11

I am making a javascript bookmarklet that resizes all images, periodically.

javascript: function x(){
    for(i=0;i<=document.getElementsByTagName('img').length;i++)
        document.getElementsByTagName('img')[i].width+=1;
};
t = window.setTimeout("x()",100);
void(0);

But it runs only once. What is the problem here??

Shubham
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  • What are you trying to accomplish? Also your for loop at the end probably meant to be `i++`. – BoltClock Dec 24 '10 at 04:08
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    You shouldn't do `i<=document.getElementsByTagName('img').length`. You're reselecting all the images in the document as many times as there are images. You should cache the selection. `var images = document.getElementsByTagName('img'); for( var i = 0, len = images.length; i < len; i++ )` Notice also that I did `<` instead of `<=` as well. – user113716 Dec 24 '10 at 04:21
  • seems kind of useless because if you don't stop the timer images will grow forever. – Michiel van der Blonk Feb 24 '18 at 12:59

6 Answers6

31

Are you looking for setInterval() instead of setTimeout() by any chance?

t = window.setInterval("x()",100);
BoltClock
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0

Here is the same code properly indented for clarity

function x() {
    for(i=0;i<=document.getElementsByTagName('img').length;++)
        document.getElementsByTagName('img')[i].width+=1;
};
t = window.setTimeout("x()",100);
void(0);

window.setTimout() executes the passed code only once because it is meant to. If you want to execute code more often, use window.setInterval().

Oswald
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0

Shouldn't it be i++ at the end of your for loop?

epaps
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0

Also there is a syntax error.

for(i=0;i<=document.getElementsByTagName('img').length;i++)

miqbal
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0

You need to put the... 

t = window.setTimeout("x()",100);

inside the function x() brackets { } and it works with SetTimeout()

function x() {
    for(i=0;i<=document.getElementsByTagName('img').length;i++)
        document.getElementsByTagName('img')[i].width+=1;

t = window.setTimeout("x()",100);
};

    x();

void(0);

You can only call x() after all the images have been loaded on the page or there is an error.

gravityboy
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-2

It might be window.setTimeOut("x",100)

Edit : correct the answer to this window.setTimeout(x,100).

PS: thats what happens if you simply work with a IDEs.

user113716
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Biswanath
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