如何在没有 root 权限的情况下创建多分区 SD 磁盘映像？

是否可以在没有 root 权限（即无环回挂载）的情况下在 linux 中创建完整的 SD 映像？我正在寻找一种自动化嵌入式系统映像创建的方法。该映像应包含特定的分区结构以及格式化为 FAT 和 ext2 的分区，并填充来自构建系统的文件。

最小可运行sfdisk + mke2fs没有的例子sudo

在此示例中，我们将创建不带sudo or setsuid，一个包含两个 ext2 分区的映像文件，每个分区都填充了主机目录中的文件。

然后我们将使用sudo losetup只是为了安装分区来测试 Linux 内核实际上可以读取它们，如下所述：如何在 Linux 上挂载包含多个分区的镜像文件中的一个分区？ https://stackoverflow.com/questions/1419489/how-to-mount-one-partition-from-an-image-file-that-contains-multiple-partitions/39675265#39675265

欲了解更多详情，请参阅：

• sfdisk：处理分区表：https://superuser.com/questions/332252/how-to-create-and-format-a-partition-using-a-bash-script/1132834#1132834 https://superuser.com/questions/332252/how-to-create-and-format-a-partition-using-a-bash-script/1132834#1132834
• mke2fs：处理分区的 EXT 格式化：https://superuser.com/questions/605196/how-to-create-ext2-image-without-superuser-rights/1366762#1366762 https://superuser.com/questions/605196/how-to-create-ext2-image-without-superuser-rights/1366762#1366762

这个例子：

#!/usr/bin/env bash

# Input params.
root_dir_1=root1
root_dir_2=root2
partition_file_1=part1.ext2
partition_file_2=part2.ext2
partition_size_1_megs=32
partition_size_2_megs=32
img_file=img.img
block_size=512

# Calculated params.
mega="$(echo '2^20' | bc)"
partition_size_1=$(($partition_size_1_megs * $mega))
partition_size_2=$(($partition_size_2_megs * $mega))

# Create a test directory to convert to ext2.
mkdir -p "$root_dir_1"
echo content-1 > "${root_dir_1}/file-1"
mkdir -p "$root_dir_2"
echo content-2 > "${root_dir_2}/file-2"

# Create the 2 raw ext2 images.
rm -f "$partition_file_1"
mke2fs \
  -d "$root_dir_1" \
  -r 1 \
  -N 0 \
  -m 5 \
  -L '' \
  -O ^64bit \
  "$partition_file_1" \
  "${partition_size_1_megs}M" \
;
rm -f "$partition_file_2"
mke2fs \
  -d "$root_dir_2" \
  -r 1 \
  -N 0 \
  -m 5 \
  -L '' \
  -O ^64bit \
  "$partition_file_2" \
  "${partition_size_2_megs}M" \
;

# Default offset according to
part_table_offset=$((2**20))
cur_offset=0
bs=1024
dd if=/dev/zero of="$img_file" bs="$bs" count=$((($part_table_offset + $partition_size_1 + $partition_size_2)/$bs)) skip="$(($cur_offset/$bs))"
printf "
type=83, size=$(($partition_size_1/$block_size))
type=83, size=$(($partition_size_2/$block_size))
" | sfdisk "$img_file"
cur_offset=$(($cur_offset + $part_table_offset))
# TODO: can we prevent this and use mke2fs directly on the image at an offset?
# Tried -E offset= but could not get it to work.
dd if="$partition_file_1" of="$img_file" bs="$bs" seek="$(($cur_offset/$bs))"
cur_offset=$(($cur_offset + $partition_size_1))
rm "$partition_file_1"
dd if="$partition_file_2" of="$img_file" bs="$bs" seek="$(($cur_offset/$bs))"
cur_offset=$(($cur_offset + $partition_size_2))
rm "$partition_file_2"

# Test the ext2 by mounting it with sudo.
# sudo is only used for testing, the image is completely ready at this point.

# losetup automation functions from:
# https://stackoverflow.com/questions/1419489/how-to-mount-one-partition-from-an-image-file-that-contains-multiple-partitions/39675265#39675265
loop-mount-partitions() (
  set -e
  img="$1"
  dev="$(sudo losetup --show -f -P "$img")"
  echo "$dev" | sed -E 's/.*[^[:digit:]]([[:digit:]]+$)/\1/g'
  for part in "${dev}p"*; do
    if [ "$part" = "${dev}p*" ]; then
      # Single partition image.
      part="${dev}"
    fi
    dst="/mnt/$(basename "$part")"
    echo "$dst" 1>&2
    sudo mkdir -p "$dst"
    sudo mount "$part" "$dst"
  done
)
loop-unmount-partitions() (
  set -e
  for loop_id in "$@"; do
    dev="/dev/loop${loop_id}"
    for part in "${dev}p"*; do
      if [ "$part" = "${dev}p*" ]; then
        part="${dev}"
      fi
      dst="/mnt/$(basename "$part")"
      sudo umount "$dst"
    done
    sudo losetup -d "$dev"
  done
)

loop_id="$(loop-mount-partitions "$img_file")"
sudo cmp /mnt/loop0p1/file-1 "${root_dir_1}/file-1"
sudo cmp /mnt/loop0p2/file-2 "${root_dir_2}/file-2"
loop-unmount-partitions "$loop_id"

在 Ubuntu 18.04 上测试。GitHub 上游 https://github.com/cirosantilli/linux-cheat/blob/77dd7baa68dac3534a50f8a084f732f378113c59/mke2fs-multi.sh.