用 php SQL CE 在桌子上显示图像

Question

在过去的两周里，我一直在从事这个项目，但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（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);
}

这是带有 enctype="multipart/form-data" 的表单，它使我可以将图像插入数据库。

<form action="" method="post" enctype="multipart/form-data">
    <input type="file" name="image" />
    <button>Upload</button>
</form>

数据库中图片的字段类型为：longtext

<?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);
?>

这就是我必须在 my.php 页面中显示数据库表的内容

Answer 1

请将图片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);
}

Answer 2

<?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);
?>

谢谢您的帮助！

最好的问候，布鲁诺