如何在新窗口弹出窗口中打开php插件中的此链接

Question

I have this code for pinterest which I want to open in a popup window:

<a target="_blank" href="http://pinterest.com/pin/create/button/?url='.get_permalink().'&media='.wp_get_attachment_url( get_post_thumbnail_id() ).'&description='.get_the_title(  $post->ID ).'" class="sharebtn_pinterest" count-layout="horizontal"></a>';

It is a plugin for WordPress that I am editing.

Edit: Here is the code I would like to work but no popup opens

 {
    echo '<div class="bla"> 
    <a onClick="MyWindow=window.open("http://pinterest.com/pin/create/button/?url='.get_permalink().'&media='.wp_get_attachment_url( get_post_thumbnail_id() ).'&description='.get_the_title(  $post->ID ).',"MyWindow","width=600,height=500") class="sharebtn_pinterest" "/></a></div>';
}

Answer 1

Some javascript to add:

<script language="javascript" type="text/javascript">
<!--
function popitup(url) {
    newwindow=window.open(url,'name','height=200,width=150');
    if (window.focus) {newwindow.focus()}
    return false;
}

// -->
</script>

Edit your link to:

<a onclick="return popitup("http://pinterest.com/pin/create/button/?url='.get_permalink().'&media='.wp_get_attachment_url( get_post_thumbnail_id() ).'&description='.get_the_title(  $post->ID ).'")"
    >Link to popup</a>

Copied from here. 1st result. Did you try searching this on Google first?