如何防止反应本机中的 Algolia 即时搜索初始加载?
How to prevent initial load from Algolia instant search in react native?
我有一个用户集合,我想在用户输入时查询它,但是当我切换到“搜索用户”屏幕时,即使搜索为空,所有用户都已填充在屏幕上。
如何防止这种情况发生并仅在 he/she 在搜索框中输入时才将结果发送给客户端??
我找到了一些其他链接和资源,但它们要么太旧,对我来说难以理解,要么使用 React
而不是 React Native
。
我是一名初级开发人员,如果您能详细说明解决方案,那就太好了。
这是我的代码:
SEARCH_USER.JS:
import React, {Component} from 'react';
import {
View,
Text,
TextInput,
Alert,
FlatList,
ActivityIndicator,
} from 'react-native';
import firestore from '@react-native-firebase/firestore';
import algoliasearch from 'algoliasearch';
import {
InstantSearch,
connectSearchBox,
connectInfiniteHits,
} from 'react-instantsearch-native';
import PropTypes from 'prop-types';
import InfiniteHits from '../components/Algolia Components/InfiniteHits';
class SearchAddBuddyScreen extends Component {
constructor(props) {
super(props);
this.state = {
loading: false,
value: null,
};
}
searchClient = algoliasearch(
'########',
'####################',
);
render() {
return (
<View>
<SecondaryHeader secondaryHeaderTitle="Add Buddies" />
<InstantSearch
searchClient={this.searchClient}
indexName="prod_USERSCOLLECTION"
onSearchStateChange={searchState =>
console.log('=====> ', searchState)
}>
<ConnectedSearchBox />
<InfiniteHits navigation={this.props.navigation} />
</InstantSearch>
</View>
);
}
}
class SearchBox extends Component {
render() {
console.log('Connected Search Box called');
return (
<View
style={[
textInput.generalTextInput,
{
marginBottom: 24,
alignSelf: 'center',
justifyContent: 'center',
},
]}>
<TextInput
placeholderTextColor="#333647"
style={[textInput.generalTextInput, {alignSelf: 'center'}]}
onChangeText={text => this.props.refine(text)}
value={this.props.currentRefinement}
placeholder={'Search'}
clearButtonMode={'always'}
spellCheck={false}
autoCorrect={false}
autoCapitalize={'none'}
/>
</View>
);
}
}
SearchBox.propTypes = {
refine: PropTypes.func.isRequired,
currentRefinement: PropTypes.string,
};
const ConnectedSearchBox = connectSearchBox(SearchBox);
export default SearchUserScreen;
InfiniteHits.js:
import React, {Component} from 'react';
import {StyleSheet, Text, View, FlatList} from 'react-native';
import {connectInfiniteHits} from 'react-instantsearch-native';
import AddFriendComponent from '../AddFriend';
class InfiniteHits extends Component {
constructor(props) {
super(props);
this.navigation = this.props.navigation;
}
_renderItem = ({item}) => {
return (
<AddFriendComponent
username={item.username}
fullName={item.fullName}
onPressUserProfile={() =>
this.props.navigation.navigate('UserProfile', {profileId: item.uid})
}
/>
);
};
render() {
return (
<FlatList
data={this.props.hits}
keyExtractor={item => item.objectID}
// ItemSeparatorComponent={() => <View style={styles.separator} />}
onEndReached={() => this.props.hasMore && this.props.refine()}
renderItem={this._renderItem}
/>
);
}
}
export default connectInfiniteHits(InfiniteHits);
阅读更多内容后 material 我发现了这个 https://www.algolia.com/doc/guides/building-search-ui/going-further/conditional-requests/react/
并对我的代码进行了以下更改并成功。
conditionalQuery = {
search(requests) {
if (
requests.every(({params}) => !params.query) ||
requests.every(({params}) => params.query === ' ') ||
requests.every(({params}) => params.query === ' ') ||
requests.every(({params}) => params.query === ' ')
) {
// Here we have to do something else
console.log('Empty Query');
return Promise.resolve({
results: requests.map(() => ({
hits: [],
nbHits: 0,
nbPages: 0,
processingTimeMS: 0,
})),
});
}
const searchClient = algoliasearch(
'########',
'###################',
);
return searchClient.search(requests);
},
};
render() {
return (
<View>
<SecondaryHeader secondaryHeaderTitle="Add Buddies" />
<InstantSearch
searchClient={this.conditionalQuery}
indexName="prod_USERSCOLLECTION"
onSearchStateChange={searchState =>
console.log('=====> ', searchState)
}>
<ConnectedSearchBox />
<InfiniteHits navigation={this.props.navigation} />
</InstantSearch>
</View>
);
}
此解决方案仍可针对键入的空格进行改进,但我不知道该怎么做。
我有一个用户集合,我想在用户输入时查询它,但是当我切换到“搜索用户”屏幕时,即使搜索为空,所有用户都已填充在屏幕上。
如何防止这种情况发生并仅在 he/she 在搜索框中输入时才将结果发送给客户端??
