我有一个jsp页面,通过它我从下拉列表中选择一个具有数据时间格式的值,我有一个日历,从中我选择数据和时间。当我提交此表单时,然后调用另一个调用方法的JSP页面在java类中定义。 该方法具有以下代码
public LinkedHashMap<Double, String> ClosestToMultiplesOfTen_User(String start,String end) throws SQLException {
int row_id ;
int bIdx = 0;
double[] vals=null;
int rowIndex = 0 ;
int i=0;
try
{
con = getConnection();
stmt = con.createStatement(ResultSet.TYPE_SCROLL_SENSITIVE,ResultSet.CONCUR_READ_ONLY);
String sql="select distinct beam_current from INDUS2_BDS.dbo.DCCT where logtime between '"+start+"' and '"+end+"'"+
"and (beam_current like '%9.95' or beam_current like '%9.96' or beam_current like '%9.97' or beam_current like '%9.98' or beam_current like '%9.99' or beam_current like '%0' or beam_current like '%_0.01' or beam_current like '%_0.02' or beam_current like '%_0.03' or beam_current like '%_0.04' or beam_current like '%_0.05' or beam_current like '%_0.06') ";
stmt.executeQuery(sql);
rs = stmt.getResultSet();
rs.last();
int row_cnt=rs.getRow();
System.out.println("row_count of closest " +row_cnt);
vals = new double[row_cnt];
rs.beforeFirst();
while(rs.next())
{
for(int j=0; j<1; j++)
{
vals[i] = rs.getDouble(1);
}
i++;
}
}
catch( Exception e )
{
System.out.println("\nException "+e);
}
// get the max value, and its multiple of ten to get the number of buckets
double max = java.lang.Double.MIN_VALUE;
for (double v : vals) max = Math.max(max, v);
Arrays.sort(vals);
System.out.println("value at vals[0] c "+vals[0]);
double min=vals[0];
int m2=(int) Math.round(min);
int m3=(int) Math.round(max);
int bucketCount = 1+((m3-m2)/10);
double[] bucket =new double[bucketCount];
// initialise the buckets array to store the closest values
double[][] buckets = new double[bucketCount][3];
for (int i1 = 0; i1 < bucketCount; i1++){
// store the current smallest delta in the first element
buckets[i1][0] = java.lang.Double.MAX_VALUE;
// store the current "closest" index in the second element
buckets[i1][1] = -1d;
// store the current "closest" value in the third element
buckets[i1][2] = java.lang.Double.MAX_VALUE;
}
// iterate the rows
for (row_id=1 ; row_id < vals.length; row_id++)
{
// get the value from the row
double v = vals[row_id];
// get the closest multiple of ten to v
double mult = getMultipleOfTen(v);
// get the absolute distance of v from the multiple of ten
double delta = Math.abs(mult - v);
// get the bucket index based on the value of `mult`
bIdx = (int)(mult / 10d);
// System.out.println("value of bidx for bucket index is"+bIdx);
// test the last known "smallest delta" for this bucket
if (buckets[bIdx][0] > delta)
{
// this is closer than the last known "smallest delta"
buckets[bIdx][0] = delta;
buckets[bIdx][1] = row_id;
buckets[bIdx][2] = v;
}
}
// print out the result
for (int i1 =1; i1 < buckets.length; i1++)
{
bucket = buckets[i1];
rowIndex = (int) bucket[1];
double rowValue = bucket[2];
DecimalFormat twoDForm = new DecimalFormat("#.##");
System.out.println("row index closeset "+rowIndex+ "value is closest "+rowValue);
rs.absolute(rowIndex+1);
user_current_map.put(java.lang.Double.valueOf(twoDForm.format(rs.getDouble(1))),"");
}
System.out.println("user_current_map "+user_current_map);
return user_current_map;
}
public double getMultipleOfTen(double v)
{
System.out.println(10d * Math.round(v / 10d));
return 10d * Math.round(v / 10d);
}
检索到的bucketcount值是正确的。但是我在行(例如 java.lang.ArrayIndexOutOfBoundsException:21 的
答案 0 :(得分:1)
我明白了,**bIdx = (int)(mult / 10d);**
的值从m2 / 10值开始而不是从0开始,因此当它尝试访问大于桶大小的索引时等于桶数,数组索引超出绑定异常。
解决方案= bIdx = (int)(mult / 10d) - m2/10;
除以检索到的最小值/ 10,以便它从0开始直到桶大小。
答案 1 :(得分:0)
数组索引从0开始直到n-1。在您的情况下,您可以访问索引0中的元素,直到21-1 = 20。所以你应该使用:
if (buckets[bIdx-1][0] > delta)