Reputation: 345
Here response data contains{name ,details} fields. I have a upload button to upload single/multi files. After uploading files, I don't refresh the page. Now for ex, I upload single file and displayed the response. Again I upload 2 files. Now I should be able to display response of current upload as well as previous one. It should be in the following format: 1.Name1 Details1 2. Name2 Details2 3. Name3 Details 3
this.state.list = this.state.list.concat(details);
this.state.list_name = this.state.list_name.concat(name);
< form onSubmit={this.handleSubmit} >
<label>
Upload a file: <br /><br />
<input type="file" name="file" multiple onChange{this.onChangeHandler}/>
</label>
<br /><br />
<button type="submit">
Upload</button>
</form >
<ol>
// {this.state.list_name.map((k) =>
// <li>{k} :</li>)}
//{this.state.list.map((k) =>
// <li>{k}</li>
// )}
</ol>
handleSubmit = (event) => {
event.preventDefault();
const formData = new FormData();
for (var x = 0; x < this.state.selectedFile.length; x++) {
formData.append('inputFile', this.state.selectedFile[x])
fetch('url', {
method: 'POST',
body: formData
}).then(res => {
return res.json()
})
.then((res) => {
this.setState({
details: res.data.details,
name: res.data.name
})
console.log("res data", res)
//console.log("Data is", res.data.name) //displays name in the console
})
.catch(error => console.log('ERROR message'))
}
};
I have commented the code that I have tried. Thanks in advance.
Upvotes: 0
Views: 72
Reputation: 325
you can do this
remove this
this.state.list = this.state.list.concat(details);
this.state.list_name = this.state.list_name.concat(name);
add this
fetch('url', {
method: 'POST',
body: formData
}).then(res => {
return res.json()
})
.then((res) => {
const list = this.state.list;
list.concat(res.data.details);
const list_name = this.state.list_name;
list_name.concat(name);
this.setState({
list,
list_name,
details: res.data.details,
name: res.data.name
})
console.log("res data", res)
//console.log("Data is", res.data.name) //displays name in the console
})
.catch(error => console.log('ERROR message'))
}
Upvotes: 1