我已经准备了这个php webservice
<?php
$user_name = "root";
$user_pass = "";
$host_name = "localhost";
$db_name = "dbupload";
$con = mysqli_connect($host_name,$user_name,$user_pass,$db_name);
if ($con) {
$image = $_POST["image"];
$name = $_POST["name"];
$sql = "insert into imageinfo(name) values ('$name')";
$upload_path = "B:/Smart_Music/Emotion_Model/images/$name.jpg";
if (mysqli_query($con,$sql)) {
file_put_contents($upload_path,base64_decode($image));
$reponse=array();
$reponse["response"]="Image Uploaded Successfully";
echo json_encode($reponse);
}
else {
$reponse=array();
$reponse["response"]="Failure";
echo json_encode($reponse);
}
}
else {
$reponse=array();
$reponse["response"]="Failure";
echo json_encode($reponse);
}
mysqli_close($con);
?>
我想要实现的是在发送响应后自动执行一些Windows命令。
这是命令:
cd /d B:\Smart_Music\Emotion_Model\src
activate Emotion
python image_emotion_demo.py ../images/Selected_Photo.jpg
您可以使用exec or shell_exec
<?php
// Use ls command to shell_exec
// function
$output = shell_exec('ls');
// Display the list of all file
// and directory
echo "<pre>$output</pre>";
?>
输出将是对我来说
gfg.php
index.html
geeks.php
为了安全起见,请确保未在本地计算机上禁用此功能
我尝试过了,正如您提到的,但是没有用系统中没有发生移动到新目录的情况,该目录显示htdocs文件夹的内容
$command_one='cd /d B:\Smart_Music\Emotion_Model\src';
shell_exec($command_one);
$output = shell_exec('dir');
echo "<pre>$output</pre>";