Bash 脚本中的字符串连接
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String concatenation in Bash script
提问by olidev
I am writing this Bash script:
我正在编写这个 Bash 脚本：
count=0
result
for d in `ls -1 $IMAGE_DIR | egrep "jpg$"`
do
if (( (count % 4) == 0 )); then
result="abc $d"
if (( count > 0 )); then
echo "$result;"
fi
else
result="$result $d"
fi
(( count++ ))
done
if (( (count % 4) == 0 )); then
echo $result
fi
The script is to concate part strings into a string when the value is divided by 4 and it should be larger than 0.
该脚本是将部分字符串在值除以 4 并且应该大于 0 时连接成一个字符串。
In the IMAGE_DIR, I have 8 images,
在 IMAGE_DIR 中，我有 8 张图片，
I got outputs like this:
我得到了这样的输出：
abc et004.jpg
abc et008.jpg
But I would expect to have:
但我希望有：
abc et001.jpg et002.jpg et003.jpg et004.jpg;
abc et005.jpg et006.jpg et007.jpg et008.jpg;
How can I fix this?
我怎样才能解决这个问题？
回答by Kilian Foth
The =operator must always be written without spaces around it:
该=操作必须写入的周围没有空格：
result="$result $d"
(Pretty much the most important difference in shell programming to normal programming is that whitespace matters in places where you wouldn't expect it. This is one of them.)
（在 shell 编程与普通编程中最重要的区别几乎是空白在你意想不到的地方很重要。这是其中之一。）
回答by Kusalananda
Something like this?
像这样的东西？
count=0
find $IMAGE_DIR -name "*.jpg" |
while read f; do
if (( (count % 4) == 0 )); then
result="abc $f"
if (( count > 0 )); then
echo $result
fi
else
result="$result $d"
fi
(( count++ ))
done
回答by Dimitre Radoulov
Something like this (untested, of course):
像这样的东西（当然未经测试）：
count=0 result=
for d in "$IMAGE_DIR"/*jpg; do
(( ++count % 4 == 0 )) &&
result="abc $d"
(( count > 0 )) &&
printf '%s\n' "$result" ||
result+=$d
done
