Question

我刚刚开始编程，但是我不知道如何将POST响应转换为PHP文件，

Request URL: http://localhost/getEmployees.php
Request Method: POST
Status Code: 200 OK
Remote Address: 127.0.0.1:63342
Referrer Policy: no-referrer-when-downgrade
access-control-allow-credentials: true
access-control-allow-origin: http://localhost:63342
content-length: 4850
Content-type: text/html; charset=UTF-8
server: PhpStorm 2020.1
vary: origin
X-Powered-By: PHP/7.3.6
Accept: application/json, text/javascript, */*; q=0.01
Accept-Encoding: gzip, deflate, br
Accept-Language: en-US,en;q=0.9,es-US;q=0.8,es;q=0.7,sm;q=0.6
Cache-Control: no-cache
Connection: keep-alive
Content-Length: 105
Content-Type: application/x-www-form-urlencoded; charset=UTF-8
Cookie: Phpstorm-7e92279c=4f7ff7c6-1a41-4c9e-90dd-dcc6680dcb42; loginAuthorised=loginAuthorised; companyId=1; userId=10; userName=Claudia+Najera; PHPSESSID=nequ49622cdstv6oh0ahsmis10
Host: localhost:63342
Origin: http://localhost:63342
Pragma: no-cache
Referer: http://localhost:63342/htdocs/flavorite.io/tools/tests.php?_ijt=buaa7pmm8icavombffnis1htfu
Sec-Fetch-Dest: empty
Sec-Fetch-Mode: cors
Sec-Fetch-Site: same-origin
User-Agent: Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/81.0.4044.138 Mobile Safari/537.36
X-Requested-With: XMLHttpRequest
query[generalSearch]: Mario
selectedAllRows: false
requestIds: true
sort[field]: 
sort[sort]: asc

我想做的是获取查询[generalSearch]字符串。我在另一个PHP文件中可以找到结果：

$dbConnection = connection();

$generalSearchJSON = @$_POST['generalSearch'];

$generalSearch = json_decode($generalSearchJSON);

我想使用$ generalSearch变量搜索数据库。但这似乎没有得到结果。有什么建议吗？

Answer 1

您有错字。请从@删除$_POST['generalSearch'];

所以你会得到这个：

$dbConnection = connection();

$generalSearchJSON = $_POST['generalSearch'];

$generalSearch = json_decode($generalSearchJSON);

无法从PHP中的AJAX响应获取数据

1 个答案: