I have this code:
If (string1 Like string2) AND string3.Contains(string4) Then
What is the difference of both?
I thought like is a contains but I am not sure... being a C# code.
Taking a look at the documentation, it would appear that the Like keyword has a bit more comparison logic than a simple .Contains() operation. The second string in the Like operation isn't just a string, but a pattern (like a regular expression). For example:
testCheck = "F" Like "[A-Z]"
In this operation testCheck will evaluate to True because the first string matches (or is included in) the pattern identified by the second string.
Like is more powerful as using pattern: http://msdn.microsoft.com/de-de/library/swf8kaxw.aspx (Compares a string against a pattern)
? Any single character
* Zero or more characters
# Any single digit (0–9)
[ charlist ] Any single character in charlist
[! charlist ] Any single character not in charlist
Related
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);
I try to match a pattern which is year-year. For example "2019-20 ggtt", "1990-91 ggcc" and etc.
I used function PatINDEX to look for if the pattern exists in my field, however the below pattern I created did not give me the correct result. Someone please help?
PATINDEX('%[0-9]^4\-[0-9]^2%', fieldName)
Thanks,
Try :
PATINDEX('%[0-9][0-9][0-9][0-9]-[0-9][0-9]%', fieldName)
PATINDEX ( '%pattern%' , expression )
pattern
Is a character expression that contains the sequence to be
found. Wildcard characters can be used; however, the % character must
come before and follow pattern (except when you search for first or
last characters). pattern is an expression of the character string
data type category. pattern is limited to 8000 characters.
It seems, pattern only supports Wildcard characters, not regex
I want to judge if a positive number string is end with ".0", so I wrote the following sql:
select '12310' REGEXP '^[0-9]*\.0$'. The result is true however. I wonder why I got the result, since I use "\" before "." to escape.
So I write another one as select '1231.0' REGEXP '^[0-9]\d*\.0$', but this time the result is false.
Could anyone tell me the right pattern?
Dot (.) in regexp has special meaning (any character) and requires escaping if you want literally dot:
select '12310' REGEXP '^[0-9]*\\.0$';
Result:
false
Use double-slash to escape special characters in Hive. slash has special meaning and used for characters like \073 (semicolon), \n (newline), \t (tab), etc. This is why for escaping you need to use double-slash. Also for character class digit use \\d:
hive> select '12310.0' REGEXP '^\\d*?\\.0$';
OK
true
Also characters inside square brackets do not need double-slash escaping: [.] can be used instead of \\.
If you know it is a number string, why not just use:
select ( val like '%.0' )
You need regular expression if you want to validate that the string has digits everywhere else. But if you only need to check the last two characters, like is sufficient.
As for your question . is a wildcard in regular expressions. It matches any character.
I'm trying to figure out the base regex to capture the middle of a google url out of a sql database.
For example, a few links:
https://www.google.com/cars/?year=2016&model=dodge+durango&id=1234
https://www.google.com/cars/?year=2014&model=jeep+cherokee+crossover&id=6789
What would be the regex to capture the text to get dodge+durango , or jeep+cherokee+crossover ? (It's alright that the + still be in there.)
My Attempts:
1)
\b[=.]\W\b\w{5}\b[+.]?\w{7}
, but this clearly does not work as this is a hard coded scenario that would only work like something for the dodge durango example. (would extract "dodge+durango)
2) Using positive lookback ,
[^+]( ?=&id )
but I am not fully sure how to use this, as this only grabs one character behind the & symbol.
How can I extract a string of (potentially) any length with any amount of + delimeters between the "model=" and "&id" boundaries?
seems like you could use regexp_replace and access match groups:
regexp_replace(input, 'model=(.*?)([&\\s]|$)', E'\\1')
from here:
The regexp_replace function provides substitution of new text for
substrings that match POSIX regular expression patterns. It has the
syntax regexp_replace(source, pattern, replacement [, flags ]). The
source string is returned unchanged if there is no match to the
pattern. If there is a match, the source string is returned with the
replacement string substituted for the matching substring. The
replacement string can contain \n, where n is 1 through 9, to indicate
that the source substring matching the n'th parenthesized
subexpression of the pattern should be inserted, and it can contain \&
to indicate that the substring matching the entire pattern should be
inserted. Write \ if you need to put a literal backslash in the
replacement text. The flags parameter is an optional text string
containing zero or more single-letter flags that change the function's
behavior. Flag i specifies case-insensitive matching, while flag g
specifies replacement of each matching substring rather than only the
first one
I may be misunderstanding, but if you want to get the model, just select everything between model= and the ampersand (&).
regexp_matches(input, 'model=([^&]*)')
model=: Match literally
([^&]*): Capture
[^&]*: Anything that isn't an ampersand
*: Unlimited times
examples:
"""Romeo and Juliet"""
'another string that is quoted with '' single quotes'
the problem is that the string can have characters used for db escaping even in it's beginning and end, so regex should look if given char is used in sequence of odd length that is 1,3,5... at the end of matched string
Try this regex: '(?:[^']|'')*' for single quotes. The same for double quotes, i.e. full regex:
'(?:[^']|'')*'|"(?:[^"]|"")*"
In string hello 'my ''beautiful''' 'world'! """Romeo and Juliet""" it will find:
'my ''beautiful'''
'world'
"""Romeo and Juliet"""
Where your string includes (or might include) characters that cause problems in your sql, you should always escape.
If you're using PHP with a mysql database, use mysql_real_escape_string($text); (docs)
For other databases and languages, you'll have to check for your specific purposes, but there's likely to be an existing method.