JSON.parse：無法在代碼中找到錯誤

Question

我有一個從PHP接收JSON的腳本。 我必須解析數據。 不幸的是，我無法將JSON解析為一個對象並獲取不同的數組。 我試圖驗證JSON，但也無法在其中找到錯誤。

碼：

$.ajax({
  type: "POST",
  dataType: "html",
  url: "/sqldblink.php",
  data: data,
  success: function(data) {
    var recdata=data;
    console.log("Received data from listregistered:");
    console.log("Server reports:" + recdata);
    ListRegisteredResults(recdata);
  },
  error: function () {
    console.log("Failed to list!");
   }
});

function ListRegisteredResults(recdata) {
    console.log(typeof recdata);
    var data = JSON.parse(recdata);
    console.log(data);
    console.log(typeof data);
    console.log(data.Address.length);
}

輸出：

Server reports:"{\"Address\":[\"Home\",\"\",\"Home,\nHoover House 85\",\"\",\"Home\",\"\",\"\",\"\",\"\",\"\"],\"BloodGroup\":[\"o+\",\"\",\"B+\",\"B+\",\"AB+\",\"o+\",\"o-\",\"\",\"\",\"\"],\"Occupation\":[\"Vendor\",\"\",\"Carpenter\",\"Playing\",\"Nurse\",\"IT professional\",\"Engineer\",\"Doctor\",\"\",\"\"],\"Alternate\":[\"0\",\"0\",\"925\",\"0\",\"0\",\"0\",\"0\",\"0\",\"0\",\"0\"],\"Email\":[\"some@site.com\",\"\",\"jim@site.com\",\"\",\"\",\"\",\"arun@site.com\",\"\",\"\",\"\"],\"Mobile\":[\"90000006\",\"90000005\",\"90000005\",\"34344444\",\"902w0w05\",\"90002005\",\"900020w5\",\"90000005\",\"90002105\",\"90000005\"],\"Marital\":[\"Married\",\"Married\",\"Married\",\"Unmarried\",\"Unmarried\",\"Married\",\"Married\",\"Married\",\"Married\",\"Married\"],\"Gender\":[\"1\",\"2\",\"\",\"1\",\"1\",\"2\",\"2\",\"1\",\"1\",\"1\"],\"Age\":[\"28\",\"65\",\"35\",\"2\",\"25\",\"34\",\"31\",\"28\",\"60\",\"58\"],\"Name\":[\"Tracy Jim\",\"George Jose\",\"Jim G Mathew\",\"Cary jim\",\"Becky Mathew\",\"Cary Guy\",\"Arun Mose\",\"Tracy Kelly\",\"Dr Kim\",\"Steven Ludwig\"],\"HospitalID\":[\"3\",\"5\",\"6\",\"7\",\"8\",\"9\",\"10\",\"11\",\"12\",\"16\"]}"
string
{"Address":["Home","","Home,\nHoover House 85","","Home","","","","",""],"BloodGroup":["o+","","B+","B+","AB+","o+","o-","","",""],"Occupation":["Vendor","","Carpenter","Playing","Nurse","IT professional","Engineer","Doctor","",""],"Alternate":["0","0","925","0","0","0","0","0","0","0"],"Email":["some@site.com","","jim@site.com","","","","arun@site.com","","",""],"Mobile":["90000006","90000005","90000005","34344444","902w0w05","90002005","900020w5","90000005","90002105","90000005"],"Marital":["Married","Married","Married","Unmarried","Unmarried","Married","Married","Married","Married","Married"],"Gender":["1","2","","1","1","2","2","1","1","1"],"Age":["28","65","35","2","25","34","31","28","60","58"],"Name":["Tracy Jim","George Jose","Jim G Mathew","Cary jim","Becky Mathew","Cary Guy","Arun Mose","Tracy Kelly","Dr Kim","Steven Ludwig"],"HospitalID":["3","5","6","7","8","9","10","11","12","16"]}
string
userjs/main.js:347 Uncaught TypeError: Cannot read property 'length' of undefined
    at ListRegisteredResults (userjs/main.js:347)
    at Object.success (userjs/main.js:295)
    at i (jquery-3.2.1.min.js:2)
    at Object.fireWith [as resolveWith] (jquery-3.2.1.min.js:2)
    at A (jquery-3.2.1.min.js:4)
    at XMLHttpRequest.<anonymous> (jquery-3.2.1.min.js:4)

在將字符串解析為JSON之后，為什么仍將其報告為字符串？

Answer 1

從您的代碼中刪除dataType: "html"以及所有JSON.parse() ，它將起作用。 如果您沒有設置錯誤的dataTypes或json ，jQuery會自動解析數據

$.ajax({
  type: "POST",
  url: "/sqldblink.php",
  data: data,
  success: function(data) {
    var recdata=data;
    console.log("Received data from listregistered:");
    console.log("Server reports:" + recdata);
    ListRegisteredResults(recdata);
  },
  error: function () {
    console.log("Failed to list!");
   }
});

function ListRegisteredResults(recdata) {
    console.log(typeof recdata);
    var data = recdata;
    console.log(data);
    console.log(typeof data);
    console.log(data.Address.length);
}

Answer 2

您的JSON無效。 \\n在位置28正在破壞它，您需要對其進行轉義。

"Address":["Home","","Home,\\nHoover House 85"

應該

"Address":["Home","","Home,\\\\nHoover House 85"

如果您想保留該\\n 。

 let jsonString = `{"Address":["Home","","Home,\\\\nHoover House 85","","Home","","","","",""],"BloodGroup":["o+","","B+","B+","AB+","o+","o-","","",""],"Occupation":["Vendor","","Carpenter","Playing","Nurse","IT professional","Engineer","Doctor","",""],"Alternate":["0","0","925","0","0","0","0","0","0","0"],"Email":["some@site.com","","jim@site.com","","","","arun@site.com","","",""],"Mobile":["90000006","90000005","90000005","34344444","902w0w05","90002005","900020w5","90000005","90002105","90000005"],"Marital":["Married","Married","Married","Unmarried","Unmarried","Married","Married","Married","Married","Married"],"Gender":["1","2","","1","1","2","2","1","1","1"],"Age":["28","65","35","2","25","34","31","28","60","58"],"Name":["Tracy Jim","George Jose","Jim G Mathew","Cary jim","Becky Mathew","Cary Guy","Arun Mose","Tracy Kelly","Dr Kim","Steven Ludwig"],"HospitalID":["3","5","6","7","8","9","10","11","12","16"]}`; let json = JSON.parse(jsonString); console.log(json.Address); 

Answer 3

這僅是由於dataType: "html"與您的郵寄請求一起發送。 因此，您使用'\\'得到的響應，因此JSON.parse()無法處理它們。 從發布請求中刪除dataType ，然后嘗試，然后您將獲得正確的響應和所需的結果。

碼：

$.ajax({
  type: "POST",
  url: "/sqldblink.php",
  data: data,
  success: function(data) {
    var recdata=data;
    console.log("Received data from listregistered:");
    console.log("Server reports:" + recdata);
    ListRegisteredResults(recdata);
  },
  error: function () {
    console.log("Failed to list!");
   }
});

function ListRegisteredResults(recdata) {
    console.log(typeof recdata);
    var data = JSON.parse(recdata);
    console.log(data);
    console.log(typeof data);
    console.log(data.Address.length);
}