Question

我有一个PHP生成的XML文件，我正在尝试使用Javascript。我的主要错误是我接触到Javascript它说“Uncaught TypeError：无法调用方法'getElementsByTagName'为null”我现在很多东西都是在线尝试复制/粘贴几个教程。我将包括三个块，首先在我的PHP脚本中，第二个是它输出的内容，第三个是javascript（这是我真的不知道的部分）。该项目的目标是将数据库表放入Javascript小程序。

if($id == "poll"){
//Need to report back everything
$check = mysql_query("SELECT * FROM `lightstatus`");

$xml = new XMLWriter();

$xml->openURI("php://output");
$xml->startDocument();
$xml->setIndent(true);

$xml->startElement('doc');

while ($row = mysql_fetch_assoc($check)) {
$xml->startElement("setting"); 
 $xml->startElement("id");
  $xml->writeRaw($row['id']);
  $xml->endElement();
  $xml->startElement("status");
  $xml->writeRaw($row['status']);
  $xml->endElement();
$xml->endElement();
}

$xml->endElement();

header('Content-type: text/xml');
$xml->flush();

}

^^这很好用

<?xml version="1.0"?>
<doc>
 <setting>
  <id>1</id>
  <status>0</status>
 </setting>
 <setting>
  <id>2</id>
  <status>0</status>
 </setting>
 <setting>
  <id>3</id>
  <status>0</status>
 </setting>
 <setting>
  <id>4</id>
  <status>0</status>
 </setting>
 <setting>
  <id>5</id>
  <status>0</status>
 </setting>
 <setting>
  <id>6</id>
  <status>0</status>
 </setting>
 <setting>
  <id>7</id>
  <status>0</status>
 </setting>
 <setting>
  <id>8</id>
  <status>1</status>
 </setting>
 <setting>
  <id>9</id>
  <status>0</status>
 </setting>
 <setting>
  <id>10</id>
  <status>0</status>
 </setting>
 <setting>
  <id>11</id>
  <status>0</status>
 </setting>
 <setting>
  <id>12</id>
  <status>0</status>
 </setting>
 <setting>
  <id>13</id>
  <status>0</status>
 </setting>
 <setting>
  <id>14</id>
  <status>0</status>
 </setting>
 <setting>
  <id>15</id>
  <status>0</status>
 </setting>
 <setting>
  <id>16</id>
  <status>0</status>
 </setting>
 <setting>
  <id>17</id>
  <status>0</status>
 </setting>
 <setting>
  <id>18</id>
  <status>0</status>
 </setting>
 <setting>
  <id>19</id>
  <status>0</status>
 </setting>
 <setting>
  <id>20</id>
  <status>0</status>
 </setting>
 <setting>
  <id>21</id>
  <status>0</status>
 </setting>
 <setting>
  <id>22</id>
  <status>0</status>
 </setting>
 <setting>
  <id>47</id>
  <status>1</status>
 </setting>
</doc>

^^^看起来很好（我认为）

<html>
<head>
<body>

<script>
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.open("GET",'status.php?id=poll',false);
xmlhttp.send();
xmlDoc=xmlhttp.responseXML; 
//alert(responseXML);
document.write("<table border='1'>");
var x=xmlDoc.getElementsByTagName("setting");

for (i=0;i<x.length;i++)
  { 
  document.write("<tr><td>");
  document.write(x[i].getElementsByTagName("id")[0].childNodes[0].nodeValue);
  document.write("</td><td>");
  document.write(x[i].getElementsByTagName("status")[0].childNodes[0].nodeValue);
  document.write("</td></tr>");
  }
document.write("</table>");
</script>

</body>
</html>

^^^只是不确定，Javascript是我的弱点

同样，问题出在我认为的Javascript中，错误显示在Chrome的开发者视图中。

javascript和PHP，xmlhttp＆amp; XMLwritter“Uncaught TypeError：无法调用方法'getElementsByTagName'为null”

0 个答案: