首页 \ 问答 \ 用递归进行XSLT多字符串替换(XSLT multiple string replacement with recursion)

用递归进行XSLT多字符串替换(XSLT multiple string replacement with recursion)

我一直在试图用递归执行多个(不同的)字符串替换,并且遇到了障碍。 我已经成功地获得了第一个替代品,但随后的替代品从未开火。 我知道这与递归有关,以及with-param字符串如何传递回调用模板。 我看到我的错误,为什么下一个xsl:当永远不会触发,但我似乎无法弄清楚如何将完整的修改后的字符串从第一个xsl:何时传递到第二个xsl:when。 任何帮助是极大的赞赏。

<xsl:template name="replace">
    <xsl:param name="string" select="." />
    <xsl:choose>
        <xsl:when test="contains($string, '&#13;&#10;')">
            <xsl:value-of select="substring-before($string, '&#13;&#10;')" />
            <br/>
            <xsl:call-template name="replace">
                <xsl:with-param name="string" select="substring-after($string, '&#13;&#10;')"/>
            </xsl:call-template>
        </xsl:when>
        <xsl:when test="contains($string, 'TXT')">
            <xsl:value-of select="substring-before($string, '&#13;TXT')" />
            <xsl:call-template name="replace">
                <xsl:with-param name="string" select="substring-after($string, '&#13;')" />
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$string"/>
        </xsl:otherwise>

    </xsl:choose>
</xsl:template>

I have been attempting to perform multiple (different) string replacement with recursion and I have hit a roadblock. I have sucessfully gotten the first replacement to work, but the subsequent replacements never fire. I know this has to do with the recursion and how the with-param string is passed back into the call-template. I see my error and why the next xsl:when never fires, but I just cant seem to figure out exactly how to pass the complete modified string from the first xsl:when to the second xsl:when. Any help is greatly appreciated.

<xsl:template name="replace">
    <xsl:param name="string" select="." />
    <xsl:choose>
        <xsl:when test="contains($string, '&#13;&#10;')">
            <xsl:value-of select="substring-before($string, '&#13;&#10;')" />
            <br/>
            <xsl:call-template name="replace">
                <xsl:with-param name="string" select="substring-after($string, '&#13;&#10;')"/>
            </xsl:call-template>
        </xsl:when>
        <xsl:when test="contains($string, 'TXT')">
            <xsl:value-of select="substring-before($string, '&#13;TXT')" />
            <xsl:call-template name="replace">
                <xsl:with-param name="string" select="substring-after($string, '&#13;')" />
            </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
            <xsl:value-of select="$string"/>
        </xsl:otherwise>

    </xsl:choose>
</xsl:template>

原文:https://stackoverflow.com/questions/5213644
更新时间:2022-11-19 09:11

最满意答案

不,矢量的容量不会因reserveresize而减少。

std::vector::reserve

如果new_cap是[不大于capacity ] ...没有迭代器或引用无效。

std::vector::resize

调整大小到更小时,向量容量永远不会减少,因为这会使所有迭代器无效,而不是只有那些由等效的pop_back()调用序列无效的迭代器。

标准的相关部分


No, the vector's capacity will not be reduced by reserve or resize.

std::vector::reserve:

If new_cap is [not greater than capacity] ... no iterators or references are invalidated.

std::vector::resize

Vector capacity is never reduced when resizing to smaller size because that would invalidate all iterators, rather than only the ones that would be invalidated by the equivalent sequence of pop_back() calls.

Relevant part of the standard

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