使用YQL多查询&；XPath解析HTML，如何转义嵌套引号？

标题比它必须要复杂得多，这是问题所在

SELECT * 
FROM query.multi 
WHERE queries="
    SELECT * 
        FROM html 
        WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
        AND xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span';
    SELECT * 
        FROM xml 
        WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
    SELECT * 
        FROM xml 
        WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"

特别是这条线,

xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span'

这是有问题的，因为引用，我必须把它们嵌套三层，我已经用光了引用字符。我尝试了以下几种变体，但没有成功：

//no attribute quoting
xpath='//li[@class=listLi]/div[@class=views]/a/span' 

//try to quote attribute w/ backslash & single quote
xpath='//li[@class=\'listLi\']/div[@class=\'views\']/a/span'

//try to quote attribute w/ backslash & double quote
xpath='//li[@class=\"listLi\"]/div[@class=\"views\"]/a/span'

//try to quote attribute with double single quotes, like SQL
xpath='//li[@class=''listLi'']/div[@class=''views'']/a/span'

//try to quote attribute with double double quotes, like SQL
xpath='//li[@class=""listLi""]/div[@class=""views""]/a/span'

//try to quote attribute with quote entities
xpath='//li[@class="listLi"]/div[@class="views"]/a/span'

//try to surround XPath with backslash & double quote
xpath=\"//li[@class='listLi']/div[@class='views']/a/span\"

//try to surround XPath with double double quote
xpath=""//li[@class='listLi']/div[@class='views']/a/span""

一切都没有成功

我没有看到太多关于转义XPath字符串的内容，但我发现的一切似乎都是关于使用concat（因为“或”都不可用）或html实体的变体。不使用属性引号不会引发错误，但会失败，因为它不是我需要的实际XPath字符串

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我在YQL文档中没有看到任何关于如何处理转义的内容。我知道edge casey的情况，但希望他们能提供某种转义指南。

我提出了一个解决方案，它并没有真正回答我最初的问题，但确实解决了问题

该表将使用CSS选择器&将其解析为XPath，从而避免了令人讨厌的转义问题

SELECT *
FROM query.multi 
    WHERE queries="
        SELECT * 
            FROM data.html.cssselect 
            WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
            AND css='li.listLi div.views a span';
        SELECT * 
            FROM xml 
            WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
        SELECT * 
            FROM json 
            WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
        SELECT * 
            FROM xml 
            WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
        SELECT * 
            FROM json 
            WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"

---

您需要用双反斜杠转义XPath查询中的任何字符，换句话说：

SELECT * FROM query.multi 
WHERE queries="
    SELECT * 
        FROM html 
        WHERE url='http://www.stumbleupon.com/url/http://www.guildwars2.com' 
        AND xpath='//li[@class=\\'listLi\\']/div[@class=\\'views\\']/a/span';
    SELECT * 
        FROM xml 
        WHERE url='http://services.digg.com/1.0/endpoint?method=story.getAll&link=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://api.tweetmeme.com/url_info.json?url=http://www.guildwars2.com';
    SELECT * 
        FROM xml 
        WHERE url='http://api.facebook.com/restserver.php?method=links.getStats&urls=http://www.guildwars2.com';
    SELECT * 
        FROM json 
        WHERE url='http://www.reddit.com/button_info.json?url=http://www.guildwars2.com'"

---

（）

默认情况下，当我尝试在我的页面中使用它时，JS正在吃\\。我不得不做这些废话来让它工作，啊，明白了。但是很简陋。“从html中选择*，其中url='stumbleupon.com/url/%url%'和xpath='//li[@class=\\“+”\\'listLi\\\\“+”\']/div[@class=\\\“+”\\\'views\\\\\\\+']/a/span”奇怪的是，看起来data.html.cssselect比使用xpath从html中选择要快，尽管data.htmlcsselect只是转换为带xpath的从html中选择。古怪的