<form action="login.php" method="post" enctype="multipart/form-data">
<br><br><br>
select image:<input type="file" name="image" size="40" id="image">
<br><small> must be less than 512kb </small>
<br><br> <input type="submit" name="submit" value="submit">
</form>
没有收到任何错误..仍然无法正常工作...
<?php
if(isset($_POST['submit']))
{
if(getimagesize($_FILES['image'] ['tmp_name'])==FALSE)
{
echo "<script> please insert the image </script>";
exit();
}
else
{
$image=addslashes($_FILES['image'] ['tmp_name']);
$name=addslashes($_FILES['image'] ['name']);
$image= file_get_contents($image);
$image= base64_encode($image);
saveimage($name,$image);
}
}
displayimage();
function saveimage($name,$image)
{
$con=@mysqli_connect("localhost","root","","work");
$qry="insert into pics( name, image) values('$name','$image')";
$result= mysqli_query($con,$qry);
if($result)
{
echo"<br> image uploaded";
}
else
{
echo"<br> image not uploaded";
}
}
function displayimage()
{
$con=@mysqli_connect("localhost","root","","work");
$qry= "select* from pics";
$result= mysqli_query($con,$qry);
**我的问题是这一段时间..但是我不确定.. **
while($row = mysqli_fetch_array($result))
{
$img_type = 'png or jpg';
echo"<img height='250' width='250' src='data:image/".$img_type.";base64', '".$row[1]."' >";
}
mysqli_close($con);
}
?>
不知道该怎么办..请帮助我
最佳答案
echo "<img height='250' width='250' src='data:image/" . $img_type . ";base64," . $row[1] . "' >";
尝试将其替换为@showdev建议的代码
关于php - 基本的64位代码不起作用…无法以html php mysql形式显示数据库中的图像,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/31815043/