以XML格式返回的样式Feed内容

我有从我的数据库中提取信息然后使用API检查它的代码。我试图将返回信息自定义到表中。我会添加一张照片来显示我想要的内容。我到处搜索，我不知道如何编码。不要担心CSS只是获取信息来填充表格。以XML格式返回的样式Feed内容

<?php 
     $key = "********************"; 
     $address = urlencode($CustomFields->field('jr_address',$listing,false,false)); 
     $city = $listing['Category']['title']; 
     $zip = $CustomFields->field('jr_zipcode',$listing,false,false); 
     $url = "http://api.greatschools.org/schools/nearby?key={$key}&address={$address}&city={$city}&state=MI&zip={$zip}&schoolType=public-charter&levelCode=elementary-schools&minimumSchools=50&radius=10&limit=5"; 
     $curl = curl_init(); 
     curl_setopt($curl, CURLOPT_URL, $url); 
     curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); 
     $response = curl_exec($curl);  
     curl_close($curl); 
     print_r($response); 
?><style type='text/css'>gsId{display:none;}name{color:#9BB055; font-size:18px;}type{display:none;}gradeRange{}enrollment{}gsRating{}city{display:none;}state{display:none;}district{}districtNCESId{display:none;}address{display:none;}phone{display:none; }fax{display:none;}ncesId{display:none;}lat{display:none;}lon{display:none;}</style>

返回代码：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?> 
<schools> 
<school> 
<gsId>6350</gsId> 
<name>Chinese Education Center</name> 
<type>public</type> 
<gradeRange>K-5</gradeRange> 
<enrollment>63</enrollment> 
<gsRating>3</gsRating> 
<city>San Francisco</city> 
<state>CA</state> 
<district>San Francisco Unified School District</district> 
<districtNCESId>0634410</districtNCESId> 
<address>657 Merchant St., San Francisco, CA 94111</address> 
<phone>(415) 291-7918</phone> 
<fax>(415) 291-7965</fax> 
<ncesId>063441005596</ncesId> 
<lat>37.795</lat> 
<lon>-122.4042</lon> 
<overviewLink>http://www.greatschools.org/california/san-francisco/6350-Chinese-Education-Center/?s_cid=gsapi</overviewLink> 
<ratingsLink>http://www.greatschools.org/school/rating.page?state=CA&amp;id=6350&amp;s_cid=gsapi</ratingsLink> 
<reviewsLink>http://www.greatschools.org/school/parentReviews.page?state=CA&amp;id=6350&amp;s_cid=gsapi</reviewsLink> 
</school> 
<school> 
<gsId>6389</gsId> 
<name>Gordon J. Lau Elementary School</name> 
<type>public</type> 
<gradeRange>K-5</gradeRange> 
<enrollment>667</enrollment> 
<gsRating>7</gsRating> 
<city>San Francisco</city> 
<state>CA</state> 
<district>San Francisco Unified School District</district> 
<districtNCESId>0634410</districtNCESId> 
<address>950 Clay St., San Francisco, CA 94108</address> 
<phone>(415) 291-7921</phone> 
<fax>(415) 291-7952</fax> 
<website>http://www.gjles.org/</website> 
<ncesId>063441005599</ncesId> 
<lat>37.794</lat> 
<lon>-122.4086</lon> 
<overviewLink>http://www.greatschools.org/california/san-francisco/6389-Gordon-J.-Lau-Elementary-School/?s_cid=gsapi</overviewLink> 
<ratingsLink>http://www.greatschools.org/school/rating.page?state=CA&amp;id=6389&amp;s_cid=gsapi</ratingsLink> 
<reviewsLink>http://www.greatschools.org/school/parentReviews.page?state=CA&amp;id=6389&amp;s_cid=gsapi</reviewsLink> 
</school> 
</schools>

我不知道该怎么写脚本，使它看起来像这样：链接到图片： link text

所以新文件看起来像这样？

<?php 
    $key = "xxxxxxxxxxxxxxxxxxxxxxxx"; 
    $address = urlencode($CustomFields->field('jr_address',$listing,false,false)); 
    $city = $listing['Category']['title']; 
    $zip = $CustomFields->field('jr_zipcode',$listing,false,false); 
    $url = "http://api.greatschools.org/schools/nearby?key={$key}&address={$address}&city={$city}&state=MI&zip={$zip}&schoolType=public-charter&levelCode=elementary-schools&minimumSchools=50&radius=10&limit=5"; 
    $curl = curl_init(); 
    curl_setopt($curl, CURLOPT_URL, $url); 
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); 
    $response = curl_exec($curl);  
    curl_close($curl); 
    echo (xml2html($response, "/xslt/schools.xsl");

？>

我得到回音线的错误。

这是哪里去的？

<?php 
function xml2html($xmldata, $xslPath) 
{ 
    /* $xmldata -> your XML */ 
    /* $xsl -> XSLT file */ 

    $arguments = array('/_xml' => $xmldata); 
    $xsltproc = xslt_create(); 
    xslt_set_encoding($xsltproc, 'ISO-8859-1'); 
    $html = 
     xslt_process($xsltproc, 'arg:/_xml', $xslPath, NULL, $arguments); 

    if (empty($html)) { 
     die('XSLT processing error: '. xslt_error($xsltproc)); 
    } 
    xslt_free($xsltproc); 
    return $html; 
} 
?>

来源

2011-01-23 Carl V

也许有不止一个学校一个XML文档将更有帮助。另外，每所学校都有经纬度，但是应该计算与学校距离的点的坐标是什么？这不是在你的问题中指定的。 – 2011-01-23 06:58:04

我在上面的代码中包含了另一所学校。就纬度和经度而言，我不知道你是否看过我发布的图像（在原始文章的底部）。我只会隐藏它。当我将地址信息发送给API时，它将排序哪些学校更接近。我的挑战是我的知识是非常有限的，我迷失于如何编码任何这些。 – 2011-01-23 16:56:52

浏览器无法看到自定义XML（即您的学校xml）。所以你应该手动将其转换为HTML。我知道2级的解决方案：

解析XML与xml parser functions
通过XSLT转换板XML（对我来说，更好的办法）。

对于XML提供了我写的下一个XSLT：link text（对不起，外部链接，我在这里粘贴XSLT代码无法解释的烦恼）

转型可以由功能xslt_process应用。（你应该在你的服务器的（例如“XSLT/schools.xsl”文件夹中存放的地方XSL文件）和转换XML响应）：

<?php 
function xml2html($xmldata, $xslPath) 
{ 
    /* $xmldata -> your XML */ 
    /* $xsl -> XSLT file */ 

    $arguments = array('/_xml' => $xmldata); 
    $xsltproc = xslt_create(); 
    xslt_set_encoding($xsltproc, 'ISO-8859-1'); 
    $html = 
     xslt_process($xsltproc, 'arg:/_xml', $xslPath, NULL, $arguments); 

    if (empty($html)) { 
     die('XSLT processing error: '. xslt_error($xsltproc)); 
    } 
    xslt_free($xsltproc); 
    return $html; 
} 

// making request... 
$response = curl_exec($curl);  
curl_close($curl); 
echo(xml2html($response, "/xslt/schools.xsl"));

来源

2011-01-23 08:36:21

以XML格式返回的样式Feed内容

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