如何在函数/类中传递图像作为参数，或者这是不可能的，如果它是如何解决它？例如：
var tree = new Image();
tree.src = "img/statobj/tree.png"
function additem(dimage){
document.getElementById("myitems").rows[0].insertCell(0).innerHTML ='<div id="invetoryitem" >'+ this.dimage + '</div>'
console.log(dimage) //gets undefined
}
我已经尝试过dimage.src和其他方法，但我使用的似乎没有用：/
像你一样传递它，但要显示它，使用appendChild而不是innerHTML：
var tree = new Image();
tree.src = "https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Ash_Tree_-_geograph.org.uk_-_590710.jpg/220px-Ash_Tree_-_geograph.org.uk_-_590710.jpg";
function showTree(dimage) {
document.getElementById("div").appendChild(dimage);
console.log(dimage);
}
showTree(tree);<div id="div"></div>如果你想使用innerHTML，请使用<img>制作src为dimage.src：
var tree = new Image();
tree.src = "https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Ash_Tree_-_geograph.org.uk_-_590710.jpg/220px-Ash_Tree_-_geograph.org.uk_-_590710.jpg";
function showTree(dimage) {
document.getElementById("div").innerHTML = "<img src='" + dimage.src + "'>";
console.log(dimage);
}
showTree(tree);<div id="div"></div>您必须将该函数命名并通过将图像作为参数传递来调用它。如果您正在使用innerHTML，则必须创建图像标记并将接收到的图像源添加到函数中作为参数
var tree = new Image();
tree.src = "img/statobj/tree.png"
function a(dimage){
document.getElementById("myitems").innerHTML ='<div id="invetoryitem" ><img src="'+dimage.src+ '"></div>'
console.log(dimage) //gets undefined
}
a(tree);<body id="myitems"></body>