bash中的子字符串失败：意外令牌“完成”附近的语法错误。

Question

I'm trying to loop through a log file and grab a portion of a string.

The log file contains:

Variable_name Value
Slave_running ON

First I need to see if the current row contains the substring Slave_running (13 characters long starting at position 0). If I get a match, then I need to test what is a character to the right of the same string (14:2).

Here's my feeble attempt to first just print the substring to console:

while read p; do
  echo ${$p:0:13}
done <slaverunning.log

This returns:

syntax error near unexpected token `done'.   

What am I getting wrong with the bash substring syntax wrong?

Answer 1

Firstly, the issue with your code is just a syntactical one, you don't need the $ before p , only before { . The below should work as expected:

while read p; do
  echo ${p:0:13}
done < slaverunning.log

That said, awk seems to be a better fit for this problem, something like this:

awk '/^Slave_running/ { if ($2 == "ON") { print "running"; } }' slaverunning.log

A great walkthrough of Bash's string manipulation functions: http://www.thegeekstuff.com/2010/07/bash-string-manipulation/
A good place to get familiar with Awk: http://www.grymoire.com/Unix/Awk.html

Answer 2

There's a syntax error, but it's not about unexpected token...

The $p inside the ${...} is wrong, this should give a bad substitution error:

  echo ${$p:0:13} 

With that fixed, your script should work:

while read p; do
  echo ${p:0:13}
done <slaverunning.log