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Bash sequence

仅包含偶数的数组

发布于 2024-11-30 12:30:07 字数 592 浏览 0 评论 0 原文

for X in {18..2500} ; 是我脚本的一行，这意味着逐一选择数字，例如：18,19,20,21,22,23....直到2500

但是我发现我现在只需要偶数：18,20,22,24.....2500

那么我应该稍微修改一下该行怎么办？

感谢

编辑：它是 bash...

我的脚本现在更改为：

#!/bin/bash



TASK=1101;

NUM=9;

TEND=1100;

for X in {18..2500};{

   if (X % 2 == 0);

   do

     echo "$X      echo \"Wait until $NUM job is done\" $NUM" ;

     NUM=$((NUM+2)) ;

     X=$((X+1)) ;

     TEND=$((TEND+100)) ;

     echo "$X      -t $TASK-$TEND jobs.sh" ;

     TASK=$((TASK+100)) ;}

done

但出现如下错误：第 15 行：意外标记“do”附近出现语法错误

原文

for X in {18..2500} ; is one line of my script, which means to pick number one by one like: 18,19,20,21,22,23....till 2500

However I find I only need even number right now: 18,20,22,24.....2500

Then what should I do by a slight modify of the line?

Thanks

edit:
It's bash...

My script is now changed to:

#!/bin/bash



TASK=1101;

NUM=9;

TEND=1100;

for X in {18..2500};{

   if (X % 2 == 0);

   do

     echo "$X      echo \"Wait until $NUM job is done\" $NUM" ;

     NUM=$((NUM+2)) ;

     X=$((X+1)) ;

     TEND=$((TEND+100)) ;

     echo "$X      -t $TASK-$TEND jobs.sh" ;

     TASK=$((TASK+100)) ;}

done

but got errors like:
line 15: syntax error near unexpected token `do'

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思念绕指尖 2024-12-07 12:30:07

您可以指定增量：

for X in {18..2500..2}

   A sequence  expression takes the form {x..y[..incr]}, where x and y are
   either integers or single characters, and incr, an optional increment, is
   an integer.

或者

for X in `seq 18 2 2500`

You can specify the increment:

for X in {18..2500..2}

   A sequence  expression takes the form {x..y[..incr]}, where x and y are
   either integers or single characters, and incr, an optional increment, is
   an integer.

for X in `seq 18 2 2500`

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唯憾梦倾城 2024-12-07 12:30:07

这不是 C++。这是一个 bash 脚本。

您的 for 循环需要以 do 开头：

for X in {18..2500}; do

您的 if-语句语法看起来不对。它可能应该是这样的，注意then：

    if [[ $((X % 2)) == 0 ]]; then

if-blocks以：

fi

和for-do 块结尾为：

done

更好的是...取消 if 语句并使用 Bash 的 for-循环构造仅生成偶数：

for ((X = 18; X <= 2500; X += 2)); do
     echo "$X      echo \"Wait until $NUM job is done\" $NUM" ;
     # ... 
done

This is not C++. This is a bash script.

Your for-loop needs to start with a do:

for X in {18..2500}; do

Your if-statement syntax looks off. It should probably be something like this, note the then:

    if [[ $((X % 2)) == 0 ]]; then

if-blocks end with:

fi

And the for-do block ends with:

done

Better still... do away with the if statement and use Bash's for-loop construct to generate only even numbers:

for ((X = 18; X <= 2500; X += 2)); do
     echo "$X      echo \"Wait until $NUM job is done\" $NUM" ;
     # ... 
done

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☆獨立☆ 2024-12-07 12:30:07

尝试模数运算符。在几乎所有语言中，它看起来都是这样的：

if (x % 2 == 0)    // …Do something

这是有效的 C 代码，但可以轻松应用于其他语言。

您可以将 mod 运算符视为放置在同一位置的除法符号，但它不返回除法结果，而是返回余数。因此，在这段代码中，如果除以二的余数为 0，则它能被二整除，因此根据定义它是偶数。

Try the modulus operator. In almost all languages, it'll look something like this:

if (x % 2 == 0)    // …Do something

That is valid C code, but can easily be applied to other languages.

You can think of the mod operator as a division sign placed in the same location, but rather than returning the results of the division it returns the remainder. Therefore in this code, if the remainder of a divide-by-two is 0, then it divides evenly by two, and so it's even by definition.

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