PHP使用xmllint命令处理xml与html的方法_PHP
本文实例讲述了PHP使用xmllint命令处理xml与html的方法。分享给大家供大家参考。具体分析如下:
xmllint是一个很方便的处理及验证xml、处理html的工具,linux下只要安装libxml2就可以使用这个命令。首先看下其结合--html 、--xpath参数处理html时的例子:
示例如下:
代码如下:
curl http://www.bitsCN.com /ip/?q=8.8.8.8 2>/dev/null | xmllint --html --xpath "//ul[@id='csstb']" - 2>/dev/null | sed -e 's/]*>//g'
上例中主要是通过在123cha上查询的IP地址的归属情况后,通过提取结果(ul#csstb),只获取文本部分的内容。上面的脚本语句执行后的结果如下:
[您的查询]:8.8.8.8
本站主数据:
美国
本站辅数据:Google Public DNS提供:hypo
美国 Google免费的Google Public DNS提供:zwstar参考数据一:美国
参考数据二:美国
下面再结合示例看下其他主要参数的用法。
1、 --format
此参数用于格式化xml,使其具有良好的可读性。
假设有xml(person.xml)内容如下:
代码如下:
执行如下操作后其输出为更易读的xml格式:
代码如下:
#xmllint --format person.xml
2、 --noblanks
与--format相反,有时为了节省传输量,我们希望去掉xml中的空白,这时我们可以使用--noblanks命令。
假设xml(person.xml)内容如下
代码如下:
执行该参数操作后,其输出结果为:
代码如下:
#xmllint --noblanks person.xml
3、--schema
使用scheam验证xml文件的正确性(XML Schema 是基于 XML 的 DTD 替代者)
假设有xml文件(person.xml)和scheam文件(person.xsd)文件,内容分别如下
person.xml
代码如下:
person.xsd
代码如下:
按如下命令执行后的结果是:
代码如下:
#xmllint --schema person.xsd person.xml
person.xml validates
注:默认情况下,验证后会输出验证的文件内容,可以使用 --noout选项去掉此输出,这样我们可以只得到最后的验证结果。
代码如下:
#xmllint --noout --schema person.xsd person.xml
person.xml validates
下面我们改动person.xml,使这份文件age字段和sex都是不符合xsd定义的。
代码如下:
#xmllint --noout --schema person.xsd person.xml
person.xml:4: element age: Schemas validity error : Element 'age': 'not age' is not a valid value of the atomic type 'xs:integer'.
person.xml:5: element sex: Schemas validity error : Element 'sex': [facet 'enumeration'] The value 'test' is not an element of the set {'male', 'female'}.
person.xml:5: element sex: Schemas validity error : Element 'sex': 'test' is not a valid value of the local atomic type.
person.xml fails to validate
可以看到xmllint成功的报出了错误!
4、 关于--schema的输出
在讲输出之前先看下面一个场景,假如你想通过php执行xmllint然后拿到返回结果,你的代码通常应该是这个样子valid.php
代码如下:
$command = "xmllint --noout --schema person.xsd person.xml";
exec($command, $output, $retval);
//出错时返回值不为0
if ($retval != 0){
var_dump($output);
}
else{
echo "yeah!";
}
我们保持上文中person.xml的错误。
执行此代码,你会发现,你拿到的output不是错误,而是array(0) {}, amazing!
为什么会这样呢?
因为xmllint --schema,如果验证出错误,错误信息并不是通过标准输出(stdout)显示的,而是通过标准错误(stderr)进行显示的。
而exec的output参数拿到的,只能是标准输出(stdout)显示的内容。
所以,为了拿到出错信息,我们需要将标准错误重定向到标准输出,对应修改代码:
代码如下:
$command = "xmllint --noout --schema person.xsd person.xml 2>$1";
再次执行valid.php,错误信息顺利拿到!
例子如下:
首先建立一份 xml 文档,命名为 po.xml,其内容如下:
代码如下:
然后为 po.xml 写的 schema 文件,取名为 po.xsd,内容如下:
代码如下:
Purchase order schema for Example.com.
Copyright 2000 Example.com. All rights reserved.
使用 xmllint 对 po.xml 文件进行校验:
代码如下:
$ xmllint -schema po.xsd po.xml
如果无出错信息,就说明校验通过了。希望本文所述对大家的PHP程序设计有所帮助。

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