我想編輯悉尼主題(wordpress)的滑塊按鈕,爲每個按鈕分別設置鏈接,並找到以下JQuery代碼。此代碼運行良好,但打開當前頁面中的鏈接。現在,我想編輯此代碼以打開新選項卡中的鏈接。我怎麼能這樣做?JQuery代碼打開新標籤中的鏈接
jQuery(function($) {
//Define the new buttons. Delete the lines you don't need
var button1 = '<a href="http://www.google.com" class="roll-button button-slider">Read More</a>'; //Slide 1
var button2 = '<a href="http://www.yahoo.com" class="roll-button button-slider">Read More</a>'; //Slide 2
var button3 = '<a href="http://example.org" class="roll-button button-slider">Button 3</a>'; //Slide 3
var button4 = '<a href="http://example.org" class="roll-button button-slider">Button 4</a>'; //Slide 4
var button5 = '<a href="http://example.org" class="roll-button button-slider">Button 5</a>'; //Slide 5
//Hide the default button
$('.slide-inner a').hide();
//Add the new buttons. Delete the lines you don't need
$('.slide-item:nth-of-type(1) .slide-inner').append(button1); //Slide 1
$('.slide-item:nth-of-type(2) .slide-inner').append(button2); //Slide 2
$('.slide-item:nth-of-type(3) .slide-inner').append(button3); //Slide 3
$('.slide-item:nth-of-type(4) .slide-inner').append(button4); //Slide 4
$('.slide-item:nth-of-type(5) .slide-inner').append(button5); //Slide 5
});
的可能的複製[如何打開在HTML新標籤的鏈接?](http://stackoverflow.com/questions/17711146/how-to-open-link-in-new-tab-on- html) – Sojtin
使用屬性target = _blank在標籤 –
http://stackoverflow.com/questions/17711146/how-to-open-link-in-new-tab-on-html – lordkain