XPath XML 查询

发布于 2024-11-26 06:29:09 字数 826 浏览 3 评论 0原文

我在使用 xml 查询时遇到问题。下面是xml文档。我需要做的就是显示 cdata 部分内的文本。我写了这个查询：/item/title/text()，但它崩溃了并且不起作用。

<title>
 <![CDATA[Envious at Envie! 50% off anything you want, anytime you want!]]>
</title>

我正在为 iphone xcode 使用 hpple 库。这是我的代码：

    NSData *urlData = [[NSData alloc]initWithContentsOfURL:[NSURL URLWithString:[@"http://www.twangoo.com/userfiles/rss/rss_hong-kong.xml" stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]]];

    TFHpple *xpathParser = [[TFHpple alloc] initWithHTMLData:urlData];

    NSArray *title = [xpathParser search:@"//rss/channel/item[1]/title/text()"];

   TFHppleElement *element2 = [title objectAtIndex:0];

    NSString *titleString = [element2 content];

    NSLog(@"@%", titleString)

它返回零。

原文

I have trouble working with xml query. Below is the xml document. All I need to do is to display the text inside the cdata section. I wrote this query: /item/title/text(), but it crashed and doesn't work.

<title>
 <![CDATA[Envious at Envie! 50% off anything you want, anytime you want!]]>
</title>

I am using hpple library for iphone xcode. Here is my code:

    NSData *urlData = [[NSData alloc]initWithContentsOfURL:[NSURL URLWithString:[@"http://www.twangoo.com/userfiles/rss/rss_hong-kong.xml" stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]]];

    TFHpple *xpathParser = [[TFHpple alloc] initWithHTMLData:urlData];

    NSArray *title = [xpathParser search:@"//rss/channel/item[1]/title/text()"];

   TFHppleElement *element2 = [title objectAtIndex:0];

    NSString *titleString = [element2 content];

    NSLog(@"@%", titleString)

It returns nil.

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讽刺将军 2024-12-03 06:29:09

如果您可以提供更多有关您的环境、语言、正在使用的库以及生成的实际错误消息的信息，那将会有很大帮助。

但是，我认为 xsl 样式表中的有效 xslt 语句将是

……

<xsl:value-of select="/item/title"/>

如果您以编程方式在 PHP 中访问该数据，

它将类似于：

$doc = new DOMDocument;
$doc->loadXML("<item><title>My name is Kevin</title></item>");
$xpath = new DOMXPath($doc);
$query = "/item/title";
$entries = $xpath->query($query);
foreach ($entries as $entry) { print $entry->nodeValue; }

If you could provide a bit more information as to your environment, language, which libraries use are using and the actual error message produced that would help a lot.

However, I think a valid xslt statement in an xsl stylesheet would be

...

<xsl:value-of select="/item/title"/>

...

If you were programmatically accessing that data in PHP it would be something like:

$doc = new DOMDocument;
$doc->loadXML("<item><title>My name is Kevin</title></item>");
$xpath = new DOMXPath($doc);
$query = "/item/title";
$entries = $xpath->query($query);
foreach ($entries as $entry) { print $entry->nodeValue; }

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