Question

我有一个脚本，可以切换选择表/表单中的所有复选框-但我需要修改为仅切换/选中未禁用的复选框。非常感谢您的帮助。

    <?php

    namespace Tests\Browser;
    use Facebook\WebDriver\Chrome\ChromeOptions;
    use Facebook\WebDriver\Remote\RemoteWebDriver;
    use Facebook\WebDriver\Remote\DesiredCapabilities;

    use Tests\DuskTestCase;
    use Laravel\Dusk\Browser;


    use Illuminate\Foundation\Testing\DatabaseMigrations;

    class ExampleTest extends DuskTestCase
    {

        private $ua = 'Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/60.0.3112.90 Safari/537.36';

        /**
         * A Dusk test example.
         *
         * @return void
         */
        public function testExample()
        {

            $this->browse(function (Browser $browser ) {
                $browser->visit('https://www.whatismybrowser.com/');
            dd("Stop");

            });
        }

        protected function driver()
        {
            $ua = $this->ua;
            $capabilities = DesiredCapabilities::chrome();
            $options = (new ChromeOptions)->addArguments([
                '--disable-gpu',
               // '--headless',
                '--force-device-scale-factor=0.8',
                '--window-size=1920,1080',
            ]);
            $options->setExperimentalOption('mobileEmulation', ['userAgent' => $ua]);

            return RemoteWebDriver::create(
                'http://localhost:9515', $options->toCapabilities()

            );

        }
}

Answer 1

您可以使用querySelectorAll()定位所有允许任何有效CSS selector的 enabled 输入。您还可以使用forEach()遍历更简洁的元素：

function toggle(source) {
  var checkboxes = document.querySelectorAll('input[name="chkInvID"]:enabled');
  [].slice.call(checkboxes).forEach(function(chk){
    chk.checked = source.checked;
  });
}

document.querySelector('input[name="source"]').addEventListener('change', function(){toggle(this)});

<input name="source" type="checkbox"/>source<br>
<input name="chkInvID" type="checkbox" />
<input name="chkInvID" type="checkbox" disabled/>
<input name="chkInvID" type="checkbox" />

Answer 2

您可以通过一个简单的条件来实现：

function toggle(source) {
  checkboxes = document.getElementsByName('chkInvID');
  for(var i=0, n=checkboxes.length;i<n;i++) {
    if(!checkboxes[i].disabled){
        checkboxes[i].checked = source.checked;
    }
  }
}

切换选择所有复选框-未禁用

2 个答案: