我想知道是否有人能够帮助我。使用Xpath检索所选记录
我正在使用下面的脚本创建一个“图库”页面。
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<?php
//This variable specifies relative path to the folder, where the gallery with uploaded files is located.
$galleryPath = 'UploadedFiles/';
$thumbnailsPath = $galleryPath . 'Thumbnails/';
$absGalleryPath = realpath($galleryPath) . DIRECTORY_SEPARATOR;
$descriptions = new DOMDocument('1.0');
$descriptions->load($absGalleryPath . 'files.xml');
path="files[userid=1]/originalname/folder/description/source/thumbnail";
// code for IE
if (window.ActiveXObject)
{
var nodes=xml.selectNodes(path);
for (i=0;i<nodes.length;i++)
{
document.write(nodes[i].childNodes[0].nodeValue);
}
}
?>
<head>
<title>Gallery</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link href="Libraries/fancybox/jquery.fancybox-1.3.1.css" rel="stylesheet" type="text/css" />
<link href="Styles/style.css" rel="stylesheet" type="text/css" />
<!--[if IE]>
<link href="Styles/ie.css" rel="stylesheet" type="text/css" />
<![endif]-->
<script src="Libraries/jquery/jquery-1.4.3.min.js" type="text/javascript"></script>
<script src="Libraries/fancybox/jquery.fancybox-1.3.1.pack.js" type="text/javascript"></script>
<script type="text/javascript">
$(function() { $('a.fancybox').fancybox(); });
</script>
<style type="text/css">
<!--
.style1 {
font-size: 14px;
margin-right: 110px;
}
.style4 {font-size: 12px}
-->
</style>
</head>
<body style="font-family: Calibri; color: #505050; font-size: 9px; border-bottom-width: thin; margin-top: 5px; margin-left: -475px; margin-right: 1px; margin-bottom: -10px;">
<div align="right" class="style1"> <a href = "imagefolders.php" /> View Uploaded Images In Folder Structure <a/> ← View All Uploaded Images </div>
<form id="gallery" class="page">
<div id="container">
<div id="center">
<div class="aB">
<div class="aB-B">
<?php if ('Uploaded files' != $current['title']) :?>
<?php endif;?>
<label>
<input name="userid" type="text" id="userid" value="1" />
<input name="locationid" type="text" id="locationid" value="1" />
</label>
<div class="demo">
<div class="inner">
<div class="container">
<div class="gallery">
<ul class="gallery-image-list">
<?php for ($i = 0; $i < $descriptions->documentElement->childNodes->length; $i++) :
$xmlFile = $descriptions->documentElement->childNodes->item($i);
$name = htmlentities($xmlFile->getAttribute('originalname'), ENT_COMPAT, 'UTF-8');
$description = htmlentities($xmlFile->getAttribute('description'), ENT_COMPAT, 'UTF-8');
$folder = htmlentities($xmlFile->getAttribute('folder'), ENT_COMPAT, 'UTF-8');
$source = $galleryPath . rawurlencode($xmlFile->getAttribute('source'));
$thumbnail = $thumbnailsPath . rawurlencode($xmlFile->getAttribute('thumbnail'));
?>
<li class="item">
<a class="fancybox" target="_blank" rel="original" href="<?php echo $source; ?>"><img class="preview"
alt="<?php echo $name; ?>" src="<?php echo $thumbnail; ?>" /></a> </li>
<li class="item"></li>
<p><span class="style4"><b>Image Description:</b> <?php echo htmlentities($xmlFile->getAttribute('description'));?> <br />
<b>Image contained in folder:</b> <?php echo htmlentities($xmlFile->getAttribute('folder'));?> </span><br />
<?php endfor; ?>
</li>
</p>
</ul>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
<div class="aB-a"> </div>
</div>
</div>
</div>
</form>
</body>
</html>
此页面的数据来自文件'files.xml',其中的摘录可以在下面找到。
<?xml version="1.0" encoding="utf-8" ?>
- <files>
<file name="Test 1/article-0-07D01B74000005DC-138_468x617.jpg" source="article-0-07D01B74000005DC-138_468x617.jpg" size="143110" originalname="article-0-07D01B74000005DC-138_468x617.jpg" thumbnail="article-0-07D01B74000005DC-138_468x617.jpg" description="No description provided" userid="1" locationid="1" folder="Test_1" />
</files>
我想要做的是过滤返回的记录,是由“用户ID”和“locationid”和用于测试目的更具体的这些有“1”的值。
我从XML 'Select Where' Statement这个网站获得了一些有用的指导,并且从我读过的后续'XPath'教程中,我将我认为是正确的xpath代码放在一起,这是我的顶部'画廊'页脚本。
然而,当我试图运行此我收到以下错误:
Parse error: syntax error, unexpected '=' in /homepages/2/d333603417/htdocs/development/gallery.php on line 16
很显然,这个问题是与“等于”,但是当我又回到了教程,这似乎是正确的语法使用。
我只是想知道是否有人可以看看这个请让我知道我哪里出了问题。
非常感谢
修订后的XPath代码
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<?php
$galleryPath = 'UploadedFiles/';
$thumbnailsPath = $galleryPath . 'Thumbnails/';
$absGalleryPath = realpath($galleryPath) . DIRECTORY_SEPARATOR;
$descriptions = new DOMDocument('1.0');
$descriptions->load($absGalleryPath . 'files.xml');
/files/file[@userid="1" and @locationid="1"];
?>
你好,非常感谢这一点。我必须承认,在这里我的深度已经相当大,因为除了我已经有的脚本之外,我不确定我需要包括什么。我最初包含的代码来自我一直在努力的一个教程。我刚刚尝试使用'$ path ='文件[userid = 1]/originalname/folder/description/source/thumbnail';'by itslef但不幸的是它仍然检索所有的记录。亲切的问候 – IRHM 2012-03-17 18:13:36
不清楚你是如何使查询“文件[userid = 1] /原始名称/文件夹/说明/源/缩略图”。 – 2012-03-17 23:08:09
从XML中,它看起来像“原始名称”,“文件夹”,“描述”等是“文件”的属性,在这种情况下,查询可能应该像“/ files/file [@ userid =”1“ ]“只是用于userid 1.如果你想通过更多的属性进行搜索,然后”/文件/文件[@ userid =“1”和@ originalname =“somename”]“等等 – 2012-03-17 23:16:49