解析文本文件以匹配字符串并提取值（在 Golang 中）

Question

我正在摸索一个似乎可能很常见的需求，但我无法在网络上找到任何示例。

我有一个这样的文件：

  answer VNET_1_DHCP yes

  answer VNET_1_DHCP_CFG_HASH 4CF2C196E368CE83B9D1895C5E05301CDFDEBCA0  

  answer VNET_1_HOSTONLY_NETMASK 255.255.255.0

  answer VNET_1_HOSTONLY_SUBNET 192.168.224.0

  answer VNET_1_VIRTUAL_ADAPTER yes

  answer VNET_8_DHCP yes

  answer VNET_8_DHCP_CFG_HASH D326C0BC7FF6C38C57AF341F9075E576C175B250

  answer VNET_8_HOSTONLY_NETMASK 255.255.255.0

  answer VNET_8_HOSTONLY_SUBNET 172.16.102.0

  answer VNET_8_NAT yes

  answer VNET_8_VIRTUAL_ADAPTER yes

我需要提取特定子网 (192.168.224.0) 的 VNET 编号。VNET 数量可能会有所不同（理论上，子网甚至可能不存在）。所以我需要匹配子网是否存在，如果存在则提取网络号（1在本例中）。

我发现在 BASH 中实现它非常容易：

 if grep -q 192.168.224.0 ./networking; then

      echo "The ${VMNET_SUBNET} network already exists"

      NETWORK_NUMBER=$(grep ${VMNET_SUBNET} ./networking | cut -d'_' -f 2)

      echo NETWORK_NUMBER  

 else <do something to create it.....> 

我试图找到使用 Go 实现这一点的最简单方法。

Answer 1

您可以使用正则表达式：re := regexp.MustCompile(`.*VNET_(\d+)_.*192.168.224.0`)matches := re.FindStringSubmatch(text)fmt.Println(matches[1])游乐场：http : //play.golang.org/p/NQlA2BObtU。

Answer 2

这是一种基于@ainar-g 的答案解析这些数据的可靠方法：http://play.golang.org/p/6-PELcLvVz此处的目标是使用以下类型将每个 VNET 的属性存储在映射中：type vnet map[int]map[string]string这段代码：var re = regexp.MustCompile(`.*VNET_(\d+)_([^\s]+) (.*)`)func ReadVnet(r io.Reader) vnet {&nbsp; &nbsp; s := bufio.NewScanner(r)&nbsp; &nbsp; v := make(vnet)&nbsp; &nbsp; for s.Scan() {&nbsp; &nbsp; &nbsp; &nbsp; matches := re.FindStringSubmatch(s.Text())&nbsp; &nbsp; &nbsp; &nbsp; id, err := strconv.Atoi(matches[1])&nbsp; &nbsp; &nbsp; &nbsp; if err != nil {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; continue&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; if _, ok := v[id]; !ok {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; v[id] = make(map[string]string)&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; v[id][matches[2]] = matches[3]&nbsp; &nbsp; }&nbsp; &nbsp; return v}创建有问题的地图：map[1:map[DHCP:yes DHCP_CFG_HASH:4CF2C196E368CE83B9D1895C5E05301CDFDEBCA0 HOSTONLY_NETMASK:255.255.255.0 HOSTONLY_SUBNET:192.168.224.0 VIRTUAL_ADAPTER:yes] 8:map[DHCP:yes DHCP_CFG_HASH:D326C0BC7FF6C38C57AF341F9075E576C175B250 HOSTONLY_NETMASK:255.255.255.0 HOSTONLY_SUBNET:172.16.102.0 NAT:yes VIRTUAL_ADAPTER:yes]]现在您可以在地图上迭代以找到感兴趣的项目：func main() {&nbsp; &nbsp; v := ReadVnet(bytes.NewBufferString(text))&nbsp; &nbsp; for id, properties := range v {&nbsp; &nbsp; &nbsp; &nbsp; if ip, ok := properties["HOSTONLY_SUBNET"]; ok && ip == "192.168.224.0" {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; fmt.Println(id)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }}

Answer 3

这是一个没有正则表达式的版本：&nbsp; idxEnd := strings.Index(txt, "192.168.224.0")&nbsp; idxVNET := strings.LastIndex(txt[:idxEnd], "VNET_")&nbsp; beginNumber := idxVNET + 5&nbsp; length := strings.Index(txt[beginNumber:idxEnd], "_")&nbsp; number, _ := strconv.Atoi(txt[beginNumber : beginNumber+length])&nbsp; fmt.Printf("number: %T %v\n", number, number)你会尝试在一个非常（非常）大的字符串上这样做，它应该更快。