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CasperJS - How to open up all links in an array of links

I'm trying to make it so that CasperJS will open up every link in an array of links. I have it so that after I open a link, it will display the title of that page. Yet when I run it, nothing is displayed.

I can use a for loop to display the links and it works perfectly.

This is the code for what I just explained:

var x;

casper.start(URL, function() {

    x = links.split(" "); // now x is an array of links

    for (var i = 0; j < x.length; i++) // for every link...
    {
        casper.thenOpen(partialURL + x[i], function() { // open that link
            console.log(this.getTitle() + '\n'); // display the title of page
        });
    }

    this.exit();
});

casper.run();

This is another method I tried:

var x;

casper.start(URL, function() {
    x = links.split(" "); // now x is an array of links
    this.exit();
});

for (var i = 0; j < x.length; i++) // for every link...
{
    casper.thenOpen(partialURL + x[i], function() { // open that link
        console.log(this.getTitle() + '\n'); // display the title of page
    });
}

casper.run();

It says that 'x' in undefined. Notice that I set x to be a global variable though. Any modifications that you could make would be great. Thanks.

like image 650
Michael Yaworski Avatar asked Jul 29 '13 14:07

Michael Yaworski


1 Answers

var x; var i = -1;

casper.start(URL, function() {
    x = links.split(" "); // now x is an array of links
});

casper.then(function() {
    this.each(x, function() { 
        i++; // change the link being opened (has to be here specifically)
        this.thenOpen((partialURL + x[i]), function() {
            this.echo(this.getTitle()); // display the title of page
        });
    });
});

casper.run();
like image 140
Michael Yaworski Avatar answered Oct 23 '22 18:10

Michael Yaworski