对chrome进行了简单的扩展。 将“querySelector”更改为“querySelectorAll”时,第二个函数不起作用。querySelectorAll不起作用,但querySelector做了
var ele2 = document.querySelectorAll(".view-count ");
window.onload = function(){
func2();
}
function func2(){
ele2.innerHTML = ele2.innerHTML.trim();
ele2.innerHTML = ele2.innerHTML.slice(0, -14);
ele2.textContent = "$" + ele2.innerHTML.replace(/ /g,' ');
}
解决方案:
for (x=0;x<ele2.length;x++){ele2[x].innerHTML = ele2[x].innerHTML.trim();}
for (x=0;x<ele2.length;x++){ele2[x].innerHTML = ele2[x].innerHTML.slice(0, -14);}
for (x=0;x<ele2.length;x++){ele2[x].textContent = "$" + ele2[x].innerHTML.replace(/ /g,' ');}
谢谢。但我如何选择所有节点?不只是0或1或4,但所有,querySelectorAll可以找到? – pi1yau1u
@ pi1yau1u你将不得不循环他们。 'for(x = 0; x
Musa
是的,这是!太棒了。非常感谢您的参与! – pi1yau1u