1
我試圖複製一個圖像交換,但mouseout最終成爲圖像3在三次交換中的每一次,當我只希望它是最後一次。任何幫助我可以弄清楚我如何使互換不同,以便他們不會調用相同的圖像將不勝感激,謝謝!JS圖像交換中的多個圖像,mouseout錯誤圖像
//---imageswap1
if(document.images) {
cars1 = new Array();
cars1[1] = new Image();
cars1[1].src = "car4.png";
cars1[2] = new Image();
cars1[2].src = "car1.png";
}
function swapping_pics(picture_name, value_2) {
document.images[picture_name].src = cars1[value_2].src;
}
//---imageswap2
if(document.images) {
cars2 = new Array();
cars2[1] = new Image();
cars2[1].src = "car5.png";
cars2[2] = new Image();
cars2[2].src = "car2.png";
}
//---imageswap3
function swapping_pics(picture_name, value_2) {
document.images[picture_name].src = cars2[value_2].src;
}
if(document.images) {
cars3 = new Array();
cars3[1] = new Image();
cars3[1].src = "car6.png";
cars3[2] = new Image();
cars3[2].src = "car3.png";
}
function swapping_pics(picture_name, value_2) {
document.images[picture_name].src = cars3[value_2].src;
}
<div id="imageswap1" onMouseOver="swapping_pics('car1',1)" onMouseOut="swapping_pics('car1',2)" href="javascript:void">
<img name="car1" border=」0」 src="car1.png" alt="car1">
</div>
<div id="imageswap2" onMouseOver="swapping_pics('car2',1)" onMouseOut="swapping_pics('car2',2)" href="javascript:void">
<img name="car2" border=」0」 src="car2.png" alt="car2">
</div>
<div id="imageswap3" onMouseOver="swapping_pics('car3',1)" onMouseOut="swapping_pics('car3',2)" href="javascript:void">
<img name="car3" border=」0」 src="car3.png" alt="car3">
</div>
解決!非常感謝你和CSS的技巧也讚賞,你只是救了我三十行左右,我可以清理。 – user2056315 2013-02-09 03:38:17
+1 Toni。並且@ user2056315,別忘了選擇Toni的答案作爲正確 – 2013-02-09 03:39:45
user2056315 CSS可以是一個好夥伴,它的簡單而乾淨的代碼。 順便說一句,謝謝@MichaelPeterson ... – 2013-02-09 03:54:37