SetTimeout arguments problem

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May 2nd, 2006, 04:00 PM

saidbakr

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SetTimeout arguments problem

Hello,
As we know that setTimeout() is used to call a function spontanousely over fixed parts of time as the following manner:
setTimeout(myFunc,1000);
The problem I got is, I want to call the function myFunc() with passing an argument to it as follows,
setTimeout(myFunc('parameter'),1000);
The above code generates an error. So how can I get my goal?

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Best Regards,
Said Bakr.

May 2nd, 2006, 04:26 PM

jnsg

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setTimeout() and setInterval() can take one of two things for their first argument. The first is, as you said, a function reference. The second is a string which gets eval()'ed. So in your example, you'd want to do this:

javascript Code:

Original - javascript Code
setTimeout("myFunc('parameter');", 1000);

setTimeout("myFunc('parameter');", 1000);

If using variables as parameters to a delayed function, keep in mind the following distinction:

javascript Code:

Original - javascript Code
// calls myFunc() with the value param will have 1000ms from now setTimeout("myFunc(param);", 1000); // calls myFunc() with the current value of param, // even if it changes by the time myFunc() actually gets called setTimeout("myFunc(" + param + ");", 1000);

// calls myFunc() with the value param will have 1000ms from now
setTimeout("myFunc(param);", 1000);
 
// calls myFunc() with the current value of param,
// even if it changes by the time myFunc() actually gets called
setTimeout("myFunc(" + param + ");", 1000);

Comments on this post

wsa1971 agrees!

Last edited by jnsg : May 2nd, 2006 at 04:33 PM.

May 4th, 2006, 06:53 AM

KorRedDevil

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Furthermore, it depends on the nature of your parameter, because the expresion is evaluated.

1. If the parameter is a number or a stringed number
var param =5
or
var param ='5'

and you need to pass it as a decimal number
the syntax is:

PHP Code:


		
			
 setTimeout('myFunc('+param+')',3000)

2. But if the parameter is an alphanumeric

var param ='foo';
the parameter has to be quoted (to avoid evaluation) and the syntax becomes:

PHP Code:


		
			
 setTimeout('myFunc("'+param+'")',3000)

In the first case as the parameter is evaluated, even if the variable is in fact a string, such as '5', when passed it becomes a decimal number.

Comments on this post

wsa1971 agrees!

Last edited by KorRedDevil : May 4th, 2006 at 06:56 AM.

May 4th, 2006, 08:21 AM