我想处理对我的这个应用程序中的链接的点击:
I would like to handle a click to the link in this application of mine:
当我点击输出文件"链接时,我希望能够在我的应用程序中生成一个操作.
When I click on the "Output File" link, I would like to be able to generate an action in my application.
截至今天,链接在富文本QLabel中是这样描述的:
As of today, the link is described like this in the rich text QLabel:
<a href="http://google.fr"><span style=" text-decoration: underline; color:#0000ff;">Output File"</span></a>
(由 Qt Designer 生成)
(generated by Qt Designer)
点击后,它会打开默认的网络浏览器去谷歌.那不是我想要的;我想要类似的东西:
When clicked, it will open the default web browser to go to Google. That's not what I want; I'd like something like:
<a href="#browse_output"><span style=" text-decoration: underline; color:#0000ff;">Output File"</span></a>
并且能够检测到点击的链接并做出相应的反应:
And be able to detect the link that's clicked and react accordingly:
(pseudo code)
if( link_clicked.toString() == "#browse_output" ){
on_browse_output_clicked();
}
这在带有 QLabel(或接近的东西)的 Qt 中是可能的吗?怎么样?
Is this possible in Qt with a QLabel (or something approaching) ? How?
好的,有兴趣的朋友,我得到了答案:
Ok, for those interested, I got the answer:
QLabel
openExternalLinks
"属性QLabel
的信号 linkActivated 连接到您的处理程序.openExternalLinks
" property of the QLabel
QLabel
to your handler.就是这样:linkActivated
为您提供了链接在参数中引用的 URL,因此我的伪代码可以完美运行.
That's all: linkActivated
gives you the URL that the link refers to in argument, so my pseudo code works perfectly.
// header
private slots:
void on_description_linkActivated(const QString &link);
// cpp
void KernelBuild::on_description_linkActivated(const QString &link)
{
if( link == "#browse_output" ){
on_outfilebtn_clicked();
}
}
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