Javascript函数（参数），将参数赋值给var

Question

i am failing and i don't understand why, i hope some body could help me out.

<div id="anvil" onmouseover="show(anvil_info)">

what i want to do is when you hover over anvil it starts a function

function show(x){
x.style.backgroundImage = "image_"+x
}

My idea was to use the assigned parameter when the function was called to get the image, but if i now look try to use it it doesn't work. How do i use a variable assigned to the function?

This one still doens't work:

function show(x, y){
    x.style.zIndex = "1";
    x.style.opacity = "1";
    y.style.backgroundImage = "url('"+ y + "_hover')"
}
<div id="anvil" onmouseover="show(anvil_info, 'anvil')"></div>

Answer 1

new version :

 function show(elem, pIdToChange){ document.getElementById(pIdToChange).style.backgroundImage = "url('"+ pIdToChange + "_hover')"; // with color switch document.getElementById(pIdToChange).style.color = "red"; elem.style.color = ""; }

 <div id="anvil_info" onmouseover="show(this, 'anvil')" >Hello</div> <div id="anvil" onmouseover="show(this, 'anvil_info')">Goodbye</div>

---

The best is to use onmouseover="show(this)"

You'll get the element in your function, and then x.style.backgroundImage will work fine.

 function show(elem, pUrl){ elem.style.backgroundImage = "url('"+ pUrl + "_hover')" alert(pUrl); }

 <div id="anvil" onmouseover="show(this, 'anvil')">hello</div>

Answer 2

function show(x, y, z){
    y.style.zIndex = "1";
    y.style.opacity = "1";
    x.style.backgroundImage = "url('"+ z + "_hover.jpg')"
}

 <div id="anvil_info" onmouseover="show(anvil, this, 'anvil')" onmouseout="hide(anvil, this, 'anvil')"></div>
 <div id="anvil" onmouseover="show(this, anvil_info, 'anvil')" onmouseout="hide(this, anvil_info, 'anvil')"></div>

works great, thanks to blag.