我正在使用 react native 创建一个测试应用程序，当我做样式时对按钮没有影响。谁能告诉我我做错了什么。例如，我试图将红色置于按钮上，但它不起作用。我能做些什么来使它正确？
import React, { Component } from 'react'
import {
AppRegistry,
StyleSheet,
Text,
Button,
ScrollView,
Dimensions,
PanResponder,
Animated,
View
} from 'react-native'
import { StackNavigator } from 'react-navigation'
class Home extends Component{
static navigationOptions = {
title:'Home'
};
componentWillMount(){
this.animatedValue = new Animated.ValueXY();
this._value = {x:0 , y:0}
this.animatedValue.addListener((value) => this._value = value);
this.panResponder = PanResponder.create({
onStartShouldSetPanResponder: (evt, gestureState) => true,
onMoveShouldSetPanResponder: (evt, gestureState) => true,
onPanResponderGrant: (e, gestureState) => {
this.animatedValue.setOffset({
x:this._value.x,
y:this._value.y,
})
this.animatedValue.setValue({x:0 , …我正在尝试使用Laravel上传视频.虽然我更改控制器线时上传图像对我来说很好
echo '$file' . $file->getClientOriginalName() . '"/>';
至:
echo '<file src="uploads/' . $file->getClientOriginalName() . '"/>';
我只看到uploaded写在新页面上,但没有视频.
控制器:
<?php
namespace App\Http\Controllers;
use Illuminate\Support\Facades\Request;
class UploadController extends Controller
{
public function upload(Request $request)
{
if(Request::hasFile('file')){
echo 'Uploaded';
$file = Request::file('file');
$file->move('uploads', $file->getClientOriginalName());
echo '$file' . $file->getClientOriginalName() . '"/>';
}
}
}
路线:
Route::get('/', function () {
return view('welcome');
});
Route::post('upload', 'UploadController@upload');
视图:
<html>
<head>
<title>Laravel</title>
<link href='//fonts.googleapis.com/css?family=Lato:100' rel='stylesheet' type='text/css'>
</head>
<body>
<div class="container">
<div class="content">
<h1>File Upload</h1>
<form action="{{ URL::to('upload') …我正在创建一个可拖动的框。我可以将其拖动到屏幕上的任何位置，但我收到此错误，其中显示“您试图在一个本应不可变且已被冻结的对象上设置密钥_value”。谁能告诉我我做错了什么。
我的代码：
import React, { Component } from 'react'
import {
AppRegistry,
StyleSheet,
Text,
Button,
ScrollView,
Dimensions,
PanResponder,
Animated,
View
} from 'react-native'
import { StackNavigator } from 'react-navigation'
export default class Home extends Component{
componentWillMount(){
this.animatedValue = new Animated.ValueXY();
this.panResponder = PanResponder.create({
onStartShouldSetPanResponder: (evt, gestureState) => true,
onMoveShouldSetPanResponder: (evt, gestureState) => true,
onPanResponderGrant: (e, gestureState) => {
},
onPanResponderMove:Animated.event([
null,{dx: this.animatedValue.x , dy:this.animatedValue.y}
]),
onPanResponderRelease: (e, gestureState) => {
},
})
}
render(){
const animatedStyle …我正在使用来自react-native的Flatlist和来自react-native-elements的ListItem,
我想最初限制屏幕上加载的列表项的数量.否则它会加载我最初的所有项目.
假设我有300个列表项,但最初我只想加载10个项目,而不是300个.
我的代码:
import React, { Component } from 'react'
import {
FlatList
} from 'react-native'
import {Avatar,Tile,ListItem} from 'react-native-elements'
export default class Login extends Component{
constructor(props) {
super(props);
this.state = {
data:[],
dataSource: []
};
}
renderList(item,i){
return(
<View>
<ListItem
subtitle={
<Avatar
small
rounded
source={{uri: "https://s3.amazonaws.com/uifaces/faces/twitter/ladylexy/128.jpg"}}
/>
{<Text>{item.body}</Text>}
}
/>
<View>
)
}
render(){
return(
<View>
<List>
<FlatList
data={this.state.data}
keyExtractor={item => item.id}
renderItem ={({item,index}) => this.renderList(item,index)}
/>
</List>
</View>
)
}
}
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