Question

我需要输出一个15位数字输入到逗号分隔的十六进制字符串，但我似乎无法从第一个字符开始，这是我到目前为止：

echo "012345678912345" |awk 'BEGIN{FS=OFS=""}{for (i=1;i<=NF;i++) {if (i%1==0) $i=$i ",0x3"}}1'

0,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x31,0x32,0x33,0x34,0x35,0x3

echo 123456789123456 | awk '$1=$1' FS= OFS=",0x3"

1,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x31,0x32,0x33,0x34,0x35,0x36

如何让它在开始时插入0x3或在开始时添加一个随机字符然后将其删除？结果id想要输入012345678912345将是：

0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x31,0x32,0x33,0x34,0x35

Answer 1

awk -v OFS=",0x3" -v FS="" '$1="0x3"$1'

以你的例子：

kent$  echo "012345678912345" |awk -v OFS=",0x3" -v FS="" '$1="0x3"$1'   
0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x31,0x32,0x33,0x34,0x35

Answer 2

如果您对非awk方法持开放态度（只有每个平台awk也有sed），请尝试：

pax> echo "012345678912345" | sed 's/\(.\)/,0x3\1/g;s/^,//'
0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x31,0x32,0x33,0x34,0x35

sed命令序列由两部分组成：

第一个s/\(.\)/,0x3\1/g为每个字符添加字符串,0x3前缀。
第二个，s/^,//摆脱了第一个逗号。

将15位数字转换为十六进制

2 个答案: