我正在建立一个web API,我相信这可以更有效地完成,但这是v0.1。第一步是访问localhost/serverList/api/rest.php?action=allServers&format=xml
。这开始下面的链条。我已经删除了代码中不相关的部分,所以这个问题的更短
serverList/api/rest.php
<?php
include 'inc/restFunctions.php';
//several lines of code removed. $functionName = allserversxml
if(in_array($action,$possibleActions)){
if(in_array($format,$possibleFormats)){
$functionName = $action.$format;
$result = $functionName();
header('Content-type: text/xml');
$return->flush();
}
}
?>
服务器列表/api/inc/restFunctions.php
<?php
function getArrayOfFieldNames($queryResults){
$fieldList = array();
while($finfo = $queryResults->fetch_field()){
$fieldName = $finfo->name;
array_push($fieldList, $fieldName);
}
return $fieldList;
}
function getXMLofQuery($queryResults,$xmlTitle){
$fieldList = getArrayOfFieldNames($queryResults);
$xml = new XMLWriter();
$xml->openURI("php://output");
$xml->startDocument();
$xml->setIndent(true);
$title = $xmlTitle;
$titlePlural = $xmlTitle."s";
$xml->startElement($titlePlural);
$fieldIDName = $title."ID";
while($row = $queryResults->fetch_assoc()){
$xml->startElement($title);
$xml->writeAttribute('id', $row[$fieldIDName]);
foreach($fieldList as $field){
$xml->startElement($field);
$xml->writeRaw($row[$field]);
$xml->endElement();
}
$xml->endElement();
}
$xml->endElement();
return $xml;
}
function allserversxml(){
global $link; //from config.php file
$allServerResults = $link->query("SELECT * FROM servers");
$xml = getXMLofQuery($allServerResults,"server");
return $xml;
}
?>
问题是,当我转到URL时,我会得到错误error on line 2 at column 1: Extra content at the end of the document. Below is a rendering of the page up to the first error.
然而。。。下面没有渲染。什么东西?
编辑:根据ndm的建议,我能够通过页面的来源获得错误。
Call to a member function flush() on a non-object in C:'path'serverList'api'rest.php on line 29
所以我想我的问题是,当函数返回xml时,在页面上显示xml的最佳方式是什么?
根据错误消息和代码,假设"删除不相关的代码部分"不包括从发布的函数和逻辑流中删除代码,那么您想要调用flush()
的变量应该是$result
,而不是$return
。
...
$result = $functionName();
header('Content-type: text/xml');
$result->flush(); // like this