shell求助

xiaozhuang314

脚本
#!/bin/bash
LOG="file"
DATE="20130801 20130802"
morning=$(echo {00..05})
echo $LOG$DATE$morning
                  

输出结果
file20130801 2013080200 01 02 03 04 05


怎么让输出结果为
file2013080100  file2013080200
file2013080101  file2013080201
file2013080102   file2013080202
file2013080103   file2013080203
file2013080104   file2013080204
file2013080105   file2013080205

reyleon

1. for i in $(seq -f %02g 0 5);do echo 20130801${i};done

复制代码

huangyu_945

echo "20130801"{00..05} | sed 's/ /\n/g'

cao627

1. [url=home.php?mod=space&username=xiaozhuang314]@xiaozhuang314[/url]
   
2. [code]$ DATE="20130801"
   
3. $ morning=({00..05})
   
4. $ for i in ${morning[@]}; do echo $DATE$i; done
   
5. 2013080100
   
6. 2013080101
   
7. 2013080102
   
8. 2013080103
   
9. 2013080104
   
10. 2013080105
    

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[/code]

cao627

1. $ DATE=(20130801 20130802)
   
2. $ morning=({00..05})
   
3. $ for i in ${morning[@]}; do for j in ${DATE[@]}; do echo -n  $j$i" "; done;echo; done
   
4. 2013080100 2013080200
   
5. 2013080101 2013080201
   
6. 2013080102 2013080202
   
7. 2013080103 2013080203
   
8. 2013080104 2013080204
   
9. 2013080105 2013080205
   

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@xiaozhuang314

reyleon

回复 1# xiaozhuang314


    要求改了也不跟大伙儿说一声，你这样是不厚道的

1. echo "20130801 20130802" | awk '{for(j=0;j<=5;j++){for(i=1;i<=NF;i++){printf("%d%02d%s",$i,j,OFS)}printf RS}}'

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xiaozhuang314

回复 6# reyleon

不好意思呀  整一下午了有点晕了