解决java的Long.getLong(str)的NullPointerException错误

问题：longtemp=Long.getLong("1366937075");出现空指针NullPointerException错误

解决：用Long.parseLong(Stringstr)

疑问：Longtmp=Long.getLong("1366937075");这个返回也为null，为什么？

参考：http://stackoverflow.com/questions/7376857/long-getlong-failing-returning-null-to-valid-string

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Long.getLong()failing,returningnulltovalidstring


up
vote2down
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I've
spentthepasttwohoursdebuggingwhatseemsextremelyunlikely.I'vestrippedthemethodofasecondaryAndroidActivitytoexactlythis:[code]publicvoidonClick(Viewv){
Stringstr="25";
longmy_long=Long.getLong(str);
}//onClick(v)And
yeah,Igetacrashwiththegoodol'NullPointerException:09-11
02:02:50.444:ERROR/AndroidRuntime(1588):Uncaughthandler:threadmainexitingduetouncaughtexception09-1102:02:50.464:ERROR/AndroidRuntime(1588):java.lang.NullPointerExceptionIt
lookslike(fromothertests)thatLong.getLong(str)returnsNULL,whichisdrivingmebonkers.WHATAMIMISSING?Thanks
inadvance.I'mokaywithstupidlymissingtheobvious,butmysanityisontheline.

androidstringnullpointerexceptionlong-integerstring-parsing
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askedSep
11'11at7:17

Scott
Biggs
503617
1
use
Long.parseLong(str);insteadofLong.getLong(str);–user370305Sep
11'11at7:31
4Answers
activeoldestvotes
up
vote12down
voteaccepted
You
aremissingthefactthat
Long.getLong(String
str)
isnotsupposedtoparseaStringtoalong,butrathertoreturnalongvalueofasystem
propertyrepresentedbythatstring.Asothershavesuggested,whatyouactuallyneedis
Long.parseLong(String
str)
.
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thisanswer
editedSep
11'11at18:36
answeredSep
11'11at8:28

MeLight
1,284820
Ahh,
thatmakesperfectsense.Thanks!–Scott
BiggsSep
11'11at16:57
Very
descriptivereply.+1forthat.–Android
KillerSep
11'11at17:30
Thanks
andwelcome:)–MeLightSep
11'11at18:35


up
vote3down
vote
I
thinkyouareusingwrongfunctionuseLong.parseLong(str)then
youcangettherightanswer.
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thisanswer
answeredSep
11'11at7:24

Android
Killer
4,41921035
You
areabsolutelyright.Thanksforthehelp!–Scott
BiggsSep
11'11at18:52
My
pleasuretohelpubuddy.–Android
KillerSep
12'11at2:26
up
vote3down
vote
you
canuse
Long.parsLong(String)
instead
of
getLong(String)
it
willsolvetheproblem
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thisanswer
editedSep
11'11at7:30
answeredSep
11'11at7:20

confucius
4,3663729
Thanks,
thechangeworked!–Scott
BiggsSep
11'11at18:52
up
vote3down
vote
Long.parseLong(someString)
approved.Don'tforgettocatchNumberFormatExceptionifthere'saprobabilityofunparsablestring.
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thisanswer
answeredSep
11'11at7:53

Y.A.P.
24716
Yes,
thankyou!Wasn'tthinkingalongthoselines.–Scott
BiggsSep
11'11at16:58