bash 如何指定多行shell变量？
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How to specify a multi-line shell variable?
提问by user2219963
I have written a query:
我写了一个查询：
function print_ui_hosts
{
local sql = "select ........."
print_sql "$ sql"
}
local sql - a very long string. Query is not formatted. How can I split a string into multiple lines?
本地 sql - 一个很长的字符串。查询未格式化。如何将字符串拆分为多行？
回答by Nik O'Lai
simply insert new line where necessary
只需在必要时插入新行
sql="
SELECT c1, c2
from Table1, Table2
where ...
"
shell will be looking for the closing quotation mark
shell 将寻找右引号
回答by dogbane
Use readwith a heredoc as shown below:
read与 heredoc 一起使用，如下所示：
read -d '' sql << EOF
select c1, c2 from foo
where c1='something'
EOF
echo "$sql"
回答by islijepcevic
I would like to give one additional answer, while the other ones will suffice in most cases.
我想给出一个额外的答案，而其他答案在大多数情况下就足够了。
I wanted to write a string over multiple lines, but its contents needed to be single-line.
我想在多行上写一个字符串，但它的内容需要是单行的。
sql=" \
SELECT c1, c2 \
from Table1, ${TABLE2} \
where ... \
"
I am sorry if this if a bit off-topic (I did not need this for SQL). However, this post comes up among the first results when searching for multi-line shell variables and an additional answer seemed appropriate.
如果这有点偏离主题，我很抱歉（我不需要这个用于 SQL）。然而，当搜索多行 shell 变量时，这篇文章出现在第一个结果中，一个额外的答案似乎是合适的。
回答by Brad Parks
Thanks to dimo414's answer to a similar question, this shows how his great solution works, and shows that you can have quotes and variables in the text easily as well:
感谢dimo414 对类似问题的回答，这显示了他出色的解决方案是如何工作的，并表明您也可以轻松地在文本中使用引号和变量：
example output
示例输出
$ ./test.sh
The text from the example function is:
Welcome dev: Would you "like" to know how many 'files' there are in /tmp?
There are " 38" files in /tmp, according to the "wc" command
test.sh
测试文件
#!/bin/bash
function text1()
{
COUNT=$(\ls /tmp | wc -l)
cat <<EOF
Would you "like" to know how many 'files' there are in /tmp?
There are "$COUNT" files in /tmp, according to the "wc" command
EOF
}
function main()
{
OUT=$(text1 "Welcome dev:")
echo "The text from the example function is: $OUT"
}
main
回答by EndlosSchleife
readdoes not export the variable (which is a good thing most of the time). Here's an alternative which can be exported in one command, can preserve or discard linefeeds, and allows mixing of quoting-styles as needed. Works for bash and zsh.
read不导出变量（大多数时候这是一件好事）。这是一种可以在一个命令中导出的替代方法，可以保留或丢弃换行符，并允许根据需要混合引用样式。适用于 bash 和 zsh。
oneLine=$(printf %s \
a \
" b " \
$'\tc\t' \
'd ' \
)
multiLine=$(printf '%s\n' \
a \
" b " \
$'\tc\t' \
'd ' \
)
I admit the need for quoting makes this ugly for SQL, but it answers the (more generally expressed) question in the title.
我承认引用的需要使 SQL 变得丑陋，但它回答了标题中的（更普遍表达的）问题。
I use it like this
我像这样使用它
export LS_COLORS=$(printf %s \
':*rc=36:*.ini=36:*.inf=36:*.cfg=36:*~=33:*.bak=33:*$=33' \
...
':bd=40;33;1:cd=40;33;1:or=1;31:mi=31:ex=00')
in a file sourced from both my .bashrcand .zshrc.
在来自 my.bashrc和.zshrc.
