JSON请求返回'undefined'对于所有键/值对


JSON request returns 'undefined' for all key/value pairs

我有一个PHP文件,将谷歌电子表格解析为JSON:

<?php
header('Content-type: application/json');
// Set your CSV feed
$feed = 'https://docs.google.com/spreadsheets/d/1IwFDZp1dLBR931OnJpEgW9ojrLx063m5_dhkqKGgugc/pub?output=csv';
// Arrays we'll use later
$keys = array();
$newArray = array();
// Function to convert CSV into associative array
function csvToArray($file, $delimiter) { 
  if (($handle = fopen($file, 'r')) !== FALSE) { 
    $i = 0; 
    while (($lineArray = fgetcsv($handle, 4000, $delimiter, '"')) !== FALSE) { 
      for ($j = 0; $j < count($lineArray); $j++) { 
        $arr[$i][$j] = $lineArray[$j]; 
      } 
      $i++; 
    } 
    fclose($handle); 
  } 
  return $arr; 
} 
// Do it
$data = csvToArray($feed, ',');
// Set number of elements (minus 1 because we shift off the first row)
$count = count($data) - 1;
//Use first row for names  
$labels = array_shift($data);  
foreach ($labels as $label) {
  $keys[] = $label;
}
// Add Ids, just in case we want them later
$keys[] = 'id';
for ($i = 0; $i < $count; $i++) {
  $data[$i][] = $i;
}
// Bring it all together
for ($j = 0; $j < $count; $j++) {
  $d = array_combine($keys, $data[$j]);
  $newArray[$j] = $d;
}
// Print it out as JSON
echo json_encode($newArray);
?>

在我的HTML中,一些代码应该把JSON渲染成HTML表格:

<body>
    <div id="id01"></div>
    <script>
        var xmlhttp = new XMLHttpRequest();
        var url = "votos.json.php";
        xmlhttp.onreadystatechange=function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                myFunction(xmlhttp.responseText);
            }
        }
        xmlhttp.open("GET", url, true);
        xmlhttp.send();
        function myFunction(response) {
            var arr = JSON.parse(response);
            var i;
            var out = "<table>";
            for(i = 0; i < arr.length; i++) {
                out += "<tr><td>" +
                arr[i].Name +
                "</td><td>" +
                arr[i].City +
                "</td><td>" +
                arr[i].Country +
                "</td></tr>";
            }
            out += "</table>";
            document.getElementById("id01").innerHTML = out;
        }
    </script>
</body>

访问index.html将显示一个由'undefined'填充的表。但是,将URL从'votos.json.php'更改为其他URL,例如,从W3: [http://www.w3schools.com/website/customers_mysql.php]中提取的URL,表将被正确的数据填充并且显示得很好。我已经验证了PHP文件的JSON输出,它似乎没有任何问题——那么是什么导致了这个问题,该如何修复呢?

从该CSV文件引用了不正确的列名。试一试…

for (i = 0; i < arr.length; i++){
    out += "<tr><td>" +
    arr[i].candidato +
    "</td><td>" +
    arr[i].votos +
    "</td><td>" +
    arr[i].porcentaje +
    "</td></tr>";
}