javascript - 将电子邮件地址和 url 作为获取参数发送到 laravel

Question

我想通过路由参数将用户的电子邮件地址和图像 URL 发送到 Laravel。我正在尝试在我的 javascript 中遵循 -

var name = resp.displayName;
  var id = resp.id;
  var img = encodeURIComponent(resp.image.url);
  var email = encodeURIComponent(resp.emails[0].value);
  //save in the database now
  window.location.href = "/login/" + id + "/" + name + "/" + img + "/" + email;

在 Laravel 中我将它们视为 -

Route::get('/login/{id}/{name}/{img}/{email}', function($id, $name, $img, $email){
  return "Hello";
});

但我每次都会收到 404 未找到页面。但是，如果我确实喜欢 - localhost/login/a/b/c/d 那么事情就可以了

请告诉我是否有某种单独的方法可以将 url 作为 get 参数发送。

谢谢

Answer 1

写下这两行:-

 var BaseUrl = '<?php echo url(); ?>';

 window.location.href = BaseUrl+"/login/" + id + "/" + name + "/" + img + "/" + email;

您需要使用 base64 对您的 img url 进行编码。

试试这个代码:-

<script>
        // Create Base64 Object
        var Base64={_keyStr:"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=",encode:function(e){var t="";var n,r,i,s,o,u,a;var f=0;e=Base64._utf8_encode(e);while(f<e.length){n=e.charCodeAt(f++);r=e.charCodeAt(f++);i=e.charCodeAt(f++);s=n>>2;o=(n&3)<<4|r>>4;u=(r&15)<<2|i>>6;a=i&63;if(isNaN(r)){u=a=64}else if(isNaN(i)){a=64}t=t+this._keyStr.charAt(s)+this._keyStr.charAt(o)+this._keyStr.charAt(u)+this._keyStr.charAt(a)}return t},decode:function(e){var t="";var n,r,i;var s,o,u,a;var f=0;e=e.replace(/[^A-Za-z0-9\+\/\=]/g,"");while(f<e.length){s=this._keyStr.indexOf(e.charAt(f++));o=this._keyStr.indexOf(e.charAt(f++));u=this._keyStr.indexOf(e.charAt(f++));a=this._keyStr.indexOf(e.charAt(f++));n=s<<2|o>>4;r=(o&15)<<4|u>>2;i=(u&3)<<6|a;t=t+String.fromCharCode(n);if(u!=64){t=t+String.fromCharCode(r)}if(a!=64){t=t+String.fromCharCode(i)}}t=Base64._utf8_decode(t);return t},_utf8_encode:function(e){e=e.replace(/\r\n/g,"\n");var t="";for(var n=0;n<e.length;n++){var r=e.charCodeAt(n);if(r<128){t+=String.fromCharCode(r)}else if(r>127&&r<2048){t+=String.fromCharCode(r>>6|192);t+=String.fromCharCode(r&63|128)}else{t+=String.fromCharCode(r>>12|224);t+=String.fromCharCode(r>>6&63|128);t+=String.fromCharCode(r&63|128)}}return t},_utf8_decode:function(e){var t="";var n=0;var r=c1=c2=0;while(n<e.length){r=e.charCodeAt(n);if(r<128){t+=String.fromCharCode(r);n++}else if(r>191&&r<224){c2=e.charCodeAt(n+1);t+=String.fromCharCode((r&31)<<6|c2&63);n+=2}else{c2=e.charCodeAt(n+1);c3=e.charCodeAt(n+2);t+=String.fromCharCode((r&15)<<12|(c2&63)<<6|c3&63);n+=3}}return t}}

        // Encode the String
        var encodedImgUrl = Base64.encode(resp.image.url);
        //var encodedImgUrl = Base64.encode("https://upload.wikimedia.org/wikipedia/en/3/34/ISync_icon.png");
        var name = resp.displayName;
        var id = resp.id;          
        var email = Base64.encode(resp.emails[0].value);
        var BaseUrl = "<?php echo url(); ?>";        

        window.location.href = BaseUrl+ "/login/" + id + "/" + name + "/" + encodedImgUrl + "/" + email;
 </script>

您可以通过 echo base64_decode($img); 获取路由中的图像 url

希望这会起作用:)