获取错误说TypeError undefined不是对象

时间:2018-06-04 21:32:01

标签: reactjs react-native react-android

我正在尝试在Listview中显示数据。首先,我根据用户在上一个屏幕上做出的选择来过滤数据(JSON文件)。我通过说:

将参数传递到下一个屏幕
this.props.navigation.state.params.JSON_ListView_Clicked_Item)

基于上面的参数,我在render函数中做了ListView.DataSource。唯一的原因,我试图在render函数中做ListView.DatSource,因为据我所知这是我可以读取参数值的唯一地方。当我试图显示我的listView时,我收到一条错误说:

TypeError undefined不是对象。以下是截图:

enter image description here

以下是我的代码:

import React, { Component } from 'react';

import { StyleSheet, Text, View, ListView, ActivityIndicator, TextInput } from 'react-native';

import ServiceDetails from '../reducers/ServiceDetails';


class ServiceListDetails extends Component
{




  static navigationOptions =
  {
     title: 'SecondActivity',
  };




ListViewItemSeparator = () => {
  return (
    <View
      style={{
        height: .5,
        width: "100%",
        backgroundColor: "#000",
      }}
    />
  );
}


render()
  {


      var x =   this.props.navigation.state.params.JSON_ListView_Clicked_Item ;
      var newList = ServiceDetails.filter(obj => obj.fk === this.props.navigation.state.params.JSON_ListView_Clicked_Item)

      let ds = new ListView.DataSource({rowHasChanged: (r1, r2) => r1 !== r2});
      this.setState({
          dataSource: ds.cloneWithRows(newList)

      }
    );


    return(
      <View style={styles.MainContainer}>
             <ListView

                dataSource={this.dataSource}
                renderSeparator= {this.ListViewItemSeparator}
                renderRow={(rowData) => <Text>{rowData}</Text>}
/>

      </View>
    );

  }
}

以下是我的JSON文件:

[

    {
       "id":"1",
       "fk": "1",
       "addr": "TestAddress1",
       "phone": "(888)889-9999",
       "LatL":"33.234567",
       "Long2":"117.284725",
       "Online": "x"
    },

    {
        "id":"2",
        "fk": "1",
        "addr": "TestAddress1",
        "phone": "(999)-999-9999",
        "LatL":"33.971110",
        "Long2":"117.31111",
        "Online": ""
     },
]

1 个答案:

答案 0 :(得分:0)

定义JSON文件时出错。在数组","中的最后一个对象中,不需要它是一个有效的JSON格式。

[

    {
        "id": "1",
        "fk": "1",
        "addr": "TestAddress1",
        "phone": "(888)889-9999",
        "LatL": "33.234567",
        "Long2": "117.284725",
        "Online": "x"
    },{
        "id": "2",
        "fk": "1",
        "addr": "TestAddress1",
        "phone": "(999)-999-9999",
        "LatL": "33.971110",
        "Long2": "117.31111",
        "Online": "y"
    }
]