这个简单的bash命令有什么问题？

Question

$ i不工作......它让我疯了!

for i in {1..200}; 
do echo "/scratch/inputTest/prob/timit.test.pg.list_10_$i_prob.bin" >> longProbList; 
done

在文件中只有

/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin
/scratch/inputTest/prob/timit.test.pg.list_10_.bin

Answer 1

它认为您的变量名称是$i_prob因为下划线是变量名称的有效字符.使用${i}来解决这个问题:

for i in {1..200}; 
do echo "/scratch/inputTest/prob/timit.test.pg.list_10_${i}_prob.bin" >> longProbList; 
done