这个简单的bash命令有什么问题？

Question

$我不工作......这让我发疯了！

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为$i_prob，因为下划线是变量名称的有效字符。使用${i}来解决此问题：

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

Answer 2

你已经忘记了$ i周围的“”

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