Arbitrary Mask Question

[stepheng54012748]

I'm hoping someone can me clean up the following Arbitrary Mask javascript I'd like to use on a form.  My goal is to change it to just 3 alphabet letters.  I can't seem to get the script right.  Here is where i'm at with this and it's not working.

if(event.willCommit){
var input = event.value;
var pattern = /^(|\[A-Za-z]{3}$|$/);
if (!pattern.test(input)) {
event.rc = false;
app.alert("Input must be either 3 alphabet letters, or blank.");}}

Thank you in advance for any ideas on how to make this work!

[try67]

Use this:

var pattern = /^[a-z]{3}$/i);

[stepheng54012748]

Here is what i input and it did not work.

if(event.willCommit){
var input = event.value;
var pattern = /^[a-z]{3}$/i);
if (!pattern.test(input)) {
event.rc = false;
app.alert("Input must be either 3 alphabet letters, or blank.");}}

[try67]

Where did you place it? Under which event?

[stepheng54012748]

Right after event value.  I made the text blue below to show you.

if(event.willCommit){
var input = event.value;
var pattern = /^[a-z]{3}$/i);
if (!pattern.test(input)) {
event.rc = false;
app.alert("Input must be either 3 alphabet letters, or blank.");}}

[try67]

No, the full code. Where did you place it under the field? Format? Keystroke? Validation? Something else?

[stepheng54012748]

Oh.  I was trying to place it in the format, custom, custom keystroke but it was taking it. 

[try67]

Use just this part as the custom Validation script:

var input = event.value;
var pattern = /^[a-z]{3}$/i);
if (!pattern.test(input)) {
event.rc = false;
app.alert("Input must be either 3 alphabet letters, or blank.");}

[stepheng54012748]

First of all, i really appreciate you helping me!

I entered in as a validation script and i got a SyntaxErro:  missing : before statement 2: at line 3

[stepheng54012748]

I was playing around with this, looking at some things you've done for others and I came up with this as a custom validation script.  I think this might work.  I also included a pop up for the user instructing them to enter 3 letters.  This is the best i could come up with.

if (event.value!="") event.rc = /^[a-z]{3}$/i.test(event.value);

[JR Boulay · Accepted solution]

You don't need JavaScript to do this:

• A - Accepts only letters (A–Z, a-z).
• X - Accepts spaces and most printable characters, including all characters available on a standard keyboard and ANSI characters in 32–126 and 128–255.
• O - The letter “O” accepts alphanumeric characters (A–Z, a-z, and 0–9).
• 9 - Accepts only numeric characters (0–9).

[stepheng54012748]

Yes!  This is the answer and THANK YOU!!