API error while trying to upload File

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  • jcleblanc

    Hi akubatoor,

     

    Would you happen to have the error message that's sent with the 400 status code in the Box API response? That'll help narrow down the reason for the 400.

     

    Thanks,

    Jon

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  • akubatoor

    The response is 400 item_name_invalid

     

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  • tonyjin

    From https://community.box.com/t5/Managing-Content-Troubleshooting/File-and-Folder-Upload-Issues/ta-p/306

     

    Verify that the file's name is valid. Box only supports file or folder names that are 255 characters or less. File names containing non-printable ascii, "/" or "\", names with leading or trailing spaces, and the special names “.” and “..” are also unsupported.

     

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  • akubatoor

    I don't think I'm giving any invalid file name. I'm trying to upload a file with the name taxdoc.txt. Below the API response I'm seeing

     

    {"type":"error","status":400,"code":"bad_request","context_info":{"errors":[{"reason":"invalid_parameter","name":"entity-body","message":"Invalid value '{\"name\":\"taxdoc.txt\", \"parent\":{\"id\":\"0\"}'. Entity body should be a correctly nested resource attribute name\/value pair"}]},"help_url":"http:\/\/developers.box.com\/docs\/#errors","message":"Bad Request","request_id":"***number removed for privacy***597673fe85807"}
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  • kendomen

    I just ran your code and made one change and it seems to work fine.

    I changed pathfileName to fileName in the method call folder.uploadFile()

     

    if (!fileExists) {
        System.out.println("uploading new file: " + fileName);
        FileInputStream stream = new FileInputStream(pathFileName);
        BoxFile.Info boxInfo = folder.uploadFile(stream, fileName);  // 

     

     

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  • akubatoor

    Yeah right..I also had changed the code a little bit to get this working. I changed it something like below

     

    if (!fileExists) {
                    FileInputStream stream = new FileInputStream(pathFileName);
                    FileUploadParams fileUploadParams = new FileUploadParams();
                    fileUploadParams.setName();
                    fileUploadParams.setContent(stream);
                    BoxFile.Info boxInfo = folder.uploadFile(fileUploadParams);
                    fileId = boxInfo.getID();
                    stream.close();
                }

     

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  • lingamp

    Hey,

     

    what is the value FILE inside concat method/ where do you initialize it ?

     

     

    Thanks,

    limanp

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