我使用 Lua 5.1.我想解析以下模式的 XML 文件.我应该怎么做? [英] I use Lua 5.1. I want to parse an XML file of the following pattern. How should I go about it?

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问题描述

我尝试使用 LuaXml 库.但是它的功能是有限的,因为它只返回特定属性的第一个子表,并且不会更进一步.然后我尝试了字符串模式匹配,它有效,但我走到了死胡同,它无法完全完成任务.LuaExpat 库存在于我的 lua 的 lib 文件夹中,并且还有一个名为 lom.lua 的文件.但通常它不起作用或给我找不到模块"的错误

I tried using LuaXml library. But its functionality is limited as this returns only the first subtable of a particular attribute and does not go further than that. Then I tried string pattern matching which worked but I reached a dead end and it couldnt completely achieve the task. LuaExpat library is present in my lib folder of lua, and a file called lom.lua is also there. But often it doesnt work or gives me the error that "module not found"

我的 XML 文件如下所示:

My XML file looks like this :

<Service>
<NewInstance ref="5A">
<Std>DiscoveredElement</Std>
<Key>5A</Key>
<Attributes>
<Attribute name="TARGET_TYPE" value="weblogic_cluster" />
<Attribute name="DISCOVERED_NAME" value="/Farm_soa4_sys20_soa4_domain/soa4_domain/WSM4_Cluster" />
<Attribute name="BROKEN_REASON" value="0" />
<Attribute name="TARGET_NAME" value="/Farm_soa4_sys20_soa4_domain/soa4_domain/WSM4_Cluster" />
<Attribute name="EMD_URL" value="https://uxsys460.schneider.com:3872/emd/main/" />
</Attributes>
</NewInstance>

<NewInstance ref="6C">
<Std>DiscoveredElement</Std>
<Key>6C</Key>
<Attributes>
<Attribute name="TARGET_TYPE" value="oracle_weblogic_nodemanager" />
<Attribute name="SERVICE_TYPE" value=" " />
<Attribute name="ORG_ID" value="0" />
<Attribute name="TARGET_NAME" value="Oracle WebLogic NodeManager-uxlab090" />
</Attributes>
</NewInstance>

<NewInstance ref="98">
<Std>DiscoveredElement</Std>
<Key>98</Key>
<Attributes>
<Attribute name="TARGET_TYPE" value="composite" />
<Attribute name="SERVICE_TYPE" value=" " />
<Attribute name="TARGET_NAME" value="SYS-IMG-Grp" />
<Attribute name="EMD_URL" value="" />
</Attributes>
</NewInstance>

<NewRelationship>
<Parent>
<Instance ref="98" />
</Parent>
<GenericRelations>
<Relations type="contains">
<Instance ref="5A" />
</Relations>
</GenericRelations>
</NewRelationship>

<NewRelationship>
<Parent>
<Instance ref="5A" />
</Parent>
<GenericRelations>
<Relations type="contains">
<Instance ref="6C" />
</Relations>
</GenericRelations>
</NewRelationship>
<NewRelationship>
<Parent>
<Instance ref="5A" />
</Parent>
<GenericRelations>
<Relations type="contains">
<Instance ref="98" />
</Relations>
</GenericRelations>
</NewRelationship>
</Service>

我的议程是显示一个 NewInstance ID 及其对应的目标类型和目标名称,以及它的关系类型和与之相关的实例 ref 的 ID,以及它的目标类型和目标名称例如:

My agenda is to display a NewInstance ID and its corresponding target type and target name and also its relation type with and the ID of instance ref its related to, along with its target type and target name for eg:

NewInstance ID - 5A
Target Type - weblogic_cluster 
Target Name - /Farm_soa4_sys20_soa4_domain/soa4_domain/WSM4_Cluster
Relation Type - contains
Instance ref - 6C
Target Type - oracle_weblogic_nodemanager
Target Name - Oracle WebLogic NodeManager-uxlab090
Instance ref - 98
Target Type - composite
Target Name - SYS-IMG-Grp

现在不能使用 LuaXml 来实现这一点.我将在下面列出字符串模式匹配的代码,它可以帮助我完成任务直到关系类型但不准确

Now LuaXml cannot be used to achieve this. String pattern matching's code I'll list below and it helps me accomplish the task till relation type but not accurately

