在bash中获取mysql数据，函数读取将每个句子作为变量读取

Question

我试图从我的数据库中获取变量，我在db中有一个值我有2个单词，但下面的命令将其视为2个变量。

是否有任何改变以我的方式获取它？

   mysql -uroot -proot --database=symfony -se "SELECT cameras.source, cameras.name, investments.name, users.username FROM cameras JOIN investments ON investments.id = cameras.investment_id JOIN users ON users.id = investments.user_id;" | while read source camera investment username; do

mysql结果：

| source                                                                 | name         | name       | username |
+------------------------------------------------------------------------+--------------+------------+----------+
| rtsp://vlc:vlc@test/cam/realmonitor?channel=1&subtype=0   | camera 1     | investment |  user    |
+------------------------------------------------------------------------+--------------+------------+----------+

击：

echo "source :"$source
echo "camera :" :$camera
echo "investments: " $investments
echo "username: " $username

结果：

source :rtsp://vlc:vlc@test/cam/realmonitor?channel=1&subtype=0channel=1&subtype=0
camera : investment
investments: 1
username:  investment username

Answer 1

为了避免将camera.name中的那个空间混淆为一个saperator，你可以用另一个char替换空格，例如_

 SELECT cameras.source
      , replace(cameras.name, ' ', '_')
      , investments.name, users.username 
  FROM cameras 
  JOIN investments ON investments.id = cameras.investment_id 
  JOIN users ON users.id = investments.user_id;