Question

我正在使用代码推送来更新我的react-native应用。如果我进行更新，我不希望每次都将其上传到App Store。我想将其直接推送到用户设备。

它工作正常，应用程序已更新，但我正在尝试显示更新成功的通知。

在docs他们这样做：

    while($row = mysql_fetch_array($rs)){
        $name = $row['first_name'];
        $name2 = $row['first_name2'];
        $lastname = $row['last_name'];
        $lastname2 = $row['last_name2'];
        $mellicode = $row['melli_code'];
        $estekhdamnum = $row['estekhdam_num'];
        $email = $row['email'];
        $username = $row['username'];
        $password = $row['password'];

echo("<script type='text/javascript'>

var table = document.getElementById('show_member');

// Create an empty <tr> element and add it to the 1st position of the table:
var row = table.insertRow(i);
i = i+1;
// Insert new cells (<td> elements) at the 1st and 2nd position of the new <tr> element:
var cell1 = row.insertCell(0);
var cell2 = row.insertCell(1);
var cell3 = row.insertCell(2);
var cell4 = row.insertCell(3);
var cell5 = row.insertCell(4);
var cell6 = row.insertCell(5);
var cell7 = row.insertCell(6);
var cell8 = row.insertCell(7);
var cell9 = row.insertCell(8);

// Add some text to the new cells:
cell1.innerHTML = '$name';
cell2.innerHTML = '$name2';
cell3.innerHTML = '$lastname';
cell4.innerHTML = '$lastname2';
cell5.innerHTML = '$mellicode';
cell6.innerHTML = '$estekhdamnum';
cell7.innerHTML = '$email';
cell8.innerHTML = '$username';
cell9.innerHTML = '$password';

</script>"

);

}
?>

<script type="text/javascript">
var index = 0;
var table = document.getElementById("show_member");
var length = table.rows.length;
var row;
alert(length);
for(index=1;index<table.rows.length-2;index++){
    row = table.rows[index];
    row.onmouseover = function(){
    row.style.backgroundColor = 'red';
    }
    row.onmouseout = function(){
        row.style.backgroundColor = '';
    }
}
index = index+1;
table.rows[index].style.backgroundColor = 'green';


</script>

我正在尝试更新状态，如果更新成功并显示模态，并在按钮上单击隐藏它。因此我的代码如下：

codePush.sync({ updateDialog: true },
  (status) => {
      switch (status) {
          case codePush.SyncStatus.DOWNLOADING_PACKAGE:
              // Show "downloading" modal
              break;
          case codePush.SyncStatus.INSTALLING_UPDATE:
              // Hide "downloading" modal
              break;
      }
  },
  ({ receivedBytes, totalBytes, }) => {
    /* Update download modal progress */
  }
);

但问题是模态会显示几秒钟后会消失，而不会点击按钮。

当状态设置为true，然后在我几秒钟后变为false

constructor() {
        super();
        this.state = {
            modalVisible: false,
            status: ""
        }
    }

    componentDidMount() {
        let self = this;
        codePush.sync({
                updateDialog: true,
                installMode: codePush.InstallMode.IMMEDIATE
            }, (status) => {
                switch (status) {
                    case codePush.SyncStatus.UPDATE_INSTALLED:
                        self.setState({modalVisible: true});
                        break;
                }
            },
            ({receivedBytes, totalBytes,}) => {
                /* Update download modal progress */
                self.setState({status: "Downloaded " + receivedBytes + " of " + totalBytes});
            }
        );
    }

    renderUpdateNotificationModal(){
        return (
            <Modal
                animationType={"slide"}
                transparent={false}
                visible={this.state.modalVisible}
                onRequestClose={() => {alert("Modal has been closed.")}}
            >
                <View style={{marginTop: 22}}>
                    <View>
                        <Text>Updated</Text>
                        <TouchableHighlight onPress={() => {this.setState({modalVisible: false})}}>
                            <Text>OK</Text>
                        </TouchableHighlight>
                    </View>
                </View>
            </Modal>
        )
    }

    render() {
        return (
            <View style={styles.container}>
                {this.renderUpdateNotificationModal()}
                <Text style={styles.welcome}>
                    Welcome!
                </Text>
                <Text style={styles.instructions}>
                    To get started, edit index.ios.js
                </Text>
                <Text style={styles.instructions}>
                    Press Cmd+R to reload,{'\n'}
                    Cmd+D or shake for dev menu
                </Text>
                <Text style={{fontSize: 18, marginTop: 15, color: "blue", textAlign: "center"}}>{this.state.status}</Text>
            </View>
        );
    }

知道怎么解决吗？

Answer 1

在加载页面上尝试使用Codepush时，我也遇到了同样的问题，我的问题是在下一个屏幕上显示了CodePush的updateDialog。因为codePush.sync是Promise函数，所以您可以使用其“ then”功能，它将确保以正确的顺序进行作业。

如果要使用CodePush，

this demo可能会有所帮助。您可以在codePushStatusDidChange（）中相应地更改组件的状态。

使用代码推送更新成功更新状态

1 个答案: