[英]What is the proper quoting to assign an interpolated sqlite3 query to a variable in Bash?
I am trying to assign the output of the following to a variable 我试图将以下输出分配给变量
sqlite3 /home/user/db_fake_book_index "select id, page from fb2 where bookcode='$bookcode' and page=$page
If I run quote it like 如果我运行报价就好
echo "sqlite3 /home/user/db_fake_book_index \"select id, page from fb2 where bookcode='$bookcode' and page=$page\""
I get 我明白了
sqlite3 /home/user/db_fake_book_index "select id, page from fb2 where bookcode='557S' and page=10"
and if I copy it to a console it works fine, but if i encase it in $() in my script it does not work, sqlite believes there are too many variables, it seems my double quotes are being removed. 如果我将它复制到控制台它工作正常,但如果我把它包装在我的脚本中的$()它不起作用,sqlite认为有太多的变量,似乎我的双引号被删除。
$(printf "sqlite3 /home/user/db_fake_book_index \"select id, page from fb2 where bookcode='$bookcode' and page=$page\"")
This gives the same faliure 这给出了相同的效果
result=$( sqlite3 /home/user/db_fake_book_index "select id, page from fb2 where bookcode='$bookcode' and page=$page" )
or 要么
query="select id, page from fb2 where bookcode='$bookcode' and page=$page"
result=$( sqlite3 /home/user/db_fake_book_index "$query" )
This will save as the variable v
这将保存为变量
v
set "select id, page from fb2 where bookcode='$bookcode' and page=$page"
read v < <(sqlite3 /home/user/db_fake_book_index "$1")
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