0

XML 文件

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ew-language id="en">
    <global>
        <phrase id="actiondeleted" value="Deleted">
            <child_phrase_1 id="1234" value="numbers">
                <child id="test_id" value="test_value"/>
                <child id="test_id" value="test_value_2"/>
            </child_phrase_1>
        </phrase>
    </global>
</ew-language>

如何通过 ID 和值获取元素,以便元素是唯一的。我尝试了这些

     $parent = ($xpath->query("//*[@id='$previous_tag_id']")&& $xpath->query("//*[@value='$previous_tag_value']"))->item(0);
----------------------AND THIS ONE-----------------------------------
     $parent = $xpath->query("//*[@id='$previous_tag_id']/*@value='$previous_tag_value')"->item(0);
----------------------AND THIS ONE-----------------------------------
 $xpath->query("//*[@id='$previous_tag_id' and @value='$previous_tag_value']");

每种语法都不起作用。

4

1 回答 1

1

使用//*[@id='$previous_tag_id' and @value='$previous_tag_value']应该有效,请参阅eval.in中的演示:

$xml = <<<XML
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ew-language id="en">
    <global>
        <phrase id="actiondeleted" value="Deleted">
            <child_phrase_1 id="1234" value="numbers">
                <child id="test_id" value="test_value"/>
                <child id="test_id" value="test_value_2"/>
            </child_phrase_1>
        </phrase>
    </global>
</ew-language>
XML;
$doc = new DOMDocument;
$doc->loadXML($xml);
$xpath = new DOMXPath($doc);
$previous_tag_id = "test_id";
$previous_tag_value = "test_value_2";
$result = $xpath->query("//*[@id='$previous_tag_id' and @value='$previous_tag_value']");
foreach($result as $r)
{
    echo $doc->saveXML($r);
}

输出 :

<child id="test_id" value="test_value_2"/>
于 2015-06-17T14:26:32.860 回答