将视频上传到服务器

时间:2018-12-03 07:57:39

标签: ios objective-c react-native react-native-camera

我在我的应用程序中使用react-native-camera(^ 0.6.0),我必须将录制的视频作为多部分数据上传到服务器。在android中,他们像这样发布信息,

{
  _parts: [
    [
      'file',
      {
        name: 'VID_20181130_150959.mp4',
        uri: 'file:///storage/emulated/0/DCIM/VID_20181130_150959.mp4',
        type: 'video/mp4'
      }
    ]
  ]
}

但是在iOS中,文件路径返回为asset-library://asset/asset.mov?id = 41B76A24-1018-46C1-A658-C1EFFC552FD0&ext = mov,但如果我发布资产路径,则不会上传。

>
{
  _parts: [
    [
      'file',
      {
        name: '41B76A24-1018-46C1-A658-C1EFFC552FD0.mov',
        uri: 'assets-library://asset/asset.mov?id=41B76A24-1018-46C1-A658-C1EFFC552FD0.mov',
        type: 'video/mov'
      }
    ]
  ]
}

有人可以帮我吗??

我正在使用iPhone 6调试代码,

var url = DomainAPI.lashHostname + '/attachments?token=' + value + '&class=Quick';
fetch(url, {
      method: 'POST',
      headers: {
        'Accept': 'application/json',
        'Content-Type': 'multipart/form-data',
      },
      body: JSON.stringify(formData)
    }).then((response) => response.json())
      .then((responseData) => {
        console.log('responseData------->************$$$$$$$$$$$' + JSON.stringify(responseData));
        if (responseData.error.code == 0) {
          this.sendFiletoServer(responseData.id, value);
        } else {
          this.setState({ loading: false });
          Actions.pop();
        }
      }).catch((err) => {
        console.log('responseData-------> err ************$$$$$$$$$$$' + JSON.stringify(err));
        this.setState({ loading: false });
        Actions.pop();
      });

1 个答案:

答案 0 :(得分:1)

请参阅此代码,希望它对我有用

 let formData = new FormData();
  formData.append("videoFile", {
    name: name.mp4,
    uri: video.uri,
    type: 'video/mp4'
});
formData.append("id", "1234567");

try {
    let response = await fetch(url, {
        method: 'post',
        headers: {
            'Content-Type': 'multipart/form-data',
        },
        body: formData
    });
    return await response.json();
}
catch (error) {
    console.log('error : ' + error);
    return error;