remove downloaded image by react native fetch blob? - react-native

Question

I am using RNFetchBlob to share image as base64 format. But the image is downloading in my local folder. I want to delete it. By my research knowledge i want to unlink the image path. I don't know how to do that. Here is my code
_downloadImageAndShare(url ,title, message) {
const { fs } = RNFetchBlob.fs;
const self = this;
RNFetchBlob.config({ fileCache: true })
.fetch('GET', url)
.then(resp => resp.readFile('base64')
.then(base64 => ({ resp, base64 })))
.then(obj => {
const headers = obj.resp.respInfo.headers;
const type = headers['Content-Type'];
const dataUrl = 'data:' + type + ';base64,' + obj.base64;
return { url: dataUrl, title, message };
})
.then(options => Share.open(options)).catch(err => {err &&
console.log(err); })
}
how to use this code inside this method
RNFetchBlob.fs.unlink(path)
.then(() => { ... })
.catch((err) => { ... })
and how to specify the unlink(path) ??
thanks in advance.

Answer 1

You can specify file location with config with path param.
const filePath = RNFetchBlob.fs.dirs.DocumentDir + '/myFile';
RNFetchBlob.config({ fileCache: true, path : filePath })
.fetch('GET', url)
.then(resp => resp.readFile('base64')
.then(base64 => ({ resp, base64 })))
.then(obj => {
const headers = obj.resp.respInfo.headers;
const type = headers['Content-Type'];
const dataUrl = 'data:' + type + ';base64,' + obj.base64;
return { url: dataUrl, title, message };
})
.then(options => Share.open(options)).catch(err => {err &&
console.log(err); })
}
Then you can call unlink like this
RNFetchBlob.fs.unlink(filePath)
.then(() => { ... })
.catch((err) => { ... })