我需要帮助来显示PHP中的图像。我想显示数据库的图像和某些文本,稍后再创建一个动态链接并通过ID转到页面。在下面的代码中,您正在查找的图像未显示在浏览器中,但是数据库中的文本显示没有问题...有人可以帮助我吗?谢谢。
<?php
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
echo "<div class='container' id='fadein1'>sds";
echo "<div class='col-sm-4'>";
echo "$row{'image'}";
echo "<img src='upload/destaques/<?php echo [$row{'image'}]; ?>' id='imagem'>";
echo "<a href='testee.php?noticia={$row['id']}'>{$row['titulo']}</a><br>\n";
echo "</div>";
echo "</div>";
}
} else{
echo"<h2>Não existe</h2>";
}
?>
答案 0 :(得分:0)
修复了许多语法错误,应该可以正常工作:
<?php
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
echo "<div class='container' id='fadein1'>sds";
echo "<div class='col-sm-4'>";
echo $row['image'];
echo "<img src='upload/destaques/".$row['image']."' id='imagem'>";
echo "<a href='testee.php?noticia=".$row['id']."'>".$row['titulo']."</a><br>\n";
echo "</div>";
echo "</div>";
}
} else{
echo"<h2>Não existe</h2>";
}
?>