英文:
Go XML error: invalid character entity
问题
Go无法解析带有声明实体的正确xml文件，一直出现以下错误：
错误：第47行的XML语法错误：无效的字符实体&n;
该行代码为<pos>&n;</pos>，实体定义为<!ENTITY n "noun (common) (futsuumeishi)">
这是Go中的程序：http://play.golang.org/p/94_60srVne
英文:
Go can't parse a correct xml file with declared entities, keep getting this error:
> error: XML syntax error on line 47: invalid character entity &n;
The line being <pos>&n;</pos> and the entity defined as <!ENTITY n "noun (common) (futsuumeishi)">
Here is the program in Go: http://play.golang.org/p/94_60srVne
答案1
得分: 6
你可以在创建Decoder并修改其Entity映射时传递实体。我怀疑该包实际上并没有解析DTD，只是从查看xml.go文件中的内容来看；我看到有一个注释说它会为调用者累积实体，但没有看到它自己在d.Entity中设置条目的内容。
（即使是encoding/xml安全地提供这个功能也很棘手，因为有一个内置的共享HTML实体映射。更新一个文档的映射会影响其他文档的解析。）
创建具有自定义实体的解码器需要比常规的xml.Unmarshal多一些工作，但并不复杂：
func main() {
jmd := JMdict{}
d := xml.NewDecoder(bytes.NewReader([]byte(str)))
d.Entity = map[string]string{
"n": "(noun)",
}
err := d.Decode(&jmd)
if err != nil {
fmt.Printf("error: %v", err)
return
}
fmt.Println(jmd)
}
这里是一个Playground链接，其中包含Entity技巧和一些输出代码，以将对象显示为JSON格式。
英文:
You can pass entities in if you create a Decoder and mess with its Entity map. I suspect the package doesn't actually parse DTDs, just from poking around xml.go; I see a comment saying it accumulates entities for the caller, but nothing that itself sets entries in d.Entity.
(It would be tricky for encoding/xml to safely provide that, even, because there is a built-in shared HTML entity map. Updating it for one doc would affect parsing of others.)
There's a little more paperwork to create a Decoder with custom entities than there is for regular xml.Unmarshal, but not too much:
func main() {
jmd := JMdict{}
d := xml.NewDecoder(bytes.NewReader([]byte(str)))
d.Entity = map[string]string{
"n": "(noun)",
}
err := d.Decode(&jmd)
if err != nil {
fmt.Printf("error: %v", err)
return
}
fmt.Println(jmd)
}
Here's a Playground link with the Entity trick and some output code to show the object as JSON.
答案2
得分: 4
之前的答案是“正确”的答案，但是我相信，根据你真正想要实现的目标，一个“快速”的答案是禁用Strict。例如：
d := xml.NewDecoder(os.Stdin)
d.Strict = false
英文:
The previous answer is the "right" answer, but I believe, depending on what you are really trying to accomplish, a "quick" answer is to disable Strict. e.g.:
d := xml.NewDecoder(os.Stdin)
d.Strict = false
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