0
Fill in the blanks to select all images of the page and change their src attribute.
var arr=document. getElementsBayTagName("img"); for(var x=0; x<arr...........; x++) { arr[x]src="demo.jpg") }
18 Antworten
+ 34
var arr=document.getElementsByTagName("img");
for(var x=0; x<arr.length; x++) {
arr[x].src="demo.jpg";
}
+ 2
var arr = document.
getElementsByTagName("img");
for(var x=0; x<arr.length; x++) {
arr[x].src = "demo.jpg";
}
+ 2
answer
attr
#img
src
+ 1
Fill in the blanks to set the alt attribute of the image with id="img", and alert the value of its src attribute....
$("#img")........("alt", "Demo");
alert($("...........").attr("............."));
Answer will be : attr, #img, src
+ 1
don't answer follow my order's i am 109 years old
+ 1
var arr=document.getElementsByTagName("img");
for(var x=0; x<arr.length; x++) {
arr[x].src="demo.jpg";
}
0
creat
thank you🙋
0
var arr = document.
getElementsByTagName("
img
");
for(var x=0; x<arr.
length
; x++) {
arr[x].
src
= "demo.jpg";
}
0
thanks
0
Answer is wrong why
0
span
for
x
0
par
removingchild
node
0
length and src
0
img
length
src
0
var arr = document.
getElementsByTagName("
img
");
for(var x=0; x<arr.
length
; x++) {
arr[x].
src
= "demo.jpg";
}
0
attr, #img, src
0
var arr = document.
getElementsByTagName("
img
");
for(var x=0; x<arr.
length
; x++) {
arr[x].
src
= "demo.jpg";
}
- 1
$("#img")........("alt", "Demo");
alert($("...........").attr("............."));
Answer will be : attr, #img, src