Question

这是一个与wordpress相关的问题。我想转动通过前端提交表单提交的图像，您可以将外部网址提供给链接到其自己的源的图像。对于前者http://i.imgur.com/7qmFBPa.jpg将是点击图片链接的最终结果。下面是php看起来的样子：

<?php $image = wp_get_attachment_image_src( get_post_thumbnail_id( $post->ID ), 'single-post-thumbnail' ); ?>
    <?php if ( has_post_thumbnail() ) { ?>
        <div class = 'span4 reddit-image-single pull-left'>
            <img src = "<?php echo $image[0]; ?>" width = "700px" class="img-rounded">
        </div>
    <?php }else{ ?>
        <div class = 'span4 reddit-image-single pull-left'>
            <img src = "<?php echo get_post_meta( $post->ID, 'wpedditimage', true ); ?>" width = "700px" class="img-rounded">
        </div>
<?php } ?>

我如何修改上述代码？我对PHP很新。

Answer 1

将图片包装在标签中，其中url为href

<?php $image = wp_get_attachment_image_src( get_post_thumbnail_id( $post->ID ), 'single-post-thumbnail' ); ?>
    <?php if ( has_post_thumbnail() ) { ?>
        <div class = 'span4 reddit-image-single pull-left'>
            <a href="<?php echo $image[0];?>"><img src = "<?php echo $image[0]; ?>" width = "700px" class="img-rounded"></a>
        </div>
    <?php }else{ ?>
        <div class = 'span4 reddit-image-single pull-left'>
            <a href="<?php echo get_post_meta( $post->ID, 'wpedditimage', true ); ?>"><img src = "<?php echo get_post_meta( $post->ID, 'wpedditimage', true ); ?>" width = "700px" class="img-rounded"></a>
        </div>
<?php } ?>

这是你要看的主线

<a href="<?php echo $image[0];?>"><img src = "<?php echo $image[0]; ?>" width = "700px" class="img-rounded"></a>

将提交的图像从前端转换为指向其自己的源的链接

1 个答案: