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Golang Path.Segments方法代码示例

本文整理汇总了Golang中github.com/ipfs/go-ipfs/path.Path.Segments方法的典型用法代码示例。如果您正苦于以下问题:Golang Path.Segments方法的具体用法?Golang Path.Segments怎么用?Golang Path.Segments使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在github.com/ipfs/go-ipfs/path.Path的用法示例。


在下文中一共展示了Path.Segments方法的3个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Golang代码示例。

示例1: ResolveToKey

// ResolveToKey resolves a path to a key.
//
// It first checks if the path is already in the form of just a key (<key> or
// /ipfs/<key>) and returns immediately if so. Otherwise, it falls back onto
// Resolve to perform resolution of the dagnode being referenced.
func ResolveToKey(ctx context.Context, n *IpfsNode, p path.Path) (key.Key, error) {

	// If the path is simply a key, parse and return it. Parsed paths are already
	// normalized (read: prepended with /ipfs/ if needed), so segment[1] should
	// always be the key.
	if p.IsJustAKey() {
		return key.B58KeyDecode(p.Segments()[1]), nil
	}

	// Fall back onto regular dagnode resolution. Retrieve the second-to-last
	// segment of the path and resolve its link to the last segment.
	head, tail, err := p.PopLastSegment()
	if err != nil {
		return key.Key(""), err
	}
	dagnode, err := Resolve(ctx, n, head)
	if err != nil {
		return key.Key(""), err
	}

	// Extract and return the key of the link to the target dag node.
	link, err := dagnode.GetNodeLink(tail)
	if err != nil {
		return key.Key(""), err
	}

	return key.Key(link.Hash), nil
}
开发者ID:musha68k,项目名称:go-ipfs,代码行数:33,代码来源:pathresolver.go

示例2: ResolveToCid

// ResolveToKey resolves a path to a key.
//
// It first checks if the path is already in the form of just a key (<key> or
// /ipfs/<key>) and returns immediately if so. Otherwise, it falls back onto
// Resolve to perform resolution of the dagnode being referenced.
func ResolveToCid(ctx context.Context, n *IpfsNode, p path.Path) (*cid.Cid, error) {

	// If the path is simply a key, parse and return it. Parsed paths are already
	// normalized (read: prepended with /ipfs/ if needed), so segment[1] should
	// always be the key.
	if p.IsJustAKey() {
		return cid.Decode(p.Segments()[1])
	}

	// Fall back onto regular dagnode resolution. Retrieve the second-to-last
	// segment of the path and resolve its link to the last segment.
	head, tail, err := p.PopLastSegment()
	if err != nil {
		return nil, err
	}
	dagnode, err := Resolve(ctx, n.Namesys, n.Resolver, head)
	if err != nil {
		return nil, err
	}

	// Extract and return the key of the link to the target dag node.
	link, _, err := dagnode.ResolveLink([]string{tail})
	if err != nil {
		return nil, err
	}

	return link.Cid, nil
}
开发者ID:VictorBjelkholm,项目名称:go-ipfs,代码行数:33,代码来源:pathresolver.go

示例3: Resolve

// Resolve resolves the given path by parsing out protocol-specific
// entries (e.g. /ipns/<node-key>) and then going through the /ipfs/
// entries and returning the final merkledag node.  Effectively
// enables /ipns/, /dns/, etc. in commands.
func Resolve(ctx context.Context, n *IpfsNode, p path.Path) (*merkledag.Node, error) {
	if strings.HasPrefix(p.String(), "/ipns/") {
		// resolve ipns paths

		// TODO(cryptix): we sould be able to query the local cache for the path
		if n.Namesys == nil {
			return nil, ErrNoNamesys
		}

		seg := p.Segments()

		if len(seg) < 2 || seg[1] == "" { // just "/<protocol/>" without further segments
			return nil, path.ErrNoComponents
		}

		extensions := seg[2:]
		resolvable, err := path.FromSegments("/", seg[0], seg[1])
		if err != nil {
			return nil, err
		}

		respath, err := n.Namesys.Resolve(ctx, resolvable.String())
		if err != nil {
			return nil, err
		}

		segments := append(respath.Segments(), extensions...)
		p, err = path.FromSegments("/", segments...)
		if err != nil {
			return nil, err
		}
	}

	// ok, we have an ipfs path now (or what we'll treat as one)
	return n.Resolver.ResolvePath(ctx, p)
}
开发者ID:musha68k,项目名称:go-ipfs,代码行数:40,代码来源:pathresolver.go


注:本文中的github.com/ipfs/go-ipfs/path.Path.Segments方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。