我正在使用Angular创建具有LinkedIn身份验证的登录页面,并在登录LinkedIn后获得了该URL:
https://example.com/welcome?code=AQAewReth0-dsds333ffdsAd_HK-CssJdsflds2b1eMsFDWKL3Hu39ssffSgdTxEsum7fT9zs9MMmjY4JUN6dJ8YGnf4tClItIR0noyeW3yX_4_3YKf1yidhxzJxnwnIQ7z59Y3WpCLv4Rdsdsfsf10qUFeO1wwbtQsfsf1FkJsAi9JIxnM8w&state=dse22ds553sf
如何通过上述URL,从code
URL查询参数中获取字符串?
答案 0 :(得分:2)
您可以在code=
上拆分URL字符串,然后像这样抓住字符串的右侧:
const url = 'https://example.com/welcome?code=AQAewReth0-dsds333ffdsAd_HK-CssJdsflds2b1eMsFDWKL3Hu39ssffSgdTxEsum7fT9zs9MMmjY4JUN6dJ8YGnf4tClItIR0noyeW3yX_4_3YKf1yidhxzJxnwnIQ7z59Y3WpCLv4Rdsdsfsf10qUFeO1wwbtQsfsf1FkJsAi9JIxnM8w&state=dse22ds553sf';
const code = url.split('code=')[1]; // This will return AQAewReth0-dsds333ffdsAd_HK-CssJdsflds2b1eMsFDWKL3Hu39ssffSgdTxEsum7fT9zs9MMmjY4JUN6dJ8YGnf4tClItIR0noyeW3yX_4_3YKf1yidhxzJxnwnIQ7z59Y3WpCLv4Rdsdsfsf10qUFeO1wwbtQsfsf1FkJsAi9JIxnM8w&state=dse22ds553sf