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I'm trying to get this to work and I hope this isn't too vague. -Chris

var example = new Array();

var test1 = "one";
var test2 = "two";
var test3 = "three";

for (v = 1; v <= 3; v++) {
    alert(example[test + v])
}

I want the alert to say:

one
two
three

share|improve this question
do you want a single alert to say "one two three" or 3 alerts to say "one" then "two" then "three" ? – Matthew J Morrison Jun 11 '10 at 16:22
I'm not faulting you for not knowing the basics, we are all at different levels here, but I would suggest quickly going through the tutorial at w3schools.com/js and get some of the basics under your belt before asking questions. There is a section on there about arrays as well. – stephenbayer Jun 22 '11 at 15:50

1 Answer

up vote 1 down vote accepted

Your array example isn't initialized anywhere. It should be like this:

var example = new Array();
example[1] = "one";
example[2] = "two";
example[3] = "three";

for (v = 1; v <= 3; v++) {
    alert(example[v]);
}

This will make three separate alerts that say "one", then "two", then "three". If you want a single alert that says "one two three", you should do:

var example = new Array();
example[1] = "one";
example[2] = "two";
example[3] = "three";

var output = "";

for (v = 1; v <= 3; v++) {
    output += " " + example[v];
}

alert(output);

In response to Chris' request for follow-up:

In that case you could nest the loops, but it would be much more efficient to generate the output string once and then loop to display it. Example:

var output = "";
// This variable just makes it happen 3 times, it no longer corresponds to the array size
for (numAlerts = 1; numAlerts  <= 3; numAlerts ++) {

    // This is the actual index for the array
    for (v = 1; v <= example.length; v++) {
        output += " " + example[v];
    }
    alert(output);

    // Clean out the string or it will keep concatenating
    output = "";
}

That example has computational cost of n^2 (technically m x n), because you are nesting the loops. To make 3 alerts, each with the concatenation of 3 array elements, would take 9 significant operations. The next example:

var output = "";
// This variable is once more for array size
for (v = 1; v <= example.length; v++) {
    output += " " + example[v];
}

// This variable controls how many times the alert is printed
for (numAlerts = 1; numAlerts <= 3; numAlerts++) {    
    alert(output);
}

This one has computational cost 2n, reduced to n. In our example, 6 significant actions occur, 3 to concatenate the string, and 3 more to print it. 6 < 9, but not by much. However, imagine you have 300 items in the list, and want to print it 1000 times. Well, now 1300 << 300000.

One last note: in most languages, arrays are zero-based, so it would be example[0], [1], [2], etc. I understand you matched example[1] = "one" but it will make loops confusing if you start at 1. The standard for cycle for an array is:

var example = ["zero", "one", "two", "three"];
for (i = 0; i < example.length; i++) {
    action(example[i]);
}
share|improve this answer
Cool thanks Andy. I'll give that a try... – Chris Chiera Jun 11 '10 at 16:22
@Chris Chiera: You can initialize the array using an array-literal too. ["one", "two", "three"]; would give you the same array in a single line of code. – Andy E Jun 11 '10 at 16:23
Hey guys I have one more question. How would this work if I wanted the output to say: one, two, three one, two, three one, two, three one, two, three – Chris Chiera Jun 11 '10 at 16:41
Sorry, so it would be: line 1: one,two,three line2: one, two, three ect... – Chris Chiera Jun 11 '10 at 16:43
Andy: Thanks again for your help. This will give me something to study for the next week! Cheers! – Chris Chiera Jun 11 '10 at 17:09

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