按设定的数量在 bash 循环中增加
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18828225/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Increment in bash loop by set amount
提问by cianius
I know how to do a loop in bash that increases by one each time, but say I have a range 1 to 773 and I want to output a range from a loop so that I get two variables in each iteration. The first will be 1 and the second will be say 19. In the second iteration the first would be 20 and the second 39.
我知道如何在 bash 中做一个每次增加一个的循环，但是假设我有一个范围 1 到 773 并且我想从循环中输出一个范围，以便我在每次迭代中得到两个变量。第一个是 1，第二个是 19。在第二次迭代中，第一个是 20，第二个是 39。
Ive been playing with something like:
我一直在玩类似的东西：
for start in {1..773}
do
start=$(($start+20))
end=$(($start+20))
echo $start ##
echo $end
done
Desired loop outcome:
期望的循环结果：
1. $start = 1 and $end = 19
2. $start = 20 and $end = 39
3. $start = 40 and $end = 59
etc
But it's not right. I want to output these two variables to a series of scripts to make R run faster, so if non bash (eg awk) solutions are easier then that's cool too if a simple > will send it the file.
但这是不对的。我想将这两个变量输出到一系列脚本以使 R 运行得更快，所以如果非 bash（例如 awk）解决方案更容易，那么如果一个简单的 > 将文件发送给它，那也很酷。
Thanks!
谢谢！
回答by mnagel
your requirement are not perfectly clear, but you are re-using variable names.
您的要求不是很清楚，但您正在重复使用变量名。
if i do this:
如果我这样做：
for index in {1..773}
do
start=$(($index+20))
end=$(($start+20))
echo $start ##
echo $end
done
i get something that resembles your desired result. observe how i renamed the loop variable from start to index.
我得到的东西类似于你想要的结果。观察我如何从开始到索引重命名循环变量。
PS: if you want to change the step size (a.k.a. "increment") in your loop, simply do it like this:
PS：如果您想更改循环中的步长（又名“增量”），只需这样做：
#!/bin/bash
for i in {0..10..2}
do
echo "Welcome $i times"
done
This will increment in steps of 2, you would want to use a 20 here. That would give you 1, 21, 41, ...as value. See http://www.cyberciti.biz/faq/bash-for-loop/for more details.
这将以 2 的步长递增，您可能希望在此处使用 20。那会给你1, 21, 41, ...作为价值。有关更多详细信息，请参阅http://www.cyberciti.biz/faq/bash-for-loop/。
回答by thefourtheye
If you want to print the ranges within 773, you can do like this
如果你想打印 773 以内的范围，你可以这样做
#!env bash
start=1
end=19
for counter in {1..773}
do
echo $counter. "$start = " $start " and $end = " $end
if [[ $start -eq 1 ]];
then
start=0
fi
start=$(($start+20))
end=$(($end+20))
if [[ $end -ge 773 ]];
then
break
fi
done
Output
输出
1. $start = 1 and $end = 19
2. $start = 20 and $end = 39
3. $start = 40 and $end = 59
4. $start = 60 and $end = 79
5. $start = 80 and $end = 99
6. $start = 100 and $end = 119
7. $start = 120 and $end = 139
8. $start = 140 and $end = 159
9. $start = 160 and $end = 179
10. $start = 180 and $end = 199
11. $start = 200 and $end = 219
12. $start = 220 and $end = 239
13. $start = 240 and $end = 259
14. $start = 260 and $end = 279
15. $start = 280 and $end = 299
16. $start = 300 and $end = 319
17. $start = 320 and $end = 339
18. $start = 340 and $end = 359
19. $start = 360 and $end = 379
20. $start = 380 and $end = 399
21. $start = 400 and $end = 419
22. $start = 420 and $end = 439
23. $start = 440 and $end = 459
24. $start = 460 and $end = 479
25. $start = 480 and $end = 499
26. $start = 500 and $end = 519
27. $start = 520 and $end = 539
28. $start = 540 and $end = 559
29. $start = 560 and $end = 579
30. $start = 580 and $end = 599
31. $start = 600 and $end = 619
32. $start = 620 and $end = 639
33. $start = 640 and $end = 659
34. $start = 660 and $end = 679
35. $start = 680 and $end = 699
36. $start = 700 and $end = 719
37. $start = 720 and $end = 739
38. $start = 740 and $end = 759
回答by ztik
You could use seqcommand
你可以使用seq命令
for start in `seq 1 20 700`
do
echo $start $(($start+19))
done
The usage of seqis:
的用法seq是：
$ seq --help
Usage: seq [OPTION]... LAST
or: seq [OPTION]... FIRST LAST
or: seq [OPTION]... FIRST INCREMENT LAST
Print numbers from FIRST to LAST, in steps of INCREMENT.
回答by konsolebox
This is one way to do it with consistent outputs:
这是具有一致输出的一种方法：
for ((i = 1, start = 1, end = 19; i <= 773; ++i, start += 20, end += 20)); do
echo "$i. $start = $start and $end = $end"
done
Output:
输出：
1. $start = 1 and $end = 19
2. $start = 21 and $end = 39
3. $start = 41 and $end = 59
4. $start = 61 and $end = 79
Or
或者
for ((i = 1, start = 1, end = 19; i <= 773; ++i, start += 20, end += 20)); do
echo "$i. $start = $start and $end = $end"
done
Output:
输出：
1. $start = 1 and $end = 20
2. $start = 21 and $end = 40
3. $start = 41 and $end = 60
4. $start = 61 and $end = 80
Another:
其他：
for ((i = 1, start = 0, end = 19; i <= 773; ++i, start += 20, end += 20)); do
echo "$i. $start = $start and $end = $end"
done
Output:
输出：
1. $start = 0 and $end = 19
2. $start = 20 and $end = 39
3. $start = 40 and $end = 59
4. $start = 60 and $end = 79
That way you can have two variables.
这样你就可以有两个变量。
回答by anubhava
Unless I'm missing something you can simply do:
除非我遗漏了一些东西，否则你可以简单地做：
for ((s=0,e=19; e<773; s+=20,e+=20)); do
echo $s "-" $e
done
OUTPUT:
输出：
0 - 19
20 - 39
40 - 59
60 - 79
80 - 99
100 - 119
120 - 139
140 - 159
160 - 179
180 - 199
200 - 219
220 - 239
240 - 259
260 - 279
280 - 299
300 - 319
320 - 339
340 - 359
360 - 379
380 - 399
400 - 419
420 - 439
440 - 459
460 - 479
480 - 499
500 - 519
520 - 539
540 - 559
560 - 579
580 - 599
600 - 619
620 - 639
640 - 659
660 - 679
680 - 699
700 - 719
720 - 739
740 - 759
回答by glenn Hymanman
Nobody has suggested a whileloop
没有人建议while循环
start=0 step=20 end=$((step - 1))
while (( end < 773 )); do
echo "$start - $end"
(( start += step, end += step ))
done
Produces
生产
0 - 19
20 - 39
40 - 59
...
720 - 739
740 - 759
Of course, the while loop can be written as a for loop:
当然，while 循环可以写成 for 循环：
for ((start=0, step=20, end=step-1; end < 773; start+=step, end+=step)); do ...
回答by pgl
Here's one way to do it:
这是一种方法：
step=20
last=773
for ((i = 0; i <= $last; i += $step))
do
start=$i
end=$(($i + (($step - 1))))
if (($end > $last))
then
end=$last
fi
echo "$start: $start"
echo "$end: $end"
done
It's basically just a simple for loop.
它基本上只是一个简单的 for 循环。
