理想情况下,我希望它们都运行,但是当我激活第二个和/或第三个时,脚本 1 会失败。(它什么也不做)
我想知道,是否可以将它们合并到一个脚本中?那能解决问题吗?(我很想尝试将它们剪切并粘贴到一个脚本中,看看会发生什么)
脚本 1(重新加载器)
(function () {
"use strict";
function walkTheDOM(node, func) {
if (node && node.nodeType) {
if (typeof func === "function") {
func(node);
}
node = node.firstChild;
while (node) {
walkTheDOM(node, func);
node = node.nextSibling;
}
}
}
function filterElementsByContains(elements, string) {
var toStringFN = {}.toString,
text = toStringFN.call(elements),
result,
length,
i,
element;
if (text !== "[object NodeList]" && text !== "[object Array]" && !($() instanceof jQuery)) {
return result;
}
result = [];
if (typeof string === "string") {
string = new RegExp("^" + string + "$");
} else if (toStringFN.call(string) !== "[object RegExp]") {
return result;
}
function getText(node) {
if (node.nodeType === 3) {
text += node.nodeValue;
}
}
length = elements.length;
i = 0;
while (i < length) {
text = "";
element = elements[i];
walkTheDOM(element, getText);
if (string.test(text)) {
result.push(element);
}
i += 1;
}
return result;
}
if(!filterElementsByContains([document.getElementsByTagName("table")[0]], /We are proud to announce that the November discounts have been chosen/).length) {
location.reload();
}
}());
脚本 2(跳转到最后一张,如果是多张)
function getPreviousLink(){
var nextLink = document.getElementById('pagination-next-link');
var links = document.getElementsByClassName('v_page_nav')[0].getElementsByTagName("a");
for(var i=0; i < links.length; i++){
if(links[i] == nextLink) { return links[i-1]; }
}
}
var link = getPreviousLink();
link.target="_blank";
link.click();
脚本 3(打开上一张,如果有的话)
var link = document.getElementById('pagination-prev-link');
link.target="_blank";
link.click();