在过去的两周里，我一直在这个项目中工作，我不明白发生了什么。
这就是问题所在：https://i.stack.imgur.com/uvWiJ.jpg
这是我必须将图像插入数据库OIL的代码(它没有错，它将图像插入数据库，这很酷(
$msg = '';
if($_SERVER['REQUEST_METHOD']=='POST'){
$image = $_FILES['image']['tmp_name'];
$img = file_get_contents($image);
$con = mysqli_connect('localhost','root','','oil') or die('Unable To connect');
$sql = "insert into servico (image) values(?)";
$stmt = mysqli_prepare($con,$sql);
mysqli_stmt_bind_param($stmt, "s",$img);
mysqli_stmt_execute($stmt);
$check = mysqli_stmt_affected_rows($stmt);
if($check==1){
$msg = 'Image Successfullly UPloaded';
}else{
$msg = 'Error uploading image';
}
mysqli_close($con);
}
这是enctype＝"0"的形式；多部分/形式数据"；这让我把图像插入数据库。
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="image" />
<button>Upload</button>
</form>
数据库上的图像字段类型为：长文本
<?php
$con=mysqli_connect("localhost","root","","oil");
// Check connection
if (mysqli_connect_errno())
{
echo "Falha ao conectar a MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM servico");
echo "<table border='1'>
<tr>
<th>Obra</th>
<th>Image</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['idObra'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
这就是我必须将数据库的表显示到我的.php页面中的内容
请将图像保存到服务器上名为uploads的文件夹中，并将文件名保存在表中。
$msg = '';
if($_SERVER['REQUEST_METHOD']=='POST'){
$image_name = $_FILES['image']['name'];
$tmp_name = $_FILES['image']['tmp_name'];
move_uploaded_file($tmp_name,'uploads/'.$image_name);
$con = mysqli_connect('localhost','root','','oil') or die('Unable To connect');
$sql = "insert into servico (image) values(?)";
$stmt = mysqli_prepare($con,$sql);
mysqli_stmt_bind_param($stmt, "s",$image_name);
mysqli_stmt_execute($stmt);
$check = mysqli_stmt_affected_rows($stmt);
if($check==1){
$msg = 'Image Successfullly UPloaded';
}else{
$msg = 'Error uploading image';
}
mysqli_close($con);
}
<?php
$con=mysqli_connect("localhost","root","","oil");
// Check connection
if (mysqli_connect_errno())
{
echo "Falha ao conectar a MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM servico");
echo "<table border='1'>
<tr>
<th>Obra</th>
<th>Image</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['idObra'] . "</td>";
echo "<td><img src='uploads/" . $row['image'] . "''/></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
谢谢你的帮助！
致问候，Bruno