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我有 oai.php 页面。现在当我输入 url oai.php?verb=identify 然后想在 identify.xml 中显示内容。但是当我查看时我想要相同的 url(oai.php?verb=identify) identify.xml.Below 代码仅显示 xml 的数据。

 <?php

include "config.php";
$id= $_GET['verb'];
//echo $id;
if($id=="identify")
{
    $date_mod = gmdate("Y-m-d\TH:i:s\Z");
    $rss_txt ='<?xml version="1.0" encoding="utf-8"?>
    <OAI-PMH xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.openarchives.org/OAI/2.0/" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/ http://www.openarchives.org/OAI/2.0/OAI-PMH.xsd">
    <responseDate>' .$date_mod. '</responseDate>
    <request verb="Identify">http://mywebsite.org/oai</request>
     <Identify>

            <repositoryName>mydomain</repositoryName>
            <baseURL>mywesite.com</baseURL>
            <protocolVersion>2.0</protocolVersion>
            <adminEmail>myemail</adminEmail>
            <earliestDatestamp>'.$date_mod.'</earliestDatestamp>
            <deletedRecord>transient</deletedRecord>
            <granularity>YYYY-MM-DDThh:mm:ssZ</granularity>


     </Identify>
     </OAI-PMH>';
}
else if($id=="ListRecords")
{
    include "config.php";

    $date_mod = gmdate("Y-m-d\TH:i:s\Z");
    $viewss1=$con->query("SELECT COUNT(upload_paper_id) FROM upload_papers");
    $cview1=$viewss1->fetch_row();
    $countt1=$cview1[0];

    $rss_txt = '<?xml version="1.0" encoding="utf-8"?>
    <?xml-stylesheet type="text/xsl" href="http://localhost/oai-file/data/oai2.xsl" ?>
    <OAI-PMH xmlns="http://www.openarchives.org/OAI/2.0/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/ http://www.openarchives.org/OAI/2.0/OAI-PMH.xsd">
    <responseDate>' .$date_mod. '</responseDate>
    <request verb="ListRecords" metadataPrefix="oai_dc">http:mywebsite.org</request>
    <ListRecords>';
        $query = "SELECT * FROM upload_papers";
        $result=$con->query($query);
        while($values_query = mysqli_fetch_assoc($result))
        {
            $year= $values_query['year'];
            $upload_paper_id= $values_query['upload_paper_id'];
    $rss_txt = '<record>
    <header>

    <identifier>' .$upload_paper_id. '</identifier>
    <datestamp>' .$date_mod. '</datestamp>
    <setSpec>mydoc8:BIH</setSpec>

    </header>
    <metadata>



    <oai_dc:dc
        xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/"
        xmlns:dc="http://purl.org/dc/elements/1.1/"
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/
        http://www.openarchives.org/OAI/2.0/oai_dc.xsd">    
    <dc:title xml:lang="en-US">' .$values_query['paper_title']. '</dc:title>
    <dc:creator>' . $values_query['author_name']. '</dc:creator>    
    <dc:creator>' . $values_query['co_author_name']. '</dc:creator> 
    <dc:description xml:lang="en-US">' .$values_query['abstract']. '</dc:description>   
    <dc:publisher xml:lang="en-US">' .$values_query['journal_nam']. '</dc:publisher>    
    <dc:contributor xml:lang="en-US">' .$values_query['author_name']. '</dc:contributor>
    <dc:type xml:lang="en-US">Article</dc:type> 
    <dc:format>application/pdf</dc:format>  
    <dc:identifier>http://mywebsite'.$year.'/article.php?page='.$upload_paper_id.'</dc:identifier>
    <dc:language>en</dc:language>   
    </oai_dc:dc>
    </metadata>
            </record>';
     }  
               $rss_txt = '<resumptionToken
                completeListSize="'.$countt1.'"
                cursor="0">'.$date_mod.'</resumptionToken>


    </ListRecords>
    </OAI-PMH>';


}
else if($id=="ListSets")
{
    $rss_txt =file_get_contents("ListSets.xml");
}
header('Content-type: application/xml');
echo $rss_txt;

?>

上面的代码只显示了xml的数据。下面附加输出 的输出我想要这样的输出想要的输出 。如果我在url中输入verb=identify，那么它想显示identify.xsd或xml文件中的任何一个。然后我输入Listsets想要显示该页面中的内容。如果动词等于空想显示识别页面。请帮助我。这真的很烦我。谢谢你

4

1 回答 1

0

代码存在一些问题。主要是将您的标头设置为 XML 内容类型。这将告诉浏览器会发生什么。但还有一些其他问题。

$id= $_GET['verb'];
//echo $id;
if($id=="identify")
{
    $date_mod = gmdate("Y-m-d\TH:i:s\Z");
    $rss_txt ='<?xml version="1.0" encoding="utf-8"?>
    <OAI-PMH xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.openarchives.org/OAI/2.0/" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/ http://www.openarchives.org/OAI/2.0/OAI-PMH.xsd">
    <responseDate>' .$date_mod. '</responseDate>
    <request verb="Identify">http://mywebsite.org/oai</request>
     <Identify>

            <repositoryName>mydomain</repositoryName>
            <baseURL>mywesite.com</baseURL>
            <protocolVersion>2.0</protocolVersion>
            <adminEmail>myemail</adminEmail>
            <earliestDatestamp>'.$date_mod.'</earliestDatestamp>
            <deletedRecord>transient</deletedRecord>
            <granularity>YYYY-MM-DDThh:mm:ssZ</granularity>


     </Identify>
     </OAI-PMH>';
}
else if($id=="ListRecords")
{
    $rss_txt =file_get_contents("compend1.xml");
}
else if($id=="ListSets")
{
    $rss_txt =file_get_contents("ListSets.xml");
}
header('Content-type: application/xml');
echo $rss_txt;

你有$rss_txt .=which 与 . 意味着附加字符串，因为您没有在任何地方声明它，所以会给您关于未定义变量的错误。还有内容 - XML 需要从字符串的开头开始，所以删除空格。

主要的是有...

header('Content-type: application/xml');

但是，要使其正常工作，您需要确保注释掉echo，我认为这是一些调试代码。

当做一个if- 单=用于分配一个值，所以你需要==比较这些值。

您可以使用$_GET[]访问参数而不是您使用的方式，简单得多。

我曾经file_get_contents()从那些其他文件中获取数据并将其移到echo最后，因此它们都具有相同的逻辑。

于 2018-04-28T07:27:40.223 回答