通过javascript弹出从php传递url。

Question

url.php

<?php
include "../googlelogin/src/Google_Client.php";
include "../googlelogin/src/contrib/Google_Oauth2Service.php";
include "../fblogin/src/facebook.php";
//FOR FACEBOOK LOGIN//
    $fbconfig['appid' ]     = "329433909112888";
    $fbconfig['secret']     = "ca2bdc9990b0b2ad0763abcka79dce60c91f";
    $fbconfig['baseurl']    = "http://localhost/sbs/fblogin/index.php?rd=hm";
    $facebook = new Facebook(array(
        'appId'  => $fbconfig['appid'],
        'secret' => $fbconfig['secret'],
        'baseurl'=>$fbconfig['baseurl'], 
        'cookie' => true,
    ));
    $loginUrl   = $facebook->getLoginUrl(
        array('redirect_uri'=>$fbconfig['baseurl'], 
        'scope'  =>            'email,offline_access,publish_stream,user_birthday,user_location,user_work_history,user_about_me,   user_hometown,user_photos ,user_work_history',
    )
    );
//FOR GOOGLE LOGIN//
    $client = new Google_Client();
    $client->setApplicationName("Google UserInfo PHP Starter Application");
    $client->setApprovalPrompt('auto');
    $client->setClientId('777637661406-cp7kekelp1l5i0se9f576sqf36a0q4lc.apps.googleusercontent.com');
    $client->setClientSecret('uztEmqG1CFt06752lqtRZjZ-');   
    $client->setState('hm');
    $client->setRedirectUri('http://localhost/sbs/googlelogin/index.php');
    $client->setDeveloperKey('AIzaSyD3HiHElDciE6Pb5UfZtDdNWe_kiKNk6rg');
    $oauth2 = new Google_Oauth2Service($client);
    $authUrl = $client->createAuthUrl();

    ?>

我想把FACEBOOK，TWITTER和GOOGLE url通过java脚本传入弹出。我知道PHP部分，但我不知道JS部分

pop.js

function sin(){
    var lbox = new LadduBox();
    lbox.init({"width":495, "height":242, "HTML":'<div style="width:495px; height:242px; background-color:#ffffff; border:2px solid orange;"><table cellspacing="0" cellpadding="0" border="0" align="center" width="485" height="152" style="font-family:arial; font-size:20px; font-weight:bold;"><tr><td align="right" colspan="3"><img src="images/untitled-1.png" style="margin:10px; cursor:pointer;" id="btnClose"/></td></tr><tr><td height="30" colspan="3"> <div style="font-family:arial; font-size:20px; color:#ff6c00; padding-left:200px;">SIGN IN</div></td></tr><tr><td><div style="margin:10px; font-size:14px;">EMAIL<br><input type="text"/></div><br><div style="margin-left:10px; margin-top:-10px; font-size:14px;">PASSWORD<br><input type="text"/></div></td><td><img src="images/orbar.png" /></td><td align="center"><img src="images/redfb.png" style="margin-bottom:7px;"/><br><img src="images/redgoogle.png" style="margin-bottom:7px;"/><br><img src="images/redtwitter.png" /></td></tr></div>', 'btnCloseId':'#btnClose'});
    lbox.fire();
}

当用户点击登录按钮时，这个弹出就会显示出来.我希望弹出FACEBOOK、TWITTER和GOOGLE URL，但我不知道如何在javascript .中传递它。

Answer 1

"When the user click on the sign-in button, this pop-up will show":

现在，弹出代码将在PHP文件中。将url.php包含在此文件的顶部，然后您可以访问该文件中任何位置的$authUrl或$loginUrl。

Answer 2

现在，在sin()中，您的lbox.init中包含了HTML。在这里，您可以将这些urls放在下面。但是，从JS文件中删除sin()，并将其放入PHP文件HEAD部分。因为，JS文件不能有PHP代码。

<div style="width:495px; height:242px; background-color:#ffffff; border:2px solid orange;"><a href="<?php echo $loginUrl; ?>">Facebook Login</a><a href="<?php echo $authUrl; ?>">Google Login</a></div>

举个例子，您可以将它放在html中任何您想要的位置。如果要将PHP放入.js文件中，则必须执行以下操作：

AddType application/x-httpd-php .js

AddHandler x-httpd-php5 .js

<FilesMatch "\.(js|php)$">
 SetHandler application/x-httpd-php
</FilesMatch>

在.htaccess文件中添加上述代码，并在js文件中运行php。