Question

我有一个PHP文件，可以将Google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，有些代码应该将JSON呈现为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会在每个单元格中显示由“undefined”填充的表格。但是，将URL从'votos.json.php'更改为其他一些，比如说，例如，这个从W3中拉出来：[http://www.w3schools.com/website/customers_mysql.php]，该表由正确的数据填充并很好地显示。我已经验证了PHP文件的JSON输出，它似乎没有任何问题 - 所以可能导致这个问题以及如何修复它？

Answer 1

您正在引用该CSV文件中的错误列名称。尝试...

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>";
}

JSON请求为所有键/值对返回'undefined'

1 个答案: