Show output filename in the cli
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parent
b19eba0560
commit
5e6884af66
2 changed files with 7 additions and 6 deletions
2
VERSION
2
VERSION
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@ -1 +1 @@
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2.0.0
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2.0.1
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11
cli/cli.go
11
cli/cli.go
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@ -177,11 +177,16 @@ func CliRun() int {
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CliAvailableFormats(streamEp.Formats)
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return 1
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}
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CliShowFormat(format)
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cli.InfoMessage(fmt.Sprintf("Format: %v", format.Name))
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if args.ChapterIdx >= 0 {
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cli.InfoMessage(fmt.Sprintf("Chapter: %v. %v", cliArgs.ChapterNum, streamEp.Chapters[args.ChapterIdx].Title))
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}
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// We already set the output file correctly so we can output it
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if args.OutputFile == "" {
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args.OutputFile = streamEp.GetProposedFilename(args.ChapterIdx)
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}
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// Start Download
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cli.InfoMessage(fmt.Sprintf("Output: %v", args.OutputFile))
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fmt.Print("\n")
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if err = streamEp.Download(args, &cli, make(chan os.Signal, 1)); err != nil {
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cli.ErrorMessage(err)
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@ -205,10 +210,6 @@ func CliAvailableFormats(formats []core.VideoFormat) {
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}
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}
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func CliShowFormat(format core.VideoFormat) {
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fmt.Printf("Format: %v\n", format.Name)
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}
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type Cli struct{}
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func (cli *Cli) DownloadProgress(progress float32, rate float64, delaying bool, waiting bool, retries int, title string) {
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