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Javascript 响应本机can';拿不到_Javascript_Android_Reactjs_React Native - Fatal编程技术网

Javascript 响应本机can';拿不到

Javascript 响应本机can';拿不到,javascript,android,reactjs,react-native,Javascript,Android,Reactjs,React Native,当我在React Native(0.20.0)中获取以下url时: 获取代码: API.prototype.GET = function(url, callback) { console.log('GET>>:' + url); return fetch(url) .then(parseJSON) .then((responseData) => { if (responseData != null) { if (responseDat

当我在React Native(0.20.0)中获取以下url时:

获取代码:

API.prototype.GET = function(url, callback) {
  console.log('GET>>:' + url);
  return fetch(url)
  .then(parseJSON)
  .then((responseData) => {
      if (responseData != null) {
        if (responseData.code != 200) {
          RNManager.showMessage(responseData.message);
          return;
        }
      }
    return callback(responseData);
  })
  .catch((error) => {
    console.warn(error);
  })
  .done();
}
报告说:

(标准输入):145715:W/ReactNativeJS(13624):[TypeError:网络请求失败]

但是这个Url可以在chrome中加载。我认为是490 | 1491 | 1485 | 1的格式导致了这个问题。但如何解决这个问题呢

[已解决]问题是fetch()没有为我执行URL\u编码,所以我自己将“|”更改为“%7C”(URL\u编码),它成功了

[Solved Update]使用
encodeURI()
完成这些技巧

API.prototype.GET = function(url, callback) {
  console.log('GET>>:' + url);
  return fetch(url)
  .then(parseJSON)
  .then((responseData) => {
      if (responseData != null) {
        if (responseData.code != 200) {
          RNManager.showMessage(responseData.message);
          return;
        }
      }
    return callback(responseData);
  })
  .catch((error) => {
    console.warn(error);
  })
  .done();
}