javascript - 解析错误 JSON.parse(意外的标记 u)

Question

var storage = chrome.storage.local;
var cachedStorage = {};

这是 js 文件。它显示了意外的标记 u。即使我已经正确完成了解析。并且它还显示了其 html 源页面的意外标记。任何人都可以建议我如何解决这个问题。

<小时/>

var defaultStorage = [{
  savedPatterns: JSON.stringify([
    [{
      "en": "English"
    }, {
      "it": "Italian"
    }, "25", true],
    [{
      "en": "English"
    }, {
      "la": "Latin"
    }, "15", false]
  ]),

}];

error occurs here unexpected token u

function createPattern() {
  console.log('createPattern begin');
  var patterns = JSON.parse(S('savedPatterns'));
  var srce = [],
    trg = [],
    prb = [];
  console.log(S('savedPatterns'));
  console.debug(S('savedPatterns'));
  var translator = document.getElementById('translatorService');
  var service = translator.children[translator.selectedIndex].value;
  srce[0] = document.getElementById('sourceLanguage');
  srce[1] = srce[0].children[srce[0].selectedIndex].value;
  srce[2] = srce[0].children[srce[0].selectedIndex].text;
  trg[0] = document.getElementById('targetLanguage');
  trg[1] = trg[0].children[trg[0].selectedIndex].value;
  trg[2] = trg[0].children[trg[0].selectedIndex].text;
  prb[0] = document.getElementById('translationProbability');
  prb[1] = prb[0].children[prb[0].selectedIndex].value;

  patterns.push([
    [srce[1], srce[2]],
    [trg[1], trg[2]],
    prb[1],
    false,
    service
  ]);
  saveBulk({
    'savedPatterns': JSON.stringify(patterns)
  }, 'Saved Pattern');
  console.log('createPattern end');
}

function S(key) {
  return cachedStorage[key];
}

function loadStorageAndUpdate(callback) {
  storage.get(null, function(data) {
    console.log('data: ' + data + ' : ' + JSON.stringify(data));
    var d = {};
    if (!data || JSON.stringify(data) == '{}') { // in this case, storage was not initialized yet
      console.log('setting storage to defaultStorage (stringified): ');
      console.log(JSON.stringify(defaultStorage));
      storage.set(defaultStorage);
      d = defaultStorage;
    } else {
      d = data;
    }

    cachedStorage = d;
    if (!!callback) {
      callback(d);
    }
  });
}

Answer 1

当 JSON.parse 失败时，会出现

错误 意外的标记，并且根据字符(本例中为 u)，您可以推测其原因。

• u 表示值未定义
• o 表示值是否为对象

try {
  JSON.parse(undefined)
} catch (ex) {
  document.write(ex.message + "<br/>")
}

try {
  JSON.parse({})
} catch (ex) {
  document.write(ex.message)
}

你可以尝试这样的事情:

function s(key) {
  var obj = {
    foo: "foo",
    bar: "bar"
  }
  var v = null;

  try {
    v = JSON.parse(obj[key]);
  } catch () {
    v = obj[key];
  }

  return v;
}

function main() {
  var v = s("foo");
}

请参阅以下帖子了解更多信息。 Uncaught SyntaxError: Unexpected token with JSON.parse