Escape backslash followed by apostrophe

[Sisilla]

Hello All,

I have a String object called "cargotype" that is assigned a value of
"20' Standard Container". I want to pass the cargotype object to a
javascript function, so I need to add a backslash before the
apostrophe to escape it. Ideally, I would like to pass "20\' Standard
Container" to the javascript function. I thought the following would
work-:

cargotype = cargotype.replaceAll("'", "\\'");

What am I doing wrong here? I appreciate any effort to help me. Thank
you for your time and consideration.

Sincerely,

Sisilla

 

[Daniel Pitts]

Hello All,

I have a String object called "cargotype" that is assigned a value of
"20' Standard Container". I want to pass the cargotype object to a
javascript function, so I need to add a backslash before the
apostrophe to escape it. Ideally, I would like to pass "20\' Standard
Container" to the javascript function. I thought the following would
work-:

cargotype = cargotype.replaceAll("'", "\\'");

What am I doing wrong here? I appreciate any effort to help me. Thank
you for your time and consideration.

Sincerely,

Sisilla

replaceAll takes an Regex, and does a regex replacement. Try "\\\\'".
I don't know that it'll work, so you'll have to try that yourself.

Basically, what that does is passes in the string \\' to the regex
handler, which escapes the \, and passes the \' to the javascript.

 

[Sisilla]

Thank you, Daniel. It worked like a charm, and I really appreciate it.

Thanks,

Sisilla

 

[Christian]

replaceAll takes an Regex, and does a regex replacement. Try "\\\\'". I
don't know that it'll work, so you'll have to try that yourself.

Basically, what that does is passes in the string \\' to the regex
handler, which escapes the \, and passes the \' to the javascript.

the first argument is a regexp not the second..
cargotype = cargotype.replaceAll(Pattern.quote("'"), "\\'");
should do the trick..

 

[Daniel Pitts]

the first argument is a regexp not the second..
cargotype = cargotype.replaceAll(Pattern.quote("'"), "\\'");
should do the trick..

The second one is the regex replacement, which isn't a normal string
either. \1 represents the first captured group, etc...