我的代码中出现此错误
警告:输入中出现意外字符:第22行/home/view_album_images.php中的'''(ASCII = 39)状态= 1
这条线有什么问题? 感谢
答案 0 :(得分:2)
您需要删除最后一个单引号:
echo "<img src='upload/".$row['photothumbnail']."' width='200' height='200'/>";
答案 1 :(得分:-1)
echo "<a href='upload/' " . $row['photopreview'] . " data-lightbox='image-1' tile='My caption'";
echo "<img src='upload/' " . $row['photothumbnail'] . " width='200' ";
echo "</a>";