好吧,所以,我试图让用户看到图像的选择,然后单击一个图像来更改表格中的图像。它目前的方式只是添加数组的最后一个图像。我知道这是为什么(我想!),但努力寻找解决方案。看评论。一如既往,所有帮助表示赞赏。显示来自阵列的图像,让用户选择一个并更改节点中的图像
for (r = 0; r < howOften; r++) {
row = table.insertRow(-1);
for (c = 0; c < numDays; c++) {
col = row.insertCell(-1);
img = new Image();
img.src = "../www/images/TEST.png";
col.appendChild(img);
img.onclick = function() {
var myImages = new Array();
myImages[0] = "../www/images/TEST3.png";
myImages[1] = "../www/images/TEST4.png";
myImages[1] = "../www/images/TEST2.png";
for (var i = 0; i < myImages.length; i++) {
var allImages = new Image();
allImages.src = myImages[i]; //I want to display this array to user somehow
//and for them to be able to choose one and for this.src to point to that.
var gogetImages = allImages.src;
this.src = (gogetImages); //I know this is wrong
};
};
};
};
//LATEST VERSION STARTS FROM HERE
function addImage (col) {
var img = new Image(); // Note that a new img variable will be declared each time this function is called
img.src = "../www/images/TEST.png";
col.appendChild(img);
img.onclick = function() {
var myImages = new Array();
myImages[0] = "../www/images/TEST3.png";
myImages[1] = "../www/images/TEST2.png";
myImages[2] = "../www/images/TEST4.png";
for (var i = 0; i < myImages.length; i++) {
var allImages = new Image();
allImages.src=myImages[i];
var newList = document.createElement("ul");
var newContent = allImages;
newList.appendChild(newContent);
my_div = document.getElementById("showPics");
document.body.insertBefore(newList, my_div);
};
allImages.onclick = function(){
alert("the click is working");//it is but only for the last image...grrrrr
//this.src = ????;
};
};
};
for (r = 0; r < howOften; r++) {
row = table.insertRow(-1);
for (c = 0; c < numDays; c++) {
col = row.insertCell(-1);
addImage(col);
};
};
document.getElementById('holdTable').appendChild(table);
};
需要遍历图像数组并为每个图像创建(并追加)一个新图像。点击处理程序会改变主图像'src' – charlietfl
嗨charlietfl,我环绕数组并按照建议追加。用户现在可以看到图像但不能点击添加。请参阅上面的编辑代码。 (最新版本)。谢谢你的帮助。 – Inkers
需要将'allImages.onclcik'移动到'for'循环的正上方,以便在创建每个图像时获取clcik处理程序...因为它只是在循环之后,所以只有最后一个图像才能得到它......可能会考虑使用jQuery代码...可以在大约4-6行代码中编写整个代码并且更易于阅读 – charlietfl