Question

我有以下对象，每次我调用object [“ key”]时，由于某种原因它返回undefined

我尝试了以下操作，但仍未定义

<SectionList
  renderItem={({item, index, section}) =>
  // render the entire item in the `renderItem` call with one `TouchableOpacity`
  <TouchableOpacity onPress={ () => this.onPressOrder(item.id) }>
    <View>
      <Text>
      {item.inspector}
      </Text>
    </View>
     <View>
      <Text>
      {item.inspection_date}
      </Text>
    </View>
    ... etc.
  </TouchableOpacity>
}
  renderSectionHeader={({section: {title}}) => (
    <Text style={{fontWeight: 'bold'}}>{title}</Text>
  )}
  sections={this.state.dataSource}
  keyExtractor={item => item.id}
/>

dataSource.push({
  // each section just has one list item that is the full list item
  data: [
    listItem,
  ],
  key: keyId,
  title: listItem.address
  })

所有人都返回未定义

object.keyName

object[keyName]

object["keyName"]

让我说我想要placeName值我尝试过

let sam = {
  "facebook": "https://facebook.com",
  "instagram": "https://Instagram.com",
  "placeName": "dazzlement",
  "snapchat": "https://snapchat.com",
  "twitter": "https://twitter.com",
}

全部返回未定义。

--------更新----------

我在Parse JS SDK中使用本机反应，这是用于从Parse数据库获取值并将其作为道具发送到组件的代码，该组件将结果显示为卡片

sam.placeName
sam[placeName]
sam["placeName"]

在组件中是代码

this.state = {
    items: null,
    loaded: false
};



componentDidMount() {
    this.queryValues()
}

queryValues = async id => {

    let cy = this;
    const GameScore = Parse.Object.extend('social');
    const query = new Parse.Query(GameScore);
    query.equalTo('A', 1);
    query.equalTo('B', 2);
    const results = await query.find().then(
        results => {
            cy.setState({
                items: results
                loaded: true,
            })
        },
        error => {}
    );

};

render() {
            return (

                {
                    this.state.loaded ?
                    <FlatList
                    data = {this.state.items}
                    renderItem = {
                        ({item}) => <SocialCard social={item} /> } /
                        >
                        : <
Text> Loading </Text>
}


                    );
                }
            }

日志显示如下

render() {
    const {
      social
    } = this.props;
    console.log(social)
    return (
      <Text>{this.props.}</Text>

    );
  }

Answer 1

这应该返回所有链接

let sam = {
    facebook: 'facebook.com', // you don't need " " for facebook
    instagram: 'instagram.com'
}

for (let i in sam){
    console.log(sam[i])
}

sam.facebook // facebook.com
sam.instagram // instagram.com

按键返回的对象值未定义

1 个答案: