Question

有没有办法从greasemonkey重新加载脚本？

例如：当我进入某个特定的网站时，来自greasemonkey的脚本可以正常工作，但是当我更改页面时（我想在网站上输入asp），脚本不会重新加载生效...

我该如何解决？

Answer 1

将Greasemonkey代码包装在函数中，然后设置文档更改事件处理程序来调用它。像这样......

/*--- To "refire" our Greasemonkey code on AJAX changes, we wrap it in
    a function and call it on a DOM change event.
*/

var zGbl_DOM_ChangeTimer                = '';
var bGbl_ChangeEventListenerInstalled   = false;


/*--- Run everything after the document has loaded.  Avoids race-
      conditions and excessive "churn".
*/
window.addEventListener ("load", MainAction, false);


function MainAction ()
{
    if (!bGbl_ChangeEventListenerInstalled)
    {
        bGbl_ChangeEventListenerInstalled   = true;

        /*--- Notes:
                (1) If the ajax loads to a specific node, add this
                    listener to that, instead of the whole body.
                (2) iFrames may require different handling.
        */
        document.addEventListener ("DOMSubtreeModified", HandleDOM_ChangeWithDelay, false);
    }

    //--- ****************************
    //--- *** YOUR CODE GOES HERE. ***
    //--- ****************************
}


function HandleDOM_ChangeWithDelay (zEvent)
{
    /*--- DOM changes will come hundreds at a time, we wait a fraction
          of a second after the LAST change in a batch.
    */
    if (typeof zGbl_DOM_ChangeTimer == "number")
    {
        clearTimeout (zGbl_DOM_ChangeTimer);
        zGbl_DOM_ChangeTimer = '';
    }
    zGbl_DOM_ChangeTimer     = setTimeout (function() { MainAction (); }, 222); //-- 222 milliseconds
}

Answer 2

Greasemonkey应该适用于每个页面加载，所以我认为您遇到的问题是由于用于相关用户脚本的@include / @ exclude规则。你能指点我们这个用户脚本的来源吗？请问你在问题中提到的两个网页网址？

有没有办法重新加载greasemonkey脚本？

2 个答案: