运算

linuxbody

1. #!/bin/bash
   
2. a=1
   
3. b=999
   
4. let "c=$(expr $b+1000)"
   
5. echo $a,$c
   
6. while read line;
   
7. do
   
8. if [$line < 100000]
   
9. then
   
10. let "d=$(expr $c+1)"
    
11. let "e=$(expr $c+2000)"
    
12. echo $d,$e
    
13. fi
    
14. done

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错在哪里呢，我想要输出
1,1999
2000,3999
4000,5999

baby_神

1、let "c=$(expr $b+1000)"  这个用法不对，应为   let c=$b+1000
2、if [$line < 100000] 这个比较不对，应为 if [ $line -lt 100000 ]
3、while read line; 这个在这用的也不对

Shell_HAT

回复 2# baby_神


    if [ $line -lt 100000 ]

Shell_HAT

1. #!/bin/bash
   
2. a=1
   
3. b=999
   
4. let c=$b+1000
   
5. echo $a,$c
   
6. e=$c
   
7. while [ $e -lt 100000 ]; do
   
8.     let d=$d+2000
   
9.     let e=$e+2000
   
10.     echo $d,$e
    
11. done
    

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elu_ligao

1. #!/bin/bash
   
2. a=1
   
3. b=999
   
4. let "c=$(expr $b+1000)"
   
5. echo $a,$c
   
6. while read line;
   
7. do
   
8. if [ $line -lt 100000 ]
   
9. then
   
10. let "d+=$(expr $c+1)"
    
11. let "e=$(expr $c+$d)"
    
12. echo $d,$e
    
13. fi
    
14. done
    
15. 
    
16. 这样？

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Shell_HAT

回复 5# elu_ligao


    你自己试试能成功么？

elu_ligao

回复 6# Shell_HAT

1. 可以呀， 刚试了
   
2. $cat yunsuan.sh
   
3. #!/bin/bash
   
4. a=1
   
5. b=999
   
6. let "c=$(expr $b+1000)"
   
7. echo $a,$c
   
8. while read line;
   
9. do
   
10. if [ $line -lt 100000 ]
    
11. then
    
12. let "d+=$(expr $c+1)"
    
13. let "e=$(expr $c+$d)"
    
14. echo $d,$e
    
15. fi
    
16. done
    
17. $ sh yunsuan.sh
    
18. 1,1999
    
19. 3
    
20. 2000,3999
    
21. 4
    
22. 4000,5999
    
23. 5
    
24. 6000,7999

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baby_神

失误失误，少敲了个空格
回复 3# Shell_HAT