Javascript array problem

[ArthurH]

var Dogs = new Array();
Dogs[0] = "Raven";
Dogs[1] = "Arek";
Dogs[2] = "Demo";
Dogs[3] = "Haley";
// Images = number of images
var Images = new Array();
Images[0] = 5;
Images[1] = 7;
Images[2] = 8;
Images[3] = 4;

function initDog(dog) {
if(dog === "") { dog='Raven'; }
imgSetNum = Dogs.indexOf(dog);
imgMax = Images[imgSetNum];
imgSetNum ++;
}

in the function initDog, imgSetNum = Dogs.indexOf(dog) always return -1 (not found). I verified that the "dog" value is correctly set to a dog name in the Dogs array.

What am I doing wrong?

[007Julien]

How do you call initDog ? with intDog() or with initDog("") ?
if (!dog) { dog='Raven'; } could be better as if(dog === "") {dog='Raven';} to work with both syntaxes...

ImgMax and imgSetNum are now local variable in the function. They have to be defined with a var imgMax,imgSetNum; outside to be global variables !

[ReFreezed]

The code works perfectly for me. The error is outside the code you posted I believe.

ImgMax and imgSetNum are now local variable in the function. They have to be defined with a var imgMax,imgSetNum; outside to be global variables !

That's not true. You have to use the 'var' keyword to make a local variable, or else it will be global by default. (Variables should be declared with var somewhere though for clarity and to give it a better closure than the global scope.)

[ArthurH]

The var ingSetNum and imgMax are defined locally.
The value of 'dog' is passed using php:
onload="initDog(<?php echo "'".$_GET['dogname']."'"?>);imgNum=0;nextImg()">

I verified that 'dog' really contains the dog name listed in the array Dogs.

[ArthurH]

Problem Solved !

The onload="initDog(<?php echo "'".$_Get['dogname'."'" ?>" correctly makes the call to the function initDog as:

"initDog('dognamevalue')" ---notice the single quotes.

The function initDog(dog) stores the 'dognamevalue' with the single quotes!!!!!

The solution is to remove the single quote:

dog = dog.substr(1,dog.lastIndexOf("'")-1);

The whole function now works correctly.