zero

问题 I have an amount field which is a decimal in the database. I need to always display this amount with 10 numbers on the left of the decimal and two after. Example: Amount = 245.00 which should display as 0000000245.00 Additionally, the amount could be over 1,000 or 10,000 which should display as: 0000001245.00 and 0000011245.00 How can I format the amount to always have the appropriate number of zeros on the left side of the decimal with a variable amount size? 回答1: You should put 0's in your

问题 I'm trying to setup jacoco to get the coverage for my integration tests. I'm running my integration tests against jetty (using the maven plugin). But even if i pass the agent in the jam args when starting up the jetty server the jacoco report shows 0%. Here is my pom.xml <groupId>org.eclipse.jetty</groupId> <artifactId>jetty-maven-plugin</artifactId> <executions> <execution> <id>start-jetty</id> <phase>pre-integration-test</phase> <goals> <goal>run-forked</goal> </goals> <configuration>

问题 Using bcdiv, i can't divide with small float using scientific notation : Working code : bcscale(30); $a = '1' ; $b = '0.00000001'; $result = bcdiv($a, $b); var_dump($result); Results in : string(20) "100000000.0000000000" Non-working code : bcscale(30); $a = '1' ; $b = '1e-8'; $result = bcdiv($a, $b); var_dump($result); Results in : Warning: bcdiv() [function.bcdiv]: Division by zero in C:\wamp\www\utilitaires\test_bcdiv.php on line XX NULL How can i do this division properly, with the less

问题 Hi I'm using f2py to wrap the lapack routine dgesvd, by compiling the dgesvd.f file and linking it against llapack, as explained here according to the docstring, the dgesvd module has the signature: dgesvd - Function signature: dgesvd(jobu,jobvt,m,n,a,s,u,vt,work,lwork,info,[lda,ldu,ldvt]) Required arguments: jobu : input string(len=1) jobvt : input string(len=1) m : input int n : input int a : input rank-2 array('d') with bounds (lda,*) s : input rank-1 array('d') with bounds (*) u : input

问题 I have this file error.log [00:00:00.284],501, [00:00:00.417],5,5294100071980 [00:00:02.463],501, [00:00:05.169],501, [00:00:05.529],501, [00:00:05.730],501, so, if the field $3 its empty i want to print "No value" Im trying this code awk '{{FS=","} if($3=="") {print $1,$2,"No value"}}' But it prints >[00:00:00.284] 501 No value >[00:00:02.463] 501 No value >[00:00:05.169] 501 No value >[00:00:05.529] 501 No value >[00:00:05.730] 501 No value >[00:00:07.193] 501 No value >[00:00:09.899] 501

问题 the code: Float f = Float.parseFloat("1.80"); System.out.println(f); prints "1.8" on screen. I need to keep the 0 in the float value (Float f) for some validation. How do I do this? 回答1: You are confusing a number value and its formatting . It is not possible to actually store 1.80 as a float, however it is possible to display the number as a formatted String which forces two decimal places. Your options are: Keep the original String that the user entered, if the number of decimal places they

问题 The part I keep getting stuck on is boolean(0 % 2 !=0) == false. I mean if 2 goes into 0, 0 times then the remainder would be 2, and 2 does not equal 0. So it should be true. Yet but when I put the boolean in my java program it is treating it as false. Anyone know why? The only logical answer I can wrap my head around is that maybe integers go into 0 and infinite number of times and so are recognized as false, anyone? 回答1: There are two steps: 0 % 2 evaluates to 0 . 0 != 0 evaluates to false

问题 The part I keep getting stuck on is boolean(0 % 2 !=0) == false. I mean if 2 goes into 0, 0 times then the remainder would be 2, and 2 does not equal 0. So it should be true. Yet but when I put the boolean in my java program it is treating it as false. Anyone know why? The only logical answer I can wrap my head around is that maybe integers go into 0 and infinite number of times and so are recognized as false, anyone? 回答1: There are two steps: 0 % 2 evaluates to 0 . 0 != 0 evaluates to false

问题 I'd like to change float like this way: 10.5000 -> 10.5 10.0000 -> 10 How can I delete all zeros after the decimal point, and change it either float (if there's non-zeros) or int (if there were only zeros)? Thanks in advance. 回答1: Why not try regexp? new Float(10.25000f).toString().replaceAll("\\.?0*$", "") 回答2: Well the trick is that floats and doubles themselves don't really have trailing zeros per se; it's just the way they are printed (or initialized as literals) that might show them.

问题 I have a vector like: a = [1,2,3,4,5,6...,n] and I would like to obtain a new vector like this: a_new = [1,0,0,2,0,0,3,0,0,4,0,0,5,0,0,6,...,0,0,n] where a fixed number of zeros (2 in the above example) are inserted between the non-zero elements. If I choose zero_p=3 , the new vector would be: a_new = [1,0,0,0,2,0,0,0,3,0,0,0,4,0,0,0,5,0,0,0,6,...,0,0,0,n] etc. How can I do this? 回答1: Try this: zero_p=3; a_new=zeros(1, (zero_p+1)*length(a)-zero_p); a_new(1:(zero_p+1):end)=a; (Untested, but