我试图显示图像,但当我单击时它仅显示文件名。我想我的echo <img>
有问题..我如何在 echo 语句中写入图像标签..我还在网络上找到了一些想法,但没有运气。任何帮助将非常感激。有什么想法吗?
<?php
//Create connection and select DB
$conn= mysqli_connect("localhost","root", "", "upload_images");
if(isset($_POST["submit"]))
{
$file_name=$_FILES['fileToUpload']['name'];
$file_tmp=$_FILES['fileToUpload']['tmp_name'];
$file_size=$_FILES['fileToUpload']['size'];
$file_tmp=$_FILES['fileToUpload']['tmp_name'];
$file_error=$_FILES['fileToUpload']['error'];
$datetime=date("Y-m-d H:m:sa");
$target_path="D:\zak/";
if(move_uploaded_file($file_tmp,$target_path.$file_name))
{
echo "Move Success";
}
else
{
echo "Not Move";
}
//Insert image content into database
$query = "INSERT into image_upload (image,created) VALUES ('$file_name','$datetime')";
if (mysqli_query($conn,$query))
{
echo "Registered Sucessfully"."<br>";
}
else
{
echo"OOPS!! Try Again..";
}
$sql1="SELECT image FROM image_upload order by Id desc limit 1";
$result=mysqli_query($conn,$sql1);
if (!$result) {
echo "Error";
}else{
echo "Fetced image";
}
while ($row=mysqli_fetch_assoc($result)) {
$path= $row['image'];
echo "<img src='$path' height='200' width='200'/>";
}
}
?>
最佳答案
尝试使用表格输出图像,也许问题是在调用原始路径时使用 $path
<td><img src="<?php echo $rows['image']?>" width='200px' height='200px'></td>
关于php - 图像未显示 - PHP,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48404858/