【发布时间】：2018-05-11 10:04:49
【问题描述】：
在一个更大的应用程序的一部分中，我希望在我的 python 脚本中（单击我稍后将执行的按钮），它将特定图像发送到可通过 IP 访问的 PHP（postgresql）服务器地址（为了演示目的，它是http://12.34.56.78/images/upload.php（实际上是不同的IP）。
现在，当我单击上面的链接时，我可以将图像添加到我的数据库中：它会将我引导至此页面 see screenshot here 使用此代码显示：
<html>
<head>
</head>
<body style="border:2px solid Black; padding: 20px; border-radius: 8px; background-color: Teal;">
<center>
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="upload" />
<br/><br/>
<input type="submit" name="submit" value="upload">
<br/><br/>
<?php
echo '<a href = "http://' . $_SERVER['SERVER_NAME'] . '/images/list.php">Return to the Index.</a>';
?>
</form>
</center>
</body>
但是，如上所述，我希望我的 python 脚本自动将图像上传到我的数据库，而不是手动选择文件并单击上传。
这就是我目前的python文件：
import requests
url = 'http://12.34.56.78/images/upload.php'
files = {'file': open('The_Great_Archiver.png', 'rb')}
r = requests.post(url, files=files)
我目前上传图片的 .php 脚本是：
<?php
include "include/connect.php";
if(isset($_POST['submit']))
{
$ip = $_SERVER['SERVER_NAME'];
$filename = $_FILES['upload']['name'];
$filetmp = $_FILES['upload']['tmp_name'];
$filebase = basename($_FILES['upload']['name']);
$len = strlen($filebase);
$arr1 = str_split($filebase, 1);
$finalfilebase = "";
$filetypes = array('png','jpg','gif');
$ext = pathinfo($filename, PATHINFO_EXTENSION);
if (!in_array($ext,$filetypes)){
echo 'File Type Invalid. Try Again With An Allowed File Type. <br>';
}
else {
echo 'File Scan Valid';
}
for ($i = 0; $i <= $len - 1; $i++){
if (" " == $arr1[$i] || "?" == $arr1[$i] || "!" == $arr1[$i]){
$arr1[$i] = "_";
echo "/!\ Had to modify file name for database.";
}
$finalfilebase = $finalfilebase . $arr1[$i];
}
echo "<br><br>";
$dir = "uploads/";
$final_dir = $dir.$finalfilebase;
if(in_array($ext,$filetypes) && move_uploaded_file($filetmp , $final_dir))
{
echo "Uploading Successful <br><br>";
$size = getimagesize($final_dir);
$img_basename = pathinfo($final_dir, PATHINFO_FILENAME);
$crop_name = $finalfilebase."_crop";
$ratio = $size[0]/$size[1]; #Width / Height
if ($ratio > 1) {
$width = 100*$ratio;
$height = "100";
}
else {
$width = 100*$ratio;
$height = "100";
}
$src = imagecreatefromstring(file_get_contents($final_dir));
$dst = imagecreatetruecolor($width, $height);
imagecopyresampled($dst, $src, 0, 0, 0, 0, $width, $height, $size[0], $size[1]);
if ($ext = "png") {
$dst_addr = "previews/$crop_name.png";
imagepng($dst,$dst_addr);
echo "$dst_addr<br><br>";
}
elseif ($ext = "jpg") {
$dst_addr = "previews/$crop_name.jpg";
imagejpeg($dst,$dst_addr);
}
else {
echo "RESIZE ERROR: $finalfilebase";
}
#Insert into database
$query = pg_query($dbh , "INSERT INTO image_name (file_name , file_path, pre_path) VALUES ('$finalfilebase' , '$final_dir' , '$dst_addr')");
if ($query)
{
echo "Image Inserted <br><br>";
echo '<a href = "http://' . "$ip" . '/images/list.php">Go To Index.</a><br>';
echo '<a href = "http://' . "$ip" . '/images/img_upload_form.php">Insert Another Image.</a>';
}
else
{
echo "Insertion Failed <br><br>";
echo '<a href = "http://' . "$ip" . '/images/list.php">Return To The Index.</a><br>';
echo '<a href = "http://' . "$ip" . '/images/img_upload_form.php">Try Again.</a>';
}
}
else
{
echo "Error While Uploading. Check If Host Is Running. <br><br>";
echo '<a href = "http://';
echo "$ip";
echo '/images/list.php">Go To Index.</a>';
}
}
else
{
echo "Some Error Occurred. Check Code.";
#echo '<a href ="localhost"></a>';
}
?>
<html>
<body style="border:2px solid Black; padding: 20px; border-radius: 8px; background-color: Teal; margin: 50px 100px 50px 100px;">
</body>
</html>
但是，当我执行 python 代码时，它会暂时加载然后完成，但我的数据库中没有添加任何内容。 为了让我的 python 脚本自动将图像正确上传到其他图像已经保存的位置，我需要进行哪些更改？
顺便说一下，这里是数据库如何显示我已经上传的图片的截图：click here
感谢您的帮助！
【问题讨论】：
标签： php python database image upload