Print output of cat statement in bash script loop

I'm trying to execute a command for each line coming from a cat command. I'm basing 开发者_Go百科this on sample code I got from a vendor.

Here's the script:

for tbl in 'cat /tmp/tables'
do
   echo $tbl
done

So I was expecting the output to be each line in the file. Instead I'm getting this:

cat
/tmp/tables

That's obviously not what I wanted.

I'm going to replace the echo with an actual command that interfaces with a database.

Any help in straightening this out would be greatly appreciated.

You are using the wrong type of quotes.

You need to use the back-quotes rather than the single quote to make the argument being a program running and piping out the content to the forloop.

for tbl in `cat /tmp/tables` 
do 
    echo "$tbl"
done

Also for better readability (if you are using bash), you can write it as

for tbl in $(cat /tmp/tables) 
do 
    echo "$tbl"
done

If your expectations are to get each line (The for-loops above will give you each word), then you may be better off using xargs, like this

cat /tmp/tables | xargs -L1 echo

or as a loop

cat /tmp/tables | while read line; do echo "$line"; done

The single quotes should be backticks:

for tbl in `cat /etc/tables`

Although, this will not get you output/input by line, but by word. To process line by line, you should try something like:

cat /etc/tables | while read line
    echo $line
done

With while loop:

while read line
do
echo "$line"
done < "file"

while IFS= read -r tbl; do echo "$tbl" ; done < /etc/tables

read this.

You can do a lot of parsing in bash by redefining the IFS (Input Field Seperator), for example

IFS="\t\n"  # You must use double quotes for escape sequences. 
for tbl in `cat /tmp/tables` 
do 
    echo "$tbl"
done