[20191013]oracle number类型存储转化脚本.txt

–//测试看看是否可以利用bc obase=100的输出解决问题。另外以前脚本忘记考虑尾数的四舍五入问题。
–//也许编程就是这样，总有一些细节没有考虑到…
–//代码如下num2raw_5.sh：

#! /bin/bash
#! number convert oracle raw.

odebug=${odebug:-0}

# process input parameter ,delete “,” and all spaces. save to variable v_num. and length to variable v_len.
v_num=”$*”
v_num=${v_num//[, ]/}

# strip e or trailing 0s in decimals or 0000.000 output 0 ,
v_num=$(echo $v_num/1 + 0 | sed -e “s/[ee]+\?/*10^/” -e “s/^/scale=180;/” | bc | tr -d ‘\n\\\r’ | sed -e “s/\.$[0-9]*[1-9]$0\+$/.\1/” -e “s/\.0\+$//”)

if [[ “$v_num” =~ ^-.*$ ]]; then
   v_sign=1
   v_num=${v_num:1:180}
else
   v_sign=0
fi

if [ $odebug -eq 1 ] ; then
echo v_num=”$v_num”
fi

v_res=””
if [ “$v_num” == “0” ]; then
   v_res=”80″
   echo “$v_res”
   exit 0
fi

v_pos=$(expr index $v_num “.”)

if [ $v_pos -gt 1 ]; then
   v_exp=$(( v_pos/2 ))
elif [ $v_pos -eq 0 ]; then
   v_exp=$(( (${#v_num}+1) /2 ))
elif [ $v_pos -eq 1 ]; then
   v_tmp1=${v_num:1:180}
   v_tmp2=$(echo $v_tmp1 | sed ‘s/^0\+//g’)
   v_exp=$(( (${#v_tmp2} – ${#v_tmp1})/2 ))
fi

v_exp1=$(printf “%02x” $(( $v_exp+192 )))
if [ $v_sign -eq 1 ]; then
v_exp1=$(printf “%02x” $(( 0xff – 0x${v_exp1} )))
fi

v_res=${v_exp1}${v_res}

# oracle number type max length is 22 bytes (not 22 is 21 bytes??), 1 bytes exponent.
# bc不作四舍五入，要加5*10^-41解决问题。
v_tmp=$(echo “scale=180 ; a=$v_num / 100^($v_exp) +5*10^-41; scale=40;a/1 ” | bc | tr -d ‘\n\\\r’| sed -e “s/\.$[0-9]*[1-9]$0\+$/.\1/” -e “s/\.0\+$//” )

if [ $odebug -eq 1 ] ; then
echo v_num=”$v_num” v_len=”$v_len” v_exp=”$v_exp” v_exp1=”$v_exp1″ v_tmp=”$v_tmp”
fi

if [ $v_sign -eq 0 ]; then
v_res=${v_res}$(echo “obase=100;$v_tmp” | bc | tr -d “.” | awk ‘begin{rs=” +”}/./{printf “,%02x”, $1+1}’)
else
v_res=${v_res}$(echo “obase=100;$v_tmp” | bc | tr -d “.” | awk ‘begin{rs=” +”}/./{printf “,%02x”, 101-$1}’)
fi

if [ $v_sign -eq 1 -a ${#v_tmp} -lt 40 ]; then
v_res=${v_res}”,””66″
fi

echo “$v_res”

[20191013]oracle number类型存储转化脚本.txt

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