小编典典

为什么我总是从Chrome收到“未捕获的SyntaxError:意外的令牌u”?

json

我花了整整一整天的时间,在Google上搜索并寻找答案,但仍然不知道。

我的代码有点长,可以在Firefox中正常运行,但是从Chrome获得“ Uncaught SyntaxError:Unexpected token u”。

谁能指出我做错了什么?提前致谢!

// when the page loads, list all the current contacts
$(document).ready(function(){

    // check if localStorage database exists
    if(!localStorage.getItem("customerDatabase")){

        // define a JSON object to hold all current address
        var contacts = {
            "users":[
                {
                    "id":"1",
                    "name":"dennis",
                    "email":"dennisboys@gmail.com"
                },
                {
                    "id":"2",
                    "name":"zoe",
                    "email":"zoeisfemale@gmail.com"             
                }
            ]   
        } // end of contacts JSON object

        // stringify the object 
        var stringObject = JSON.stringify(contacts);

        // store it into localStorage database
        var storedDatabase = localStorage.setItem("customerDatabase", stringObject);

    } else {
        // list all customers upon page loads
        listJSONCustomers();        
    }

    // list all current contacts from JSON object in localStorage
    function listJSONCustomers(){

      var displayHTML = "";
      var i;

      // get the data from localStorage
      var storedDatabase = localStorage.getItem("customerDatabase");

      // parse the data from string to JSON object
      var parseObject = JSON.parse(storedDatabase);

      // access the users key of the JSON object
      var userObject = parseObject.users;

      // get the length of the object (how many customers the database has)
      var contactsLength = userObject.length;

      for(i=0; i<contactsLength; i++){
          var trElement = '<tr id="address' + (i+1) + '">';
          var tdId = '<td id="id' + (i+1) + '">' + userObject[i].id + '</td>';
          var tdName = '<td id="name' + (i+1) + '">' + userObject[i].name + '</td>';
          var tdEmail = '<td id="email' + (i+1) + '">' + userObject[i].email + '</td>';
          var tdButton = '<td id="button"><button id="editButton' + userObject[i].id + '">Edit</button> | <button id="deleteButton' + userObject[i].id + '">Delete</button></td>';

          displayHTML += trElement + tdId + tdName + tdEmail + tdButton + '</tr>';
      }

      $('#address_list').html(displayHTML);           
    }

    // add customer to database  
    $('#saveCustomer').click(function(){

       if( $('#customerName').val() !== "" && $('#customerEmail').val() !== "" ){

           var customerName = $('#customerName').val();
           var customerEmail = $('#customerEmail').val();

           // get the data from localStorage
           var storedDatabase = localStorage.getItem("customerDatabase");

           // parse the data from string to JSON object
           var parseObject = JSON.parse(storedDatabase);

           // access the users key of the JSON object
           var userObject = parseObject.users;

           // get the new entry
           var newCustomerObject = {
                                  "id": userObject.length + 1,
                                  "name": customerName,
                                  "email": customerEmail
                                  };

           // push the new entry into the object                                                            
           userObject.push(newCustomerObject);

           // convert the object into string for localStorage
           var stringObject = JSON.stringify(parseObject);

           // store the JSON object into localStorage
           var storedDatabase = localStorage.setItem("customerDatabase", stringObject);

           // list all customes again every time a database receives a new entry
           listJSONCustomers();

       } else {
          alert("Please enter customer's name and email.");  
       }

    }); // end of $('#saveCustomer').click();


});

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2020-07-27

共1个答案

小编典典

在某些时候,您确实破坏了该密钥的LocalStorage值。LocalStorage只能存储字符串,因此,如果将其他任何内容传递给它,它将把它转换为字符串。由于您的值为'undefined',这意味着在某个时候,您可能偶然地做了这样的事情:

var value;
localStorage.setItem('key', value);

在这种情况下,valueis
undefined,不是字符串。保存后,将对其进行转换。不幸的是,"undefined"不是有效的JSON。这意味着,当它尝试解析时,它将引发异常。

要解决您的问题,您应该使用清除不良价值removeItem

localStorage.removeItem("customerDatabase");
2020-07-27