di98mase schreef:
> On 12 Mar, 15:28, Janis <janis_papanag...@[EMAIL PROTECTED]
> wrote:
>> On 12 Mrz., 15:15, Luuk <L...@[EMAIL PROTECTED]
> wrote:
>>
>>
>>
>>
>>
>>> di98mase schreef:
>>>> Hi
>>>> I have a file with numbers in it. My program reads the values and
>>>> compare them and depending on the value I increment a counter. what I
>>>> have seen is that the evaluation fails. This is an extract of my
>>>> numbers:
>>>> 7
>>>> 7
>>>> 7
>>>> 34
>>>> 92
>>>> 111
>>>> 294
>>>> 121
>>>> 142
>>>> This is parts of my program that checks the value (ms) read from the
>>>> file and increments the counters
>>>> if( ms == 18) t18++;
>>>> if( ms == 19) t19++;
>>>> if( ms >= 20 && ms < 30)
>>>> {
>>>> print ms;
>>>> t20++;
>>>> }
>>>> if( ms >= 30 && ms < 40) t30++;
>>>> if( ms >= 40 && ms < 50) t40++;
>>>> if( ms >= 50 && ms < 60) t50++;
>>>> if( ms >= 60 && ms < 70) t60++;
>>>> As you can see I dont have any numbers between 20 and 30 but I still
>>>> get an increment for the t20 counter. When I print the value it is
>>>> "294". Why do I have this behaviour?
>>> its compared alphanumeric
>> The interesting question is; Why?
>>
>> And the answer might most likely be that 'ms' had been converted
>> to a string somewhere in an unquoted part of his program (i.e.
>> the opposite of the conversion you propose below). So if there's
>> some code like
>>
>> { ms = $1 ""
>> ...
>> if( ms == 18) t18++;
>> if( ms == 19) t19++;
>> if( ms >= 20 && ms < 30)
>> ...
>> }
>>
>> It might be a more appropriateto fix the original number-to-string
>> conversion.
>>
>> Janis
>>
>>
>>
>>
>>
>>> this will work:
>>> if( 0+ms >= 20 && 0+ms < 30)
>>> --
>>> Luuk- Zitierten Text ausblenden -
>>> - Zitierten Text anzeigen -- Dölj citerad text -
>> - Visa citerad text -- Dölj citerad text -
>>
>> - Visa citerad text -
>
>
> Hi all,
>
> this is the missing part that converts from a string to an int that
> was not shown in the previous post:
>
> # Try to find out the transaction times from the logs that looks like:
> # /I/TR End, Time:82433(89) Stat:0/0016e0dch:(
> /TR End, / {
>
> match($3,"Time:")
> usecStart = RSTART + RLENGTH
> match($3,"[(]")
> usecEnd = RSTART
> usecLen = usecEnd - usecStart
> us = substr($3 , usecStart, usecLen)
>
> # Make sure that all found the matching pattern
> if (usecStart != -1 && usecEnd != -1 && usecLen != -1)
> {
> # Convert to ms and truncate decimals
> temp = us/1000;
> ms = sprintf ("%d", temp)
>
> if( ms == 1) t1++;
> if( ms == 2) t2++;
> if( ms == 3) t3++;
> if( ms == 4) t4++;
> if( ms == 5) t5++;
> etc
> etc
>
> I thougth that the sprintf would convert the string to a number, or?
man awk says:
A number is converted to a string by using the value of CONVFMT as a
format string for sprintf(3), with the numeric value of the
variable as the argument.
so, if you want it to be numeric type:
ms = 0 + sprintf ("%d", temp)
but i did not test this to see if your equation will work ;-)
--
Luuk
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