Question

00000010- 00 11 50 44  00 00 00 00  00 00 00 00  00 11 58 44 [..PD..........XD]
00000011- 00 00 00 00  00 00 00 00  00 11 80 44  00 00 00 00 [...........D....]
00000012- 00 00 00 00  00 11 88 44  00 00 00 00  00 00 00 00 [.......D........]
00000013- 00 11 90 44  00 00 00 00  00 00 00 00  00 11 98 44 [...D...........D]
00000014- 00 00 00 00  00 00 00 00  00 11 C0 44  00 00 00 00 [...........D....]

需要提取下面提到的十六进制值并将其复制到新文件 -

00 11 50 44  00 00 00 00  00 00 00 00  00 11 58 44 00 00 00 00  00 00 00 00  00 11 80 44  00 00 00 00 00 00 00 00  00 11 88 44  00 00 00 00  00 00 00 00 00 11 90 44  00 00 00 00  00 00 00 00  00 11 98 44 00 00 00 00  00 00 00 00  00 11 C0 44  00 00 00 00

Answer 1

假设您已将所有十六进制数据存储在名为$input的变量中，您可以获得一个十六进制数字列表，如下所示：

foreach line [split $input \n] {
    foreach c [scan $line %*x-%x%x%x%x%x%x%x%x%x%x%x%x%x%x%x%x] {
        if {$c ne ""} {
            lappend out [format %x $c]
        }
    }
}

之后，$out包含十六进制数字列表。明智地使用它。

Answer 2

这是另一种方法，它做出以下假设：

每行以偏移量开头，我们可以将其丢弃
此外，每行以ASCII演示文稿结尾，我们也将其丢弃
这意味着，对于每一行，我们只接受项目1 .. end-1
变量$ input包含多行hex dump

没有进一步的麻烦：

set hexList {}
foreach line [split $input "\n"] {
    set hexList [concat $hexList [lrange $line 1 16]]
}   
puts $hexList; # hexList now contains all the hex digits

Answer 3

我的TCL有点生疏，但非常天真的方法是：

# Parse all hex numbers from your input variable into hexList
set hexList [regexp -all -inline -- {\d{2}(?:\s{1,2})} $input]
# Replace some spaces to get the expected output and store it into hexData
regsub -all -- {\s{3}} [join $hexList] { } hexData
# Write hexData into your file..

TCL脚本 - 仅从Hexdump文件中提取十六进制值并将其复制到新文件

3 个答案: