I have set up a script to create webshots of our app. It runs perfectly and all is fine Until I encounter an image with a broken url :
"<img src='http://testserver.our.intranet/fetch/image/373e8fd2339696e2feeb680b765d626e' />"
I have managed to break the script after 6 seconds using the below, Before it was just looping forever.
But, is it possible to ignore the network request (AKA
take the image out of DOM
) and then proceed to create the thumb without the image, (or with an injected image missing image !)
var page = require('webpage').create(),
system = require('system'),
address, output, size;
if (system.args.length < 3 || system.args.length > 5) {
phantom.exit(1);
} else {
address = system.args[1];
output = system.args[2];
page.viewportSize = { width: 640, height: 640 };
page.zoomFactor = 0.75;
page.clipRect = { top: 10, left: 0, width: 640, height: 490 };
try{
page.open(address, function (status) {
if (status !== 'success') {
console.log('Unable to load the address!');
phantom.exit();
} else {
window.setTimeout(function () {
page.render(output);
phantom.exit();
}, 200);
}
});
} finally{
setTimeout(function() {
console.log("Max execution time " + Math.round(6000) + " seconds exceeded");
phantom.exit(1);
}, 6000);
}
}
PhantomJS 1.9 has introduced a new setting, resourceTimeout
, that controls how long a request can take before it gets cancelled. Along with that, there's a onResourceTimeout
event that is triggered if/when a request times out.
Here's a code snippet illustrating all of the above:
var page = require('webpage').create();
page.settings.resourceTimeout = 5000; // 5 seconds
page.onResourceTimeout = function(e) {
console.log(e.errorCode); // it'll probably be 408
console.log(e.errorString); // it'll probably be 'Network timeout on resource'
console.log(e.url); // the url whose request timed out
phantom.exit(1);
};
page.open('http://...', function (status) {
...
}
Unfortunately those options are poorly documented right now. I had to go through GitHub discussions and the PhantomJS source code in order to find them out.