grep 返回空响应

Question

我有下面的代码，它ctakes-backlog-new.lst逐行读取文件并从文件中 grep 完整匹配的字符串without_symlink.lst并将结果加载到ctakes-backlog-new.lst.两个文件均已排序。但是，当运行下面的代码时，我得到空响应。

if [ -z "$1" ]
    then
        echo "Filename is missing."
        usage
        exit 1
    else
        fileName="$1"
fi

while read line;
#echo "$line"
  do
    grep -w "$line" 06-09-2021/without_symlink.lst >> 06-09-2021/ctakes_full_file_name.lst
  done < "$fileName"

我运行的命令：./populate_full_file_name.sh 06-09-2021/ctakes-backlog-new.lst

当我手动运行单行命令时，grep -w "77-010_M0_1_arm_1" 06-09-2021/without_symlink.lst >> 06-09-2021/ctakes_full_file_name.lst它工作正常并加载结果（77-010_M0_1_arm_1.60c7144c9ff3060521b0b825149f7d7ee60e5afd.pdf）06-09-2021/ctakes-backlog-new.lst

我尝试用"$line"双引号括起来，但没有任何区别。例如：grep -w "$line" 06-09-2021/without_symlink.lst >> 06-09-2021/ctakes_full_file_name.lst

输入示例：

13-013_M48_9_arm_1
16-039_M42_8_arm_1
77-010_M0_1_arm_1

预期输出：

13-013_M48_9_arm_1.d4b2f31683ba73f7919d46a2514d0742c77029a3.pdf
16-039_M42_8_arm_1.546b57794b71b7a17235368900f5f9c4d34a2a87.pdf
77-010_M0_1_arm_1.60c7144c9ff3060521b0b825149f7d7ee60e5afd.pdf