将文本文件解析以匹配字符串并提取值（使用Golang）

huangapple
go评论115阅读模式
英文:

Parsing of text file to match string and extract value (in Golang)

问题

我正在思考一个似乎可能很常见但在网上找不到任何示例的需求。

我有一个像这样的文件：

  1.   answer VNET_1_DHCP yes
  2.   answer VNET_1_DHCP_CFG_HASH 4CF2C196E368CE83B9D1895C5E05301CDFDEBCA0  
  3.   answer VNET_1_HOSTONLY_NETMASK 255.255.255.0
  4.   answer VNET_1_HOSTONLY_SUBNET 192.168.224.0
  5.   answer VNET_1_VIRTUAL_ADAPTER yes
  6.   answer VNET_8_DHCP yes
  7.   answer VNET_8_DHCP_CFG_HASH D326C0BC7FF6C38C57AF341F9075E576C175B250
  8.   answer VNET_8_HOSTONLY_NETMASK 255.255.255.0
  9.   answer VNET_8_HOSTONLY_SUBNET 172.16.102.0
  10.   answer VNET_8_NAT yes
  11.   answer VNET_8_VIRTUAL_ADAPTER yes

我需要提取特定子网（192.168.224.0）的VNET编号。VNET编号可能会变化（理论上，子网甚至可能不存在）。因此，我需要匹配子网是否存在，如果存在，则提取网络编号（在此示例中为1）。

我发现在BASH中实现这个非常简单：

  1.  if grep -q 192.168.224.0 ./networking; then
  2.       echo "The ${VMNET_SUBNET} network already exists"
  3.       NETWORK_NUMBER=$(grep ${VMNET_SUBNET} ./networking | cut -d'_' -f 2)
  4.       echo NETWORK_NUMBER  
  5.  else <执行创建操作.....>

我正在尝试找到使用Go实现这个的最简单方法。

谢谢。

英文:

I am scratching my head around a need that seems to be potentially common but I couldn't locate any example on the web.

I have a file like this:

  1.   answer VNET_1_DHCP yes
  2.   answer VNET_1_DHCP_CFG_HASH 4CF2C196E368CE83B9D1895C5E05301CDFDEBCA0  
  3.   answer VNET_1_HOSTONLY_NETMASK 255.255.255.0
  4.   answer VNET_1_HOSTONLY_SUBNET 192.168.224.0
  5.   answer VNET_1_VIRTUAL_ADAPTER yes
  6.   answer VNET_8_DHCP yes
  7.   answer VNET_8_DHCP_CFG_HASH D326C0BC7FF6C38C57AF341F9075E576C175B250
  8.   answer VNET_8_HOSTONLY_NETMASK 255.255.255.0
  9.   answer VNET_8_HOSTONLY_SUBNET 172.16.102.0
  10.   answer VNET_8_NAT yes
  11.   answer VNET_8_VIRTUAL_ADAPTER yes

I need to extract the VNET number of a specific subnet (192.168.224.0). VNET numbers could vary (and the subnet could, in theory, not even exist). So I need to match if the subnet exists, and if exists extract the network number (1 in this example).

I found it to be SUPER easy to implement this in BASH:

  1.  if grep -q 192.168.224.0 ./networking; then
  2.       echo &quot;The ${VMNET_SUBNET} network already exists&quot;
  3.       NETWORK_NUMBER=$(grep ${VMNET_SUBNET} ./networking | cut -d&#39;_&#39; -f 2)
  4.       echo NETWORK_NUMBER  
  5.  else &lt;do something to create it.....&gt;

I am trying to find the easiest way to implement this using Go.

Thanks.

答案1

得分: 10

你可以使用正则表达式（regular expression）来实现：

  1. re := regexp.MustCompile(`.*VNET_(\d+)_.*192.168.224.0`)
  2. matches := re.FindStringSubmatch(text)
  3. fmt.Println(matches[1])

Playground: http://play.golang.org/p/NQlA2BObtU.

英文:

You can use a regular expression:

  1. re := regexp.MustCompile(`.*VNET_(\d+)_.*192.168.224.0`)
  2. matches := re.FindStringSubmatch(text)
  3. fmt.Println(matches[1])

Playground: http://play.golang.org/p/NQlA2BObtU.

