain characters usually serve a special function - single quotes, double quotes, backslashes, and the likes. If you want to include those inside a string, the common practice is to "escape" them, or use escape characters. Escape characters are special character combinations that get replaced with the actual character you want to see inside the string; they usually start with a forward slash (\)*.
Here is a list of common escape characters used in Objective C (source: Wikipedia)
\a - Sound alert
\b - Backspace
\f - Form feed
\n - New line
\r - Carriage return
\t - Horizontal tab
\v - Vertical tab
\ - Backslash
\" - Double quote (used when placing a double quote into a string declaration)
\' - Single quote (used when placing a double quote into a string declaration)
Fun fact: I actually had to escape that forward slash in there by typing "\\".
Try this
trackLbl2.text = #"\"Nonstop Bollywood Music\"";
You have to put \ before any special character to print it out.