这是我显示数据库中的文本并将其替换为图片的代码:
echo"<div><table>";
$role = "SELECT vidRoles FROM videoinformation WHERE id=".$row['id'];
if ($result2 = mysqli_query($con, $role)) {
while ($row3 = mysqli_fetch_row($result2)) {
$values = explode(',',$row3[0]);
foreach($values as $v2)
if (!empty($v2)) {
printf ("<img src='"addVid/roles/%s.jpg'" class='"roleImg img-circle'" >", $v2);
}
}
mysqli_free_result($result2);
}
else{
echo $lang['vldErrorMaRo'];
}
echo "</table></div></div></div>";
代码一切正常没有问题,(样式,功能)但这仅适用于Chrome,而不适用于Mozilla,Opera,IE10...
问题是在其他浏览器中,图片不显示。
必须添加<tbody>
<tr>
和<td>
标签
echo "<div><table><tbody>";
$role = "SELECT vidRoles FROM videoinformation WHERE id=".$row['id'];
if ($result2 = mysqli_query($con, $role)) {
while ($row3 = mysqli_fetch_row($result2)) {
$values = explode(',',$row3[0]);
foreach($values as $v2)
if (!empty($v2)) {
printf ("<tr><td><img src='"%s.jpg'" class='"roleImg img-circle'" alt='"%s.jpg'"></td></tr>", $v2, $v2);
}
}
mysqli_free_result($result2);
}
else{
echo $lang['vldErrorMaRo'];
}
echo "</tbody></table></div></div></div>";
并且不要忘记图像中的替代属性