我正在尝试根据下拉菜单中的值更改图像源,但由于某种原因,我使用的代码似乎不起作用。
HTML
<select id="strapOptions">
<option value="black">Black Strap</option>
<option value="white">White Strap</option>
</select>
<button type="button" onclick="chooseStrapColour()">Apply Strap</button>
的Javascript
function chooseStrapColour() {
var black = "images/straps/oxford-black.png"
var white = "images/straps/oxford-white.png"
var colourRange = document.getElementById("strapOptions").value;
document.getElementById("oxStrap").src = colourRange;
}
如果我将第5行末尾的src =更改为“黑色”或“白色”,那么它确实会更改图像源网址,因此我怀疑第4行中的某些内容已经破坏了它。
答案 0 :(得分:1)
获得java.rmi.ServerException: RemoteException occurred in server thread; nested exception is:
java.rmi.RemoteException: Exception starting the process; nested exception is:
java.rmi.ServerException: RemoteException occurred in server thread; nested exception is:
java.rmi.RemoteException: Initialization error; nested exception is:
com.jacob.com.ComFailException: Can't co-create object
后,其值仅为colourRange
或"black"
,您需要使用它来进一步确定要放入src的字符串,而不是该值:
"white"
答案 1 :(得分:0)
另一种选择就是将颜色附加到图像src中,就像这样
function chooseStrapColour() {
var colourRange = document.getElementById("strapOptions").value;
document.getElementById("oxStrap").src = "images/straps/oxford-"+colourRange+".png";
}
&#13;
<select id="strapOptions">
<option value="black">Black Strap</option>
<option value="white">White Strap</option>
</select>
<button type="button" onclick="chooseStrapColour()">Apply Strap</button>
<img src='' id='oxStrap'/>
&#13;