[英]How to call a php code on html button click
i am working on a project that detects facial expressions in python.but i have to provide image to this code through php.The following php code saves image in directory.how can i call the following code in html button. i am working on a project that detects facial expressions in python.but i have to provide image to this code through php.The following php code saves image in directory.how can i call the following code in html button.
<?php
function fun(){
$img = $_POST['image'];
$folderPath = "C:/xampp/htdocs/";
$image_parts = explode(";base64,", $img);
$image_type_aux = explode("image/", $image_parts[0]);
$image_type = $image_type_aux[1];
$image_base64 = base64_decode($image_parts[1]);
$fileName = '1'. '.jpeg';
$file = $folderPath . $fileName;
file_put_contents($file, $image_base64);
print_r($fileName);
$command = escapeshellcmd("python C:/xampp/htdocs/generate_graph.py");
$output = shell_exec($command);
}
?>
HTML form should be like the blow if your form and PHP code are on the same page.如果您的表格和 PHP 代码在同一页面上,则 HTML 表格应该像打击一样。
<form action="" method="POST">
#Updading based on comment
<button type="submit" class="">Submit</button>
</form>
function fun(){}
function fun(){}
if(isset($_POST)){
$img = $_POST['image'];
$folderPath = "C:/xampp/htdocs/";
$image_parts = explode(";base64,", $img);
$image_type_aux = explode("image/", $image_parts[0]);
$image_type = $image_type_aux[1];
$image_base64 = base64_decode($image_parts[1]);
$fileName = '1'. '.jpeg';
$file = $folderPath . $fileName;
file_put_contents($file, $image_base64);
print_r($fileName);
$command = escapeshellcmd("python C:/xampp/htdocs/generate_graph.py");
$output = shell_exec($command);
}
Since your action is empty then it will hit the current page and if
condition will work because of form submitting as POST
method.由于您的操作为空,因此它将访问当前页面,并且由于表单作为
POST
方法提交, if
条件将起作用。
Hope it will help.希望它会有所帮助。
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