Shell variable value substitution

Question

Heres the description of my issue, I have a while loop that takes values from a file

while read table
do
    schema="$(echo $table | cut -d'.' -f1)";
    tabname="$(echo $table | cut -d'.' -f2)";
    echo "$schema";
    echo "$tabname";
    echo $folder/$table_space"_auto_ddl"/$tabname"_AUTO_"$schema".sql.tmp"
    echo $folder/$table_space"_auto_ddl"/${tabname}"_AUTO_"${schema}.sql
    print $schema.$tabname;
done < $folder/tables_ddl_list.log

This is an example of one value

MCLM.OPPP

Parses the values into 2 variables So After echoing out $schema I would expect MCLM echoing out $tabnameI would expect OPPP

But I will get empty string

I'm using kornshell and I think its the older version

codeforester · Answer 1 · 2021-03-01T19:42:03.117

2

You can write your loop more efficiently like this, using read, without the need for using an external command like cut for each field to be extracted:

while IFS=. read -r schema table _; do
    # your logic
done < "$folder/tables_ddl_list.log"

The third argument to read, _ is for safety - if the input has more than one dot on a line, all the extra values would be captured by _. Optionally, you could add error checking based on whether _ gets set or not.

answered Mar 08 '19 at 15:47

codeforester

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This assumes that there is exactly one dot on each line. Maybe `read -r schema table _` to discard anything after the second dot. – tripleee Mar 01 '21 at 19:31
Great point, @tripleee. Updated the answer. – codeforester Mar 01 '21 at 19:43

nullPointer · Accepted Answer · 2019-03-08T15:57:28.120

1

Try removing the double quotes when you read in the values of the variables, and use double quotes in the $table variable, eg :

schema=$(echo "$table" | cut -d'.' -f1)
tabname=$(echo "$table" | cut -d'.' -f2)

edited Mar 08 '19 at 15:57

answered Mar 08 '19 at 15:44

nullPointer

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One can have *both* the outer and inner quotes -- there's no conflict between them, as a command substitution has its own quoting context. That said, putting the quotes around `"$table"` was indeed necessary. – Charles Duffy Mar 08 '19 at 15:53
@CharlesDuffy why is quoting causing the issue here? `.` has no special meaning to word splitting, right? `var=a.b; echo $var | cut -f1 -d.` would yield `a`, right? – codeforester Mar 08 '19 at 17:48
@codeforester, necessary for the general case; didn't mean to speak to any specific data. That said, if `IFS=.`, then we'd lose it in the `echo $var`. – Charles Duffy Mar 08 '19 at 17:57

Shell variable value substitution

2 Answers2