mjhjmtu

I'm using bash shell. I have a CSV file in which each line's tokens are separated by commas. I want to take the second and third columns and forma new string out of them (an SQL statement). I thought I could use awk for this purpose, so I tried ...

localhost:mydir davea$ awk -F ',' -v OFS=',' "REPLACE INTO my_table (ID, NAME, HOURS) VALUES ('$2', '$2', '$3');" types.csv
awk: syntax error at source line 1
 context is
 REPLACE INTO my_table >>> (ID, <<<
awk: bailing out at source line 1

but as you can see I'm getting an error. Am I leaving something out? How do I form my new string from each line in the CSV file?

awk -F ',' '
 printf("REPLACE INTO my_table (ID, NAME, HOURS) VALUES ('''%s''', '''%s''', '''%s''');n", $2, $2, $3) 
' types.csv

Putting single quotes inside of a single quoted string is really tedious. Here, I'm passing a single quote into the the awk variable "q". I'm also trying to take care of SQL injection.

awk -F ',' -v q="'" '
 
 for (i=1; i<=NF; i++) gsub(q, q q, $i)
 print "REPLACE INTO my_table (ID, NAME, HOURS) VALUES (" q $1 q "," q $2 q "," q $3 q");"
 
' <<END
foo,bar,Robert');DROP TABLE students;--
END

REPLACE INTO my_table (ID, NAME, HOURS) VALUES ('foo','bar','Robert'');DROP TABLE students;--');

Might be not the most efficient solution but shall work:

awk -F, 'print "REPLACE INTO my_table (ID, NAME, HOURS) VALUES (x27"$2"x27,x27"$2"x27,x27"$3"x27);"' types.csv