1

嗨，我在 casper.js 中有以下代码

var casper = require('casper').create({
    pageSettings: {loadImages : true,loadPlugins : false}, 
    logLevel :"debug" ,
    verbose : true, 
    onTimeout : function(){ //what to do if timeout reaches?

        this.echo('Failed to load resource.').exit();       
    },
    onStepTimeout: function(){ //what to do if specific step timeout reaches.

        this.echo('timeout: step '+ this.requestUrl);
    }
    });
    //our userAgent


casper.userAgent('Mozilla/5.0 (Windows NT 6.1; Win64; x64; rv:25.0) Gecko/20100101 Firefox/25.0');
casper.echo("Will google.com load in less than 2000 ms?");
casper.options.timeout = 2400100; //4 Minutes for process to complete it self.
casper.options.stepTimeout = 24000; //24 seconds for each step to complete it self.
casper.start("http://www.google.com/", function() {
    this.echo("Google done!");
    this.clear();
});
casper.thenOpen("http://www.bing.com/", function() {
    this.echo("Bing done!");
    this.clear();
});

casper.run(function() { 
this.echo('Finished everything!');
this.exit();
});

现在在这段代码中，我将 stepTimeout 设置为 2400ms ，但是当达到此超时时，脚本会回显超时：step + step name but doesn't move to next instance or step ...我想要的是当达到那个 stepTimeout该实例应立即 exit() 并移至下一个 thenOpen() 实例...有什么办法吗？

4

1 回答 1

1

看看这个https://github.com/n1k0/casperjs/blob/master/samples/steptimeout.js#L17 尝试使用 test.fail 和 test.pass

于 2013-11-21T10:31:45.337 回答