Fix xml parsing error

go.mod

@@ -13,7 +13,7 @@ require (
 	github.com/niemeyer/pretty v0.0.0-20200227124842-a10e7caefd8e // indirect
 	github.com/satori/go.uuid v1.2.0 // indirect
 	github.com/valyala/fastjson v1.5.4
-	golang.org/x/sys v0.0.0-20200828194041-157a740278f4 // indirect
+	golang.org/x/sys v0.0.0-20200831180312-196b9ba8737a // indirect
 	golang.org/x/time v0.0.0-20200630173020-3af7569d3a1e // indirect
 	gopkg.in/check.v1 v1.0.0-20200227125254-8fa46927fb4f // indirect
 )

go.sum

@@ -38,8 +38,8 @@ golang.org/x/sys v0.0.0-20190222072716-a9d3bda3a223/go.mod h1:STP8DvDyc/dI5b8T5h
 golang.org/x/sys v0.0.0-20191026070338-33540a1f6037/go.mod h1:h1NjWce9XRLGQEsW7wpKNCjG9DtNlClVuFLEZdDNbEs=
 golang.org/x/sys v0.0.0-20200116001909-b77594299b42/go.mod h1:h1NjWce9XRLGQEsW7wpKNCjG9DtNlClVuFLEZdDNbEs=
 golang.org/x/sys v0.0.0-20200223170610-d5e6a3e2c0ae/go.mod h1:h1NjWce9XRLGQEsW7wpKNCjG9DtNlClVuFLEZdDNbEs=
-golang.org/x/sys v0.0.0-20200828194041-157a740278f4 h1:kCCpuwSAoYJPkNc6x0xT9yTtV4oKtARo4RGBQWOfg9E=
-golang.org/x/sys v0.0.0-20200828194041-157a740278f4/go.mod h1:h1NjWce9XRLGQEsW7wpKNCjG9DtNlClVuFLEZdDNbEs=
+golang.org/x/sys v0.0.0-20200831180312-196b9ba8737a h1:i47hUS795cOydZI4AwJQCKXOr4BvxzvikwDoDtHhP2Y=
+golang.org/x/sys v0.0.0-20200831180312-196b9ba8737a/go.mod h1:h1NjWce9XRLGQEsW7wpKNCjG9DtNlClVuFLEZdDNbEs=
 golang.org/x/time v0.0.0-20200630173020-3af7569d3a1e h1:EHBhcS0mlXEAVwNyO2dLfjToGsyY4j24pTs2ScHnX7s=
 golang.org/x/time v0.0.0-20200630173020-3af7569d3a1e/go.mod h1:tRJNPiyCQ0inRvYxbN9jk5I+vvW/OXSQhTDSoE431IQ=
 gopkg.in/check.v1 v1.0.0-20200227125254-8fa46927fb4f h1:BLraFXnmrev5lT+xlilqcH8XK9/i0At2xKjWk4p6zsU=

multipart.go

@@ -11,6 +11,7 @@ import (
 	"net/http"
 	"os"
 	"path/filepath"
+	"strings"
 	"time"
 
 	"github.com/aliyun/aliyun-oss-go-sdk/oss"
@@ -230,7 +231,10 @@ func multipartUploadFile(ft fastToken, file string, sp *saveProgress) (e error)
 	)
 	// EOF错误是xml的Unmarshal导致的，响应其实是json格式，所以实际上上传是成功的
 	if err != nil && !errors.Is(err, io.EOF) {
-		panicln(err)
+		// 当文件名含有 &< 这两个字符之一时响应的xml解析会出现错误，实际上上传是成功的
+		if filename := filepath.Base(file); !strings.ContainsAny(filename, "&<") {
+			panicln(err)
+		}
 	}
 	if *verbose {
 		log.Printf("CompleteMultipartUpload的响应头的值是：\n%+v", header)