I have a following bash script:
我有以下bash脚本:
1 #!/bin/bash
2 query='query= SELECT * WHERE { ?s ?p ?o } LIMIT 5'
3 cmd="curl $1 -s -d \"$query\""
4 echo "$cmd"
5 # curl $1 -s -d "$query"
6 # $cmd
5th and 6th lines must do the same. When i uncomment the 5th line, everything works fine. But with the 6th line nothing doesn't work.
第五和第六行也必须这样做。当我取消第5行注释时，一切正常。但是在第六行，什么都不行。
So i'm wondering whats the difference?
我想知道这有什么区别吗?
Thanks.
谢谢。
2 个解决方案
#1
1
Line 5 passes $query as a single argument. Line 6 passes each word of $query as a separate argument, with " at the beginning of the first and " at the end of the last. Put your arguments in an array instead.
第5行将$query作为单个参数传递。第6行将$query的每个单词作为单独的参数传递，其中“在第一个的开头”和“在最后一个的末尾”。将参数放在数组中。
#2
2
No; in line 4 you are just displaying the command by "echo"ing it; line 6 actually executes the command.
没有;在第4行中，您只是通过“echo”显示命令;第6行实际执行命令。
E.g.
如。
$eg="ls /var/www"
echo $eg #This would literally return ls /var/www"
$eg #This would return the directory listing of /var/www/ (actually run the command).
#1
1
Line 5 passes $query as a single argument. Line 6 passes each word of $query as a separate argument, with " at the beginning of the first and " at the end of the last. Put your arguments in an array instead.
第5行将$query作为单个参数传递。第6行将$query的每个单词作为单独的参数传递，其中“在第一个的开头”和“在最后一个的末尾”。将参数放在数组中。
#2
2
No; in line 4 you are just displaying the command by "echo"ing it; line 6 actually executes the command.
没有;在第4行中，您只是通过“echo”显示命令;第6行实际执行命令。
E.g.
如。
$eg="ls /var/www"
echo $eg #This would literally return ls /var/www"
$eg #This would return the directory listing of /var/www/ (actually run the command).