小编Pie*_*rre的帖子

如何将xmlns:*属性与XSLT 1.0相匹配？我尝试使用RDF文档:

<xs:template match="rdf:RDF">
(...)
<xsl:for-each select="@*">
  <xsl:value-of select="."/>
</xsl:for-each>
(...)
</xsl:template>

但它似乎不适用于xmlns属性.

谢谢.

下面的代码是对CURL C API 的测试.问题是write_callback永远不会调用回调函数.为什么？

/** compilation: g++ source.cpp -lcurl */

#include <assert.h>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <curl/curl.h>

using namespace std;

static size_t write_callback(void *ptr, size_t size, size_t nmemb, void *userp)
{
    std::cerr << "CALLBACK WAS CALLED" << endl;
    exit(-1);
    return size*nmemb;
}

static void test_curl()
{
    int any_data=1;
    CURLM* multi_handle=NULL;
    CURL* handle_curl = ::curl_easy_init();
    assert(handle_curl!=NULL);
    ::curl_easy_setopt(handle_curl, CURLOPT_URL, "http://en.wikipedia.org/wiki/Main_Page");
    ::curl_easy_setopt(handle_curl, CURLOPT_WRITEDATA, &any_data);
    ::curl_easy_setopt(handle_curl, CURLOPT_VERBOSE, 1);
    ::curl_easy_setopt(handle_curl, CURLOPT_WRITEFUNCTION, write_callback);
    ::curl_easy_setopt(handle_curl, CURLOPT_USERAGENT, "libcurl-agent/1.0");

    multi_handle = …

我正在尝试扩展/翻译java.awt.使用AffineTransform进行形状,以便在定义的边界Rectangle中绘制它.

此外,我想在具有" 缩放 "参数的绘图区域中绘制它.

我尝试了各种AffineTransform连接,但我找不到正确的序列.例如,以下解决方案是错误的:

double zoom=(...);/* current zoom */
Rectangle2D viewRect=(...)/** the rectangle where we want to paint the shape */
Shape shape=(...)/* the original shape that should fit in the rectangle viewRect */
Rectangle2D bounds=shape.getBounds2D();

double ratioW=(viewRect.getWidth()/bounds.getWidth());
double ratioH=(viewRect.getHeight()/bounds.getHeight());


AffineTransform transforms[]=
    {
    AffineTransform.getScaleInstance(zoom, zoom),
    AffineTransform.getTranslateInstance(-bounds.getX(),-bounds.getY()),
    AffineTransform.getTranslateInstance(viewRect.getX(),viewRect.getY()),
    AffineTransform.getScaleInstance(ratioW, ratioH)
    };


AffineTransform tr=new AffineTransform();
for(int i=0;i< transforms.length;++i)
    {
    tr.concatenate(transforms[i]);
    }

Shape shape2=tr.createTransformedShape(shape);
graphics2D.draw(shape2);

关于正确的AffineTransform的任何想法？

非常感谢

皮埃尔