13

我正在尝试学习 XML，我知道这是未正确导入节点的问题。但我无法完全弄清楚。我一直在环顾四周，大多数人没有像我在部门中那样拥有多个子元素。

这是我的 XML 结构： 

<SOT>  
    <DEPARTMENT name="Aviation Technology" id="AT">  
        <EMPLOYEE type="Faculty">  
            <LOGIN>jdoe1</LOGIN>  
            <NAME>John Doe</NAME>   
        </EMPLOYEE>

        <EMPLOYEE type="Faculty">  
            <LOGIN>jdoe2</LOGIN>  
            <NAME>Jane Doe</NAME>   
        </EMPLOYEE>

        <EMPLOYEE type="Faculty">  
            <LOGIN>jdoe3</LOGIN>  
            <NAME>Joe Doe</NAME>  
        </EMPLOYEE> 
    </DEPARTMENT>    

    <DEPARTMENT name="Building and Construction Management" id="BCM">  
    </DEPARTMENT>

    <DEPARTMENT name="Computer Graphics Technology" id="CGT">  
    </DEPARTMENT>  
</SOT>

我知道 SOT 是我的根元素，部门是 SOT 的“子项”，每个部门都有多个员工“子项”。我遇到的问题是当我尝试将新员工添加到某个部门时。当我尝试时，$departmentArray->item($i)->appendChild($employee);我得到了错误的文档错误。

我正在使用此 PHP 代码尝试将孩子附加到 departmentNode

<?php

    //grab form data
    $username = $_POST['username'];
    $employeeName = $_POST['employeeName'];
    $department = $_POST['department'];

    //create new DOMDocument to hold current XML data
    $doc = new DOMDocument();
    $doc->load("test.xml");
    $xpath = new DOMXpath($doc);

    //create our new DOMDocument for combining the XML data
    $newDoc = new DOMDocument();
    $newDoc->preserveWhiteSpace = false;

    //create School of Tech Node and append to new doc
    $sotElement = $newDoc->createElement("SOT");
    $newDoc->appendChild($sotElement);
    $root = $newDoc->documentElement;

    //grab the department Nodes
    $departmentArray = $doc->getElementsByTagName("DEPARTMENT");

    //create a new employee and set attribute to faculty
    $employee = $newDoc->createElement("EMPLOYEE");
    $employee->setAttribute("type", "Faculty");

    //counters (might use them later for ->item(counter) function
    $indexCounter = 0;
    $i = 0;

    foreach($departmentArray as $departmentNode){
        if(strcmp($departmentNode->getAttribute('name'),$department) == 0){//check if departments match
            //create login element
            $loginNode = $newDoc->createElement("LOGIN");
            $loginNode->appendChild($newDoc->createTextNode($username));
            $employee->appendChild($loginNode);

            //create name node
            $nameNode = $newDoc->createElement("NAME");
            $nameNode->appendChild($newDoc->createTextNode($employeeName));
            $employee->appendChild($nameNode);

            //append employee onto department node
            //$departmentArray->item($i) = $doc->importNode($departmentArray->item($i), true);
            $departmentArray->item($i)->appendChild($employee);

            //set index of department array (possibly used for appending later)
            $indexCounter = $i;
        }
        $i++;
    }

    #######################################
    /*Write out data to XML file         */
    #######################################
    //$departmentArray = $doc->getElementsByTagName("DEPARTMENT");
    foreach($departmentArray as $departmentNode){
        $tempNode = $newDoc->importNode($departmentNode, true);
        /*if(strcmp($departmentNode->getAttribute('name'),$department) == 0){
            $sotElement->appendChild($employee);

        }*/
        $sotElement->appendChild($tempNode);
    }

    $newDoc->formatOutput = true;
    $newDoc->save("test2.xml");


?>

任何解释如何正确导入所有部门节点以便能够附加到它们的帮助将不胜感激。我试过使用数组。

4

2 回答 2

22

您需要导入任何节点以将其附加到另一个文档：

$departmentArray->item($i)->appendChild( $doc->importNode( $employee, true ) );
于 2011-04-25T22:23:18.827 回答
14

我很确定这是因为您试图将来自不同文档的元素附加到输出文档中。

我在 php 网站的评论中发现了这段代码，DOMNode::cloneNode这可能是你想要的。

<?php 
    $dom1->documentElement->appendChild( 
        $dom1->importNode( $dom2->documentElement, true )
    ); 
?>

或者，您可以查看导出节点的 XML 并将其重新导入到 aDOMDocumentFragment中，但我必须通过实验才能确定。

于 2011-04-25T22:07:42.327 回答