daveVk
msg:4407552 | 3:55 am on Jan 17, 2012 (gmt 0) |
Start by removing all spaces
value=value.replace( /\s/g, "" );
And finally reconstruct it, if need be
var parts = value.split('-');
value = parts.join(' - ');
Consider doing simpler regex on parts[0] and parts[1].
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nelsonm
msg:4407553 | 4:04 am on Jan 17, 2012 (gmt 0) |
Hi,
I actually didn't think of removing all then reconstructing. I thought there was a way to do it in the expression and maybe there is but still... Nice!
What do you mean... | Consider doing simpler regex on parts[0] and parts[1]. |
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How can it be simpler?
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daveVk
msg:4407574 | 7:14 am on Jan 17, 2012 (gmt 0) |
Probably can be done in expression, but I prefer keeping it simple.
You could test both parts with
var regexp = /^(1[012]|[1-9])[a|p]m$/i;
and report which part[s] are in error.
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nelsonm
msg:4407765 | 4:20 pm on Jan 17, 2012 (gmt 0) |
I realized i made one mistake on my expression. If i want the expression to recognize a space, i have to use "\s" not " ". So the corrected expression should be:
var regexp = /^(1[012]|[1-9])[a|p]m\s-\s(1[012]|[1-9])[a|p]m$/i;
I'm not too well versed on regular expressions so with respect to...
You could test both parts with
var regexp = /^(1[012]|[1-9])[a|p]m$/i;
and report which part[s] are in error. |
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I think you mean altering my original expression to "/(1[012]|[1-9])[a|p]m/i" which will look for all occurrences of (1 thru 12) with am or pm in the input string.
So, If i understand you correctly...
i would construct the expression as follows
var regexp = /(1[012]|[1-9])[a|p]m/i;
then test for and get the parts...
parts = value.exec(regexp);
If the array is of at least a length of 2, i have at least two valid parts - then join the first two part - ignoring the anything else...
value = parts[0] + ' - ' + part[1];
At this point, i have what i need. Do i have it right?
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nelsonm
msg:4407849 | 6:38 pm on Jan 17, 2012 (gmt 0) |
i meant...
parts = value.match(regexp);
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daveVk
msg:4407961 | 12:27 am on Jan 18, 2012 (gmt 0) |
I had in mind something like below
var parts = value.split('-');
if ( parts.length !== 2 ) { ... return error }
if ( !( regexp.test( part[0] ) ) ) { ... return error }
if ( !( regexp.test( part[1] ) ) ) { ... return error }
value = parts.join(' - ');
... return Ok
In this case I think
var regexp = /^(1[012]|[1-9])[a|p]m$/i;
is Ok, no expert on these.
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g1smd
msg:4407965 | 12:48 am on Jan 18, 2012 (gmt 0) |
Yes, this is a task that should be done in several steps: remove spaces and punctuations, check the remainder is a valid string, format the data for display.
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nelsonm
msg:4407970 | 1:23 am on Jan 18, 2012 (gmt 0) |
Thanks a lot, you've been a great help!
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