在while循环外时，无法读取从while循环内存储的变量

Question

I can't for the life of me see why I can not read the postPrioity outside the while loop. I tried "export postPrioity="500"" still didn't work.

Any ideas?

-- or in plan text --

#!/bin/bash
cat "/files.txt" | while read namesInFile; do   
            postPrioity="500"
            #This one shows the "$postPrioity" varible, as '500'
            echo "weeeeeeeeee ---> $postPrioity <--- 1"
done
            #This one comes up with "" as the $postPrioity varible. GRRR
            echo "weeeeeeeeee ---> $postPrioity <--- 2"

OUTPUT: (I only have 3 file names in files.txt)

weeeeeeeeee ---> 500 <--- 1
weeeeeeeeee ---> 500 <--- 1
weeeeeeeeee ---> 500 <--- 1
weeeeeeeeee --->  <--- 2

Answer 1

The pipe operator creates a subshell, see BashPitfalls and BashFAQ . Solution: Don't use cat , it's useless anyway.

#!/bin/bash
postPriority=0
while read namesInFile
do   
    postPrioity=500
    echo "weeeeeeeeee ---> $postPrioity <--- 1"
done < /files.txt
echo "weeeeeeeeee ---> $postPrioity <--- 2"

Answer 2

As a complement to Philipp's response, in case you MUST use a pipe (and as he pointed out, in your example you don't need cat), you can put all the logic in the same side of the pipe:

command | {
  while read line; do
    variable=value
  done
  # Here $variable exists
  echo $variable
}
# Here it doesn't

Answer 3

Alternatively use process substitution:

while read line
do    
    variable=value  
done < <(command)