我需要打字机页面。键入css select。
<div class="preloader">
<div class="cssload-speeding-wheel"></div>
</div>
<section id="wrapper" class="login-register">
<div class="login-box">
<div class="white-box">
<form class="form-horizontal form-material" id="loginform" method="post">
<h3 class="box-title m-b-20">Sign In</h3>
<div class="form-group ">
<div class="col-xs-12">
<input class="form-control" type="text" required="" name="login-username" placeholder="Username">
</div>
</div>
<div class="form-group">
<div class="col-xs-12">
<input class="form-control" type="password" required="" name="login-password" placeholder="Password">
</div>
</div>
<div class="form-group">
<div class="col-xs-12">
<div class="form-group">
<div class="col-md-12">
</div>
<div class="form-group text-center m-t-20">
<div class="col-xs-12">
<button type="submit" name="doLogin" class="btn btn-info btn-lg btn-block text-uppercase waves-effect waves-light" type="submit">Log In</button>
</div>
</div>
<div class="form-group m-b-0">
我需要用户名,密码和登录名。
page.type(“ input [@ placeholder ='Username']”,'MYUSERNAME');
此代码不起作用。请帮助我
答案 0 :(得分:0)
选择器中不需要@
,并且占位符是伪元素。您可以为选择器使用另一个属性:
page.type("input[name='login-username']", 'MYUSERNAME');
尽管我建议您为您的元素提供ID,因为这样可以提高选择器的性能:
<input id="username" class="form-control" type="text" required="" name="login-username" placeholder="Username">
然后
page.type('#username', 'MYUSERNAME');
答案 1 :(得分:0)
我认为这对您解决问题会有所帮助。我试图使其更易于理解。这样您就可以阐明您的概念。
/**
* seperate your field information
*
*/
let loginAccount ={
username:"input[name='login-username']",
password:"input[name='login-password']",
loginSubmit:".btn"
}
/** Enter username */
await page.waitForSelector(loginAccount.username);
await page.click(loginAccount.username);
await page.keyboard.type('krdheeraj51');
/** Similiarly for login password */
await page.waitForSelector(loginAccount.password);
await page.click(loginAccount.password);
await page.keyboard.type('enter your password');
/** click your submit button */
await page.waitFor(2000);
await page.waitForSelector(loginAccount.loginSubmit);
await page.click(loginAccount.loginSubmit);