我找到了一些其他链接和资源,但它们要么太旧,对我来说难以理解,要么使用 React
而不是 React Native
。
我是一名初级开发人员,如果您能详细说明解决方案,那就太好了。
这是我的代码:
SEARCH_USER.JS:
import React, {Component} from 'react';
import {
View,
Text,
TextInput,
Alert,
FlatList,
ActivityIndicator,
} from 'react-native';
import firestore from '@react-native-firebase/firestore';
import algoliasearch from 'algoliasearch';
import {
InstantSearch,
connectSearchBox,
connectInfiniteHits,
} from 'react-instantsearch-native';
import PropTypes from 'prop-types';
import InfiniteHits from '../components/Algolia Components/InfiniteHits';
class SearchAddBuddyScreen extends Component {
constructor(props) {
super(props);
this.state = {
loading: false,
value: null,
};
}
searchClient = algoliasearch(
'########',
'####################',
);
render() {
return (
<View>
<SecondaryHeader secondaryHeaderTitle="Add Buddies" />
<InstantSearch
searchClient={this.searchClient}
indexName="prod_USERSCOLLECTION"
onSearchStateChange={searchState =>
console.log('=====> ', searchState)
}>
<ConnectedSearchBox />
<InfiniteHits navigation={this.props.navigation} />
</InstantSearch>
</View>
);
}
}
class SearchBox extends Component {
render() {
console.log('Connected Search Box called');
return (
<View
style={[
textInput.generalTextInput,
{
marginBottom: 24,
alignSelf: 'center',
justifyContent: 'center',
},
]}>
<TextInput
placeholderTextColor="#333647"
style={[textInput.generalTextInput, {alignSelf: 'center'}]}
onChangeText={text => this.props.refine(text)}
value={this.props.currentRefinement}
placeholder={'Search'}
clearButtonMode={'always'}
spellCheck={false}
autoCorrect={false}
autoCapitalize={'none'}
/>
</View>
);
}
}
SearchBox.propTypes = {
refine: PropTypes.func.isRequired,
currentRefinement: PropTypes.string,
};
const ConnectedSearchBox = connectSearchBox(SearchBox);
export default SearchUserScreen;
InfiniteHits.js:
import React, {Component} from 'react';
import {StyleSheet, Text, View, FlatList} from 'react-native';
import {connectInfiniteHits} from 'react-instantsearch-native';
import AddFriendComponent from '../AddFriend';
class InfiniteHits extends Component {
constructor(props) {
super(props);
this.navigation = this.props.navigation;
}
_renderItem = ({item}) => {
return (
<AddFriendComponent
username={item.username}
fullName={item.fullName}
onPressUserProfile={() =>
this.props.navigation.navigate('UserProfile', {profileId: item.uid})
}
/>
);
};
render() {
return (
<FlatList
data={this.props.hits}
keyExtractor={item => item.objectID}
// ItemSeparatorComponent={() => <View style={styles.separator} />}
onEndReached={() => this.props.hasMore && this.props.refine()}
renderItem={this._renderItem}
/>
);
}
}
export default connectInfiniteHits(InfiniteHits);
阅读更多内容后 material 我发现了这个 https://www.algolia.com/doc/guides/building-search-ui/going-further/conditional-requests/react/
并对我的代码进行了以下更改并成功。
conditionalQuery = {
search(requests) {
if (
requests.every(({params}) => !params.query) ||
requests.every(({params}) => params.query === ' ') ||
requests.every(({params}) => params.query === ' ') ||
requests.every(({params}) => params.query === ' ')
) {
// Here we have to do something else
console.log('Empty Query');
return Promise.resolve({
results: requests.map(() => ({
hits: [],
nbHits: 0,
nbPages: 0,
processingTimeMS: 0,
})),
});
}
const searchClient = algoliasearch(
'########',
'###################',
);
return searchClient.search(requests);
},
};
render() {
return (
<View>
<SecondaryHeader secondaryHeaderTitle="Add Buddies" />
<InstantSearch
searchClient={this.conditionalQuery}
indexName="prod_USERSCOLLECTION"
onSearchStateChange={searchState =>
console.log('=====> ', searchState)
}>
<ConnectedSearchBox />
<InfiniteHits navigation={this.props.navigation} />
</InstantSearch>
</View>
);
}
此解决方案仍可针对键入的空格进行改进,但我不知道该怎么做。