Create Null Catch for Count Number of Characters in String

[Bryce Poole]

I may have answered my own question, putting what I did here incase anyone else has a similar question in the future.

I used indexOf to check for the desired input characters then used an if/else statement to either count the number of desired characters or set the variable to 0.

This appears to be working, if anyone else has a more elegant solution let me know.

Thanks!

//defines the variables
txt = thisComp.layer("Overlay TXT Here").text.sourceText;

//statement body
if (txt.indexOf("e"|"f"))
a = (txt.match(/e|f/g)||[]).length
else
a=0;

//print total value
a

[Declan Smith]

Hi
I entered your expression onto one text layer and had text on a second layer and it worked as expected. i.e. it counts the number of occurrences of e and f

If e and f do not appear in the text layer it registers 0, so I can’t see the broken condition. I’m using AE CS6

//defines the variables
txt = thisComp.layer("TXT TEST").text.sourceText;
a = (txt.match(/e|f/g)||[]).length
//print total value
a;

Declan Smith
https://www.madpanic.tv
After Effects CS6/ FCS3 / Canon XLH1 / Canon 7D / Reason / Cubase

“it’s either binary or it’s not”

[Bryce Poole]

Hey Declan,

Your right that one does seem to be okay on its own, specifically I’m running into it when I’m trying to count punctuation like in the code below, that seems to be returning total string length. I’m guessing its because AE is seeing the “.” and “!” as part of the expression as opposed to string values?

Any thoughts on how to address this?

b = ((firstline.match(/.|!/g)||[]).length)

[Declan Smith]

Ah! I see now.

Ok the reason it’s returning the string length is that the matching characters are regular expressions. The ‘.’ is a special character that means any character. You have to escape it by putting a backslash before it.

b = ((firstline.match(/\.|!/g)||[]).length)

Declan Smith
https://www.madpanic.tv
After Effects CS6/ FCS3 / Canon XLH1 / Canon 7D / Reason / Cubase

“it’s either binary or it’s not”

[Bryce Poole]

Perfect, that solved it! Thank you so much Declan!

[Declan Smith]

Glad it helped. There are a bunch of other characters that will also need escaping, such as

/ * + ? | ( ) [ ] { }

Also if you want to count the backslash itself, then escape that by putting another backslash in front i.e. \\

Declan Smith
https://www.madpanic.tv
After Effects CS6/ FCS3 / Canon XLH1 / Canon 7D / Reason / Cubase

“it’s either binary or it’s not”

[Bryce Poole]

Hey Declan,

Okay cool, thank you! I seem to be having issues with counting \ and “, though both are escaped out.

Entering a \ is causing after effects to disable the expression and give an “Unterminated string constant” error and ” just isn’t counting, any thoughts?

Thanks again!

var i = (firstline.match(/f|t|\/|\\|\"/g)||[]).length;

[Declan Smith]

Hi Bryce,

Yes, some more weirdness of javascript etc I’m afraid. The first issue, i.e. \ needs an addition \n at the end of the line otherwise the regular expression sees it as an unterminated line. The next issue is with speech marks in that the UI interface seems to insert opening or closing speech marks which don’t match with what is put in the expression (by the same UI), so you need to specify the exact unicode characters you want to match. Please see below.

i.e. Opening speech marks is 8220 (\u201C) and closing speech marks is 8221 (\u201D).

If you get any other weird characters, then just use:

txt.charCodeAt(0)

to print the decimal version of the character, then convert that to Hexidecimal and then add it to a unicode match. The above assumes 1 character at position 0.

Hope this helps

var txt = thisComp.layer("TXT").text.sourceText + "\n";
var i = (txt.match(/\\|T|\u201C|\u201D|,|\//g)||[]).length;
i;

Declan Smith
https://www.madpanic.tv
After Effects CS6/ FCS3 / Canon XLH1 / Canon 7D / Reason / Cubase

“it’s either binary or it’s not”

[Bryce Poole]

Excellent that worked, thanks again Declan!