1 because $_ POST in php cannot get the json value, use the file_get_contents ("php://input") method to get it.
After2 is obtained, the value is obtained:
title=ww&imgFile=%2Fpublic%2Fupload%2Fimage%2F20180712%2F20180712024544_27715.jpg
3 I want to get the value in imgFile =% 2Fpublic%2Fupload%2Fimage%2F20180712%2F20180712024544_27715.jpg
in the value.4 PHP version is 5.5
how do I get it?
use the explode method to process strings
explode ('& 'title=ww&imgFile=%2Fpublic%2Fupload%2Fimage%2F20180712%2F20180712024544_27715.jpg');
parse related functions, handling url parse_url (), processing? The following data parse_str ()
< hr >$url = 'https://img.codeshelper.com/upload/img/2021/03/26/qkpmguhkxq013427';
$urlArray = parse_url($url);
print_r($urlArray);
parse_str($urlArray['query'], $paramList);
print_r($paramList);
exit;result
Array
(
[scheme] => http
[host] => www.abc.com
[path] => /
[query] => title=ww&imgFile=%2Fpublic%2Fupload%2Fimage%2F20180712%2F20180712024544_27715.jpg
)
Array
(
[title] => ww
[imgFile] => /public/upload/image/20180712/20180712024544_27715.jpg
)here is a screenshot of the chrome console where you send the data. If your data is json, the format should not be like this. There is & so here should be the
from post.if you submit json directly, you can convert it to an array using json_decode ('data',true).
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