代码是:

a={}
b={}
c={}
d={}
p=0
i=0
q=0

local file = io.open("oem_topology_output.xml", "rb")   -- Open file   for    reading (binary data)
  for instance in file:read("*a"):gmatch("<NewInstance ref="(.-)">") do
     a[i] = instance
     i = i+1
  end
file:close()
local files = io.open("oem_topology_output.xml", "rb")   -- Open file for  reading (binary data)
  for instances in files:read("*a"):gmatch("<NewInstance ref=".-">(.-)</NewInstance>") do
     TARGET_TYPE = instances:match('TARGET_TYPE.-value="(.-)"')
     TARGET_NAME = instances:match('TARGET_NAME.-value="(.-)"')
     b[p] = TARGET_TYPE
     c[p] = TARGET_NAME
     p =p+1
  end
local file = io.open("oem_topology_output.xml", "rb")   -- Open file   for   reading (binary data)
  for type in file:read("*a"):gmatch("<Relations type="(.-)">") do
    d[q] = type
    q = q+1
  end
files:close()
for j=0,i-1 do
print("INSTANCE ID : ", a[j])
print("TARGET TYPE : ", b[j])
print("TARGET NAME : ", c[j])
print("RELATION TYPE : ",d[j])
end

请建议我应该采用哪种方法才能以所需的方式解析 XMl 文件.哪个内置库将提供 apt 功能.如果您提出建议,LuaExpat 让我知道它对我不起作用的可能原因.

Please suggest what approach I should follow to be able to parse the XMl file in the required way. Which in-built library will provide the apt functions. In case you suggest, LuaExpat let me know the possible reasons why it does not work for me.

推荐答案

你不需要任何特殊的库来解析 lua 中的 xml,有很多内置的功能可以编写你自己的解析器.

You don't need any special libraries to parse xml in lua, there is plenty of built in functionality to write your own parser.

例如,要检索节点的属性,您可以这样写.

For instance to retrieve the attributes of a node you can write something like this.

local file = "oem_topology_output.xml"
local node = "<(%a-)%s* "
local attributes = {
    "ref",
    "name",
    "value",
    "type"
}


for line in io.lines(file) do
    for a in line:gmatch(node) do
        for x = 1, #attributes do
            n = line:match(attributes[x]..'="(.-)"')
            if n then
                print(a, attributes[x], n)
            end
        end
    end
end

产生类似的输出

NewInstance ref 5A
Attribute   name    TARGET_TYPE
Attribute   value   weblogic_cluster
Attribute   name    DISCOVERED_NAME
Attribute   value   /Farm_soa4_sys20_soa4_domain/soa4_domain/WSM4_Cluster
Attribute   name    BROKEN_REASON
Attribute   value   0
Attribute   name    TARGET_NAME
Attribute   value   /Farm_soa4_sys20_soa4_domain/soa4_domain/WSM4_Cluster
Attribute   name    EMD_URL
Attribute   value   https://uxsys460.schneider.com:3872/emd/main/
NewInstance ref 6C
Attribute   name    TARGET_TYPE
Attribute   value   oracle_weblogic_nodemanager
Attribute   name    SERVICE_TYPE
Attribute   value    
Attribute   name    ORG_ID
Attribute   value   0
Attribute   name    TARGET_NAME
Attribute   value   Oracle WebLogic NodeManager-uxlab090
NewInstance ref 98
Attribute   name    TARGET_TYPE
Attribute   value   composite
Attribute   name    SERVICE_TYPE
Attribute   value    
Attribute   name    TARGET_NAME
Attribute   value   SYS-IMG-Grp
Attribute   name    EMD_URL
Attribute   value   
Instance    ref 98
Relations   type    contains
Instance    ref 5A
Instance    ref 5A
Relations   type    contains
Instance    ref 6C
Instance    ref 5A
Relations   type    contains
Instance    ref 98

无需打开或关闭文件,因为没有写入的意图.

No need to open or close a file, since there is no intent of writing to it.

这篇关于我使用 Lua 5.1.我想解析以下模式的 XML 文件.我应该怎么做?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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