XPath 查询中的撇号 (')

Question

我使用以下 XPATH Query 列出站点下的对象。 ListObject[@Title='SomeValue'] 。 SomeValue 是动态的。只要 SomeValue 没有撇号 (‘)，此查询就有效。也尝试使用转义序列。没用。

我究竟做错了什么？

原文由 Prabhu 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

这出乎意料地难以做到。

看一下 XPath Recommendation ，您会发现它将文字定义为：

 Literal ::=   '"' [^"]* '"'
            | "'" [^']* "'"

也就是说，XPath 表达式中的字符串文字可以包含撇号或双引号，但不能同时包含两者。

你不能使用转义来解决这个问题。像这样的文字：

 'Some'Value'

将匹配此 XML 文本：

 Some'Value

这确实意味着可能存在一段您无法生成 XPath 文字来匹配的 XML 文本，例如：

 <elm att="&quot;&apos"/>

但这并不意味着不可能将该文本与 XPath 匹配，这只是棘手而已。在您尝试匹配的值同时包含单引号和双引号的任何情况下，您都可以构造一个使用 concat 的表达式来生成要匹配的文本：

 elm[@att=concat('"', "'")]

所以这导致我们这样做，这比我希望的要复杂得多：

 /// <summary>
/// Produce an XPath literal equal to the value if possible; if not, produce
/// an XPath expression that will match the value.
///
/// Note that this function will produce very long XPath expressions if a value
/// contains a long run of double quotes.
/// </summary>
/// <param name="value">The value to match.</param>
/// <returns>If the value contains only single or double quotes, an XPath
/// literal equal to the value.  If it contains both, an XPath expression,
/// using concat(), that evaluates to the value.</returns>
static string XPathLiteral(string value)
{
    // if the value contains only single or double quotes, construct
    // an XPath literal
    if (!value.Contains("\""))
    {
        return "\"" + value + "\"";
    }
    if (!value.Contains("'"))
    {
        return "'" + value + "'";
    }

    // if the value contains both single and double quotes, construct an
    // expression that concatenates all non-double-quote substrings with
    // the quotes, e.g.:
    //
    //    concat("foo", '"', "bar")
    StringBuilder sb = new StringBuilder();
    sb.Append("concat(");
    string[] substrings = value.Split('\"');
    for (int i = 0; i < substrings.Length; i++ )
    {
        bool needComma = (i>0);
        if (substrings[i] != "")
        {
            if (i > 0)
            {
                sb.Append(", ");
            }
            sb.Append("\"");
            sb.Append(substrings[i]);
            sb.Append("\"");
            needComma = true;
        }
        if (i < substrings.Length - 1)
        {
            if (needComma)
            {
                sb.Append(", ");
            }
            sb.Append("'\"'");
        }

    }
    sb.Append(")");
    return sb.ToString();
}

是的，我在所有边缘情况下都对其进行了测试。这就是为什么逻辑如此愚蠢地复杂：

     foreach (string s in new[]
    {
        "foo",              // no quotes
        "\"foo",            // double quotes only
        "'foo",             // single quotes only
        "'foo\"bar",        // both; double quotes in mid-string
        "'foo\"bar\"baz",   // multiple double quotes in mid-string
        "'foo\"",           // string ends with double quotes
        "'foo\"\"",         // string ends with run of double quotes
        "\"'foo",           // string begins with double quotes
        "\"\"'foo",         // string begins with run of double quotes
        "'foo\"\"bar"       // run of double quotes in mid-string
    })
    {
        Console.Write(s);
        Console.Write(" = ");
        Console.WriteLine(XPathLiteral(s));
        XmlElement elm = d.CreateElement("test");
        d.DocumentElement.AppendChild(elm);
        elm.SetAttribute("value", s);

        string xpath = "/root/test[@value = " + XPathLiteral(s) + "]";
        if (d.SelectSingleNode(xpath) == elm)
        {
            Console.WriteLine("OK");
        }
        else
        {
            Console.WriteLine("Should have found a match for {0}, and didn't.", s);
        }
    }
    Console.ReadKey();
}

原文由 Robert Rossney 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

我将罗伯特的答案移植到 Java（在 1.6 中测试）：

 /// <summary>
/// Produce an XPath literal equal to the value if possible; if not, produce
/// an XPath expression that will match the value.
///
/// Note that this function will produce very long XPath expressions if a value
/// contains a long run of double quotes.
/// </summary>
/// <param name="value">The value to match.</param>
/// <returns>If the value contains only single or double quotes, an XPath
/// literal equal to the value.  If it contains both, an XPath expression,
/// using concat(), that evaluates to the value.</returns>
public static String XPathLiteral(String value) {
    if(!value.contains("\"") && !value.contains("'")) {
        return "'" + value + "'";
    }
    // if the value contains only single or double quotes, construct
    // an XPath literal
    if (!value.contains("\"")) {
        System.out.println("Doesn't contain Quotes");
        String s = "\"" + value + "\"";
        System.out.println(s);
        return s;
    }
    if (!value.contains("'")) {
        System.out.println("Doesn't contain apostophes");
        String s =  "'" + value + "'";
        System.out.println(s);
        return s;
    }

    // if the value contains both single and double quotes, construct an
    // expression that concatenates all non-double-quote substrings with
    // the quotes, e.g.:
    //
    //    concat("foo", '"', "bar")
    StringBuilder sb = new StringBuilder();
    sb.append("concat(");
    String[] substrings = value.split("\"");
    for (int i = 0; i < substrings.length; i++) {
        boolean needComma = (i > 0);
        if (!substrings[i].equals("")) {
            if (i > 0) {
                sb.append(", ");
            }
            sb.append("\"");
            sb.append(substrings[i]);
            sb.append("\"");
            needComma = true;
        }
        if (i < substrings.length - 1) {
            if (needComma) {
                sb.append(", ");
            }
            sb.append("'\"'");
        }
        System.out.println("Step " + i + ": " + sb.toString());
    }
    //This stuff is because Java is being stupid about splitting strings
    if(value.endsWith("\"")) {
        sb.append(", '\"'");
    }
    //The code works if the string ends in a apos
    /*else if(value.endsWith("'")) {
        sb.append(", \"'\"");
    }*/
    sb.append(")");
    String s = sb.toString();
    System.out.println(s);
    return s;
}

希望这对某人有帮助！

原文由 Cody S 发布，翻译遵循 CC BY-SA 3.0 许可协议