答案2

得分: 5

这是一种基于 @ainar-g 的答案的稳健的解析数据的方法：

http://play.golang.org/p/6-PELcLvVz

这里的目标是使用以下类型将每个VNET的属性存储在一个映射中：

  1. type vnet map[int]map[string]string

这段代码：

  1. var re = regexp.MustCompile(`.*VNET_(\d+)_([^\s]+) (.*)`)
  2. func ReadVnet(r io.Reader) vnet {
  3.     s := bufio.NewScanner(r)
  4.     v := make(vnet)
  6.     for s.Scan() {
  7.         matches := re.FindStringSubmatch(s.Text())
  8.         id, err := strconv.Atoi(matches[1])
  9.         if err != nil {
  10.             continue
  11.         }
  13.         if _, ok := v[id]; !ok {
  14.             v[id] = make(map[string]string)
  15.         }
  16.         v[id][matches[2]] = matches[3]
  17.     }
  19.     return v
  20. }

创建了所需的映射：

  1. 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，您可以迭代映射以找到感兴趣的项：

  1. func main() {
  2.     v := ReadVnet(bytes.NewBufferString(text))
  4.     for id, properties := range v {
  5.         if ip, ok := properties["HOSTONLY_SUBNET"]; ok && ip == "192.168.224.0" {
  6.             fmt.Println(id)
  7.             return
  8.         }
  9.     }
  10. }
英文:

Here is a robust way to parse these data that builds on @ainar-g's answer:

http://play.golang.org/p/6-PELcLvVz

The goal here is to store the properties for each VNET in a map, using the following type:

  1. type vnet map[int]map[string]string

This code:

  1. var re = regexp.MustCompile(`.*VNET_(\d+)_([^\s]+) (.*)`)
  2. func ReadVnet(r io.Reader) vnet {
  3.     s := bufio.NewScanner(r)
  4.     v := make(vnet)
  6.     for s.Scan() {
  7.         matches := re.FindStringSubmatch(s.Text())
  8.         id, err := strconv.Atoi(matches[1])
  9.         if err != nil {
  10.             continue
  11.         }
  13.         if _, ok := v[id]; !ok {
  14.             v[id] = make(map[string]string)
  15.         }
  16.         v[id][matches[2]] = matches[3]
  17.     }
  19.     return v
  20. }

creates the map in question:

  1. 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]]

Now you can iterate on the map to find the item of interest:

  1. func main() {
  2. 	v := ReadVnet(bytes.NewBufferString(text))
  4. 	for id, properties := range v {
  5. 		if ip, ok := properties[&quot;HOSTONLY_SUBNET&quot;]; ok &amp;&amp; ip == &quot;192.168.224.0&quot; {
  6. 			fmt.Println(id)
  7. 			return
  8. 		}
  9. 	}
  10. }

答案3

得分: 3

这是一个没有正则表达式的版本：

  1.   idxEnd := strings.Index(txt, "192.168.224.0")
  2.   idxVNET := strings.LastIndex(txt[:idxEnd], "VNET_")
  3.   beginNumber := idxVNET + 5
  4.   length := strings.Index(txt[beginNumber:idxEnd], "_")
  5.   number, _ := strconv.Atoi(txt[beginNumber : beginNumber+length])
  6.   fmt.Printf("number: %T %v\n", number, number)

如果你尝试在一个非常（非常）大的字符串上执行这个操作，它应该会更快。

英文:

This is a version without regexp:

  1.   idxEnd := strings.Index(txt, &quot;192.168.224.0&quot;)
  2.   idxVNET := strings.LastIndex(txt[:idxEnd], &quot;VNET_&quot;)
  3.   beginNumber := idxVNET + 5
  4.   length := strings.Index(txt[beginNumber:idxEnd], &quot;_&quot;)
  5.   number, _ := strconv.Atoi(txt[beginNumber : beginNumber+length])
  6.   fmt.Printf(&quot;number: %T %v\n&quot;, number, number)

Would you try to do that on a very (very) big string, it should be faster.

huangapple
  • 本文由 发表于 2015年12月24日 00:49:27
  • 转载请务必保留本文链接：https://go.coder-hub.com/34440268.html
  • go
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定