我想知道是否有人能夠幫助我。使用Xpath檢索所選記錄
我正在使用下面的腳本創建一個「圖庫」頁面。
<!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>
此頁面的數據來自文件'files.xml',其中的摘錄可以在下面找到。
<?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>
我想要做的是過濾返回的記錄,是由「用戶ID」和「locationid」和用於測試目的更具體的這些有「1」的值。
我從XML 'Select Where' Statement這個網站獲得了一些有用的指導,並且從我讀過的後續'XPath'教程中,我將我認爲是正確的xpath代碼放在一起,這是我的頂部'畫廊'頁腳本。
然而,當我試圖運行此我收到以下錯誤:
Parse error: syntax error, unexpected '=' in /homepages/2/d333603417/htdocs/development/gallery.php on line 16
很顯然,這個問題是與「等於」,但是當我又回到了教程,這似乎是正確的語法使用。
我只是想知道是否有人可以看看這個請讓我知道我哪裏出了問題。
非常感謝
修訂後的XPath代碼
<!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,非常感謝這一點。我必須承認,在這裏我的深度已經相當大,因爲除了我已經有的腳本之外,我不確定我需要包括什麼。我最初包含的代碼來自我一直在努力的一個教程。我剛剛嘗試使用'$ path ='文件[userid = 1]/originalname/folder/description/source/thumbnail';'by itslef但不幸的是它仍然檢索所有的記錄。親切的問候 – IRHM 2012-03-17 18:13:36
不清楚你是如何使查詢「文件[userid = 1] /原始名稱/文件夾/說明/源/縮略圖」。 – 2012-03-17 23:08:09
從XML中,它看起來像「原始名稱」,「文件夾」,「描述」等是「文件」的屬性,在這種情況下,查詢可能應該像「/ files/file [@ userid =」1「 ]「只是用於userid 1.如果你想通過更多的屬性進行搜索,然後」/文件/文件[@ userid =「1」和@ originalname =「somename」]「等等 – 2012-03-17 23:16:49