仍在努力將JSON對象與url param一起傳遞給php文件。任何想法如何？

Question

  // Creating variables to hold data from textboxes. First building associated details
  myData.Token = $("#token").val();
  myData.BuildingDisplay = $("#BuildingDisplay").val();
  myData.FeatureID = $("#FeatureID").val();
  myData.BuildingID = $("#BuildingID").val();
  myData.Address = $("#Address").val();
  myData.City = $("#City").val();
  myData.District = $("#District").val();
  myData.Location = $("#Location").val();
  myData.State = $("#State").val();
  myData.StreetName = $("#StreetName").val();
  myData.Zip = $("#Zip").val();
  myData.X = $("#X").val();
  myData.Y = $("#Y").val();
$.ajax({
  type: "POST",
  url: "proxyCheck.php",
  data: "{ 'Token': '" + mytoken + "','Address': '" + myaddress + "', 'City': '" + mycity + "','Location': '" + mylocation + "','State': '" + mystate + "', 'StreetName': '" + mystreetname + "', 'Zip': '" + myzip + "', 'X': '" + myX + "', 'Y': '" + myY + "'}",
  contentType: "application/json;charset=utf-8",
  dataType: "json",
  async: false,
  success: function (response) {
    alert("Record has been added successfully.");
    window.location.reload();
  }
});

當以上數據格式化為JSON對象時，它以以下格式顯示：

{data: 
      {
        "token":"73264280-be3f-4f5b",
        "BuildingDisplay":"Fire Station 21 Fairburn - 19 East Broad St.",
        "CallerEmail":"Jim.Parker@yahoo.com",
        "CallerFax":"",
        "CallerFirstName":"Jim",
        "CallerLastName":"Parker",
        "CallerMiddleInitial":"",
        "CallerOtherPhone":"",
        "CallerState":"",
        "CallerWorkPhone":"918-354-2874"}}

我想做的是將它們傳遞給proxyCheck.php，如下所示：

 proxyCheck.php?data={data: 
      {
        "token":"73264280-be3f-4f5b",
        "BuildingDisplay":"Fire Station 21 Fairburn - 19 East Broad St.",
        "CallerEmail":"Jim.Parker@yahoo.com",
        "CallerFax":"",
        "CallerFirstName":"Jim",
        "CallerLastName":"Parker",
        "CallerMiddleInitial":"",
        "CallerOtherPhone":"",
        "CallerState":"",
        "CallerWorkPhone":"918-354-2874"}}&token=73264280-be3f-4f5b

請注意＆token = 73264280-be3f-4f5b

我似乎無法弄清楚如何做到這一點。

有人可以幫忙嗎？

$.ajax({
    type: "POST",
    url: "proxyCheck.php?token=" + myData.Token,
     data: JSON.stringify(myData)
    contentType: "application/json;charset=utf-8",
    dataType: "json",
    async: false,
    success: function (response) {
        alert("Record has been added successfully.");
        window.location.reload();
    }
});
  return false;
 }

//proxyCheck.php

<?php
  $ch = curl_init("http://domain/UserServices/Create");
  curl_setopt($ch, CURLOPT_HEADER, 0);
  curl_setopt($ch, CURLOPT_POST, 1);
  curl_setopt ($ch, CURLOPT_POSTFIELDS, $_POST);
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
  $output = curl_exec($ch);
  curl_close($ch);
  echo $output
 ?>

// * ** * ** * ** * ** * ** * ** *

// AJAX

var uname = $("#user").val();
var upass = $("#pass").val();

$.post("proxyValidate.php",
  { data: JSON.stringify({ LoginName: uname,Password: upass }) })
    .done(function(data) {
        var result = JSON.parse(data);
        switch(result.Status) {
            case 0:
                //login successful
                tokenVal = result.Value.Token;
                location.href = "http://domain/userService.php?token="+tokenVal;
                break;

            case 2:
                //invalid login
                alert(result.Message);
                break;
        }
    })
    .fail(function() {
        alert("The AJAX request failed!");
    });
  });

// proxyValidate

  <?php

    $ch = curl_init("http://domain/Validation/Validate");
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt ($ch, CURLOPT_POSTFIELDS, $_POST);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    $output = curl_exec($ch);
    curl_close($ch);
    echo $output
  ?>

Answer 1

沒有必要將javascript對象轉換為字符串--jQuery將為您執行此操作。 嘗試這個...

$.ajax({
    type: "POST",
    url: "proxyCheck.php?token=" + myData.Token,
    data: myData,
    dataType: "json",
    async: false,
    success: function (response) {
        alert("Record has been added successfully.");
        window.location.reload();
    }
});

變量名稱區分大小寫（例如，令牌和令牌），但上面的代碼應該發布您需要的內容，並在URL中包含令牌值。

或者，如果您需要將數據作為字符串傳遞，那么您可以執行以下操作...

$.ajax({
    type: "POST",
    url: "proxyCheck.php?token=" + myData.Token,
    data: {
        data: JSON.stringify(myData)
    },
    dataType: "json",
    async: false,
    success: function (response) {
        alert("Record has been added successfully.");
        window.location.reload();
    }
});

Answer 2

對於任何可能遇到與我類似問題的人來說，這是適用於我的解決方案：

$.ajax({
    type: "POST",
    url: "proxyCheck.php?token",
    data: {
        data: JSON.stringify(myData),
        token: myData.Token
    },
    dataType: "json",
    async: false,
    success: function (response) {
        alert("Record has been added successfully.");
        window.location.reload();
    }
});

Answer 3

你真的應該使用POST來獲得那種數據。 如果你堅持按照你的建議使用GET，你應該對JSON進行urlencode，但是，你可能很容易發現你的url太長而無法打開。

編輯：@Archer應該有效的解決方案

$.ajax({
    type: "POST",
    url: "proxyCheck.php?token=" + myData.Token,
    data: {
        data: JSON.stringify(myData)
    },
    dataType: "json",
    async: false,
    success: function (response) {
        alert("Record has been added successfully.");
        window.location.reload();
    }
});