Question

我有一个wordpress插件，允许我将人们发送到新的网址。而不是在同一页面上访问该URL并让用户离开我的网站。我想要一个新的窗口，当他们去网址时打开

如何编辑此代码以便打开新窗口？

  <?php
    /**
  Plugin Name: [WPJB] Application URL
   Version: 1.1
  Author: Greg Winiarski
 Description: Allows to set application URL on per job basis.
*/

// Plugin activation
register_activation_hook( __FILE__, 'wpjb_aurl_register');

function wpjb_aurl_register() {
// registers new custom field in wp_wpjb_meta table
wpjb_meta_register("job", "application_url");
}

// Plugin deactivation
register_deactivation_hook( __FILE__, 'wpjb_aurl_unregister');

function wpjb_aurl_unregister() {
// do nothing ...
}

// Adding custom fields to the form
add_filter("wpjb_form_init_job", "wpjb_aurl_init_job");
add_filter("wpja_form_init_job", "wpjb_aurl_init_job");

function wpjb_aurl_init_job($form) {

// adding new fields to the form
$e = $form->create("application_url", "text");
$e->setBuiltin(false);
$e->setLabel("Application URL");
$e->addFilter(new Daq_Filter_WP_Url);
$e->addValidator(new Daq_Validate_Url);
$e->setValue($form->getObject()->meta->application_url->value());
$form->addElement($e, "company");

return $form;
}


add_action("wpjb_front_pre_render", "wpjb_aurl_pre_render", 10, 2);

 function wpjb_aurl_pre_render($app, $file) {
 global $post_type;

try {
    $c1 = is_wpjb() && wpjb_is_routed_to("index.single");
} catch(Exception $e) {
    $c1 = false;
}
$c2 = $post_type == "job";

if(!$c1 && !$c2) {
    return;
}

$job = $app->controller->view->job;

if(!$job->meta->application_url->value()) {
    return;
}

$app->controller->view->application_url = $job->meta->application_url-    >value();

}

?>

Answer 1

你不能仅仅使用PHP来完成它，但是你可以改为重定向到新的url echo javascript，它将在加载将在新窗口或标签中打开url的页面时触发。

使用类似

的内容

Echo "<script> var win = window.open(".$job->meta->application_url->value().", '_blank'); win.focus();</script>";

在您的PHP脚本中可能位于您现在的位置

$app->controller->view->application_url = $job->meta->application_url->value();

如何让php代码弹出一个新窗口？

1 个答案: