Double-quoted strings support the full range of escape sequences, as shown below: \a Bell/alert (0x07) \b Backspace (0x08) \e Escape (0x1b) \f Formford (0x0c) \n Newline (0x0a) \r Return (0x0d) \s Space (0x20) \t Tab (0x09) \v Vertical tab (0x0b) For single-quoted strings, two consecutive backslashes are replaced by a single backslash, and a backslash … Read more
Sound like you want a backslash in text mode; since \backslash does not work, try \textbackslash. EDIT: \symbol{92} should also work.
Quick Answer If you want to sidestep all this confusion, use the much less confusing block syntax. Here is an example that replaces each backslash with 2 backslashes: “some\\path”.gsub(‘\\’) { ‘\\\\’ } Gruesome Details The problem is that when using sub (and gsub), without a block, ruby interprets special character sequences in the replacement parameter. … Read more
For security reasons, it is not possible to get the real, full path of a file, referred through an <input type=”file” /> element. This question already mentions, and links to other Stack Overflow questions regarding this topic. Previous answer, kept as a reference for future visitors who reach this page through the title, tags and … Read more
Accrding to: http://www.vias.org/linux-knowhow/bbg_sect_08_02_01.html : -r If this option is given, backslash does not act as an escape character. The backslash is considered to be part of the line. In particular, a backslash-newline pair may not be used as a line continuation. It works on my machine. $ echo ‘\&|’ | while read -r in; do … Read more
My guess is that you are missing that backslash (‘\’) characters are escape characters in Java String literals. So when you want to use a ‘\’ escape in a regex written as a Java String you need to escape it; e.g. Pattern.compile(“\.”); // Java syntax error // A regex that matches a (any) character Pattern.compile(“.”); … Read more
To answer your question directly, put r in front of the string. final= path + r’\xulrunner.exe ‘ + path + r’\application.ini’ But a better solution would be os.path.join: final = os.path.join(path, ‘xulrunner.exe’) + ‘ ‘ + \ os.path.join(path, ‘application.ini’) (the backslash there is escaping a newline, but you could put the whole thing on one … Read more
replace works for the first quote, so you need a tiny regular expression: str = str.replace(/’/g, “\\'”);
I suspect your string already actually only contains a single backslash, but you’re looking at it in the debugger which is escaping it for you into a form which would be valid as a regular string literal in C#. If print it out in the console, or in a message box, does it show with … Read more
json_encode($response, JSON_UNESCAPED_SLASHES);