我正在使用jQuery DataTables在spring 4.x应用程序中创建一个语句页面来呈现语句.
通过Jackson消息转换器解析后,服务器返回了以下对象.
{
"data":[
{
"desc":"SUBWAY 00501197 BRONX NYUS",
"amount":"-",
"date":"5.72"
},
{
"desc":"MIDTOWN COMICS NEW YORK NYUS",
"amount":"-",
"date":"73.32"
},
{
"desc":"GOOGLE *PlayDog Soft GOOGLE.COM/CHCAUS",
"amount":"-",
"date":"1.99"
},
{
"desc":"BIMG PRIMARY CARE NEW YORK NYUS",
"amount":"-",
"date":"25.00"
},
{
"desc":"GOOGLE *PlayDog Soft GOOGLE.COM/CHCAUS",
"amount":"-",
"date":"1.99"
},
{
"desc":"GOOGLE *PlayDog Soft GOOGLE.COM/CHCAUS",
"amount":"-",
"date":"1.99"
},
{
"desc":"Walgreens Speci 205 8THNEW YORK NYUS",
"amount":"-",
"date":"10.00"
},
{
"desc":"GOOGLE *SGN Games GOOGLE.COM/CHCAUS",
"amount":"-",
"date":"9.99"
},
{
"desc":"REDBOX *DVD RENTAL 866-733-2693 …Run Code Online (Sandbox Code Playgroud) 我有一个包含两个字段用户名和密码的表单。输入用户名后,下一个按钮被启用,一旦我点击它,它就会显示一个密码字段,一旦输入,它会再次启用下一个按钮。如何等待按钮在表单更新之间启用?
我尝试了以下方法,一种被评论,另一种没有。两者都不适合我。
(async () => {
const browser = await puppeteer.launch({headless: false});
const page = await browser.newPage();
await page.goto('http://localhost:9000/start#!');
await page.type('#login-form-un-field', 'xxxx')
// await page.waitForTarget('#default-next-btn:not([disabled])')
await page.$eval('#default-next-btn:not([disabled])', elem => elem.click());
// const btnNext = await page.$('#default-next-btn');
// btnNext.click();
await page.type('login-form-passcode', '1234');
await page.click('#default-next-btn');
await browser.close();
})();```
Thanks for the help in advance.
Edit: the button is always present on the page. It is just disabled while form entries are being validated.
Run Code Online (Sandbox Code Playgroud) ajax ×1
async-await ×1
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javascript ×1
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puppeteer ×1
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