如何使用 `awk` 在列中拆分文本？

Question

如何提取第一列（“”中的第一列）

"xxxxx1" {685a}
"xx2" {bcdd}
"xx3 gsdd" {29a6ff}
"sdsdf xxx" {243b9}
"sdfsdf ccc dd" {c28f2f}
"dsdsf sfsdf" {216e}
"sdfsdfsd" {48530}
"sdfsdff" {9d2afa0n}
"sdfsdfdff sdfs" {d8681a}
"sdfsdsds sdfsdf d" {5b9b8}
"sdfsdfs sdf sdfsdf" {68e08a}
"sdfsdfsdf sdf" {107fa0}

我想要的结果是：

"xxxxx1"
"xx2"
"xx3 gsdd"
"sdsdf xxx"
"sdfsdf ccc dd"
"dsdsf sfsdf"
"sdfsdfsd"
"sdfsdff"
"sdfsdfdff sdfs"
"sdfsdsds sdfsdf d"
"sdfsdfs sdf sdfsdf"
"sdfsdfsdf sdf"

任何的想法？

Answer 1

使用? {（空格，大括号）作为字段分隔符，因为您不关心第二个字段：

$ awk -F ' {' '{print $1}' foo
"xxxxx1"
"xx2"
"xx3 gsdd"
"sdsdf xxx"
"sdfsdf ccc dd"
"dsdsf sfsdf"
"sdfsdfsd"
"sdfsdff"
"sdfsdfdff sdfs"
"sdfsdsds sdfsdf d"
"sdfsdfs sdf sdfsdf"
"sdfsdfsdf sdf"