实际上,我想读取搜索查询之后的内容,当它完成时。问题是URL只接受POST方法,它不采取任何行动与GET方法…
我必须在domdocument或file_get_contents()的帮助下读取所有内容。有没有什么方法可以让我用POST方法发送参数,然后通过PHP读取内容?
实际上,我想读取搜索查询之后的内容,当它完成时。问题是URL只接受POST方法,它不采取任何行动与GET方法…
我必须在domdocument或file_get_contents()的帮助下读取所有内容。有没有什么方法可以让我用POST方法发送参数,然后通过PHP读取内容?
当前回答
我更喜欢这个:
function curlPost($url, $data = NULL, $headers = []) {
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.132 Safari/537.36');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_TIMEOUT, 5); //timeout in seconds
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
curl_setopt($ch, CURLOPT_ENCODING, 'identity');
if (!empty($data)) {
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
}
if (!empty($headers)) {
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
}
$response = curl_exec($ch);
if (curl_error($ch)) {
trigger_error('Curl Error:' . curl_error($ch));
}
curl_close($ch);
return $response;
}
使用的例子:
$response=curlPost("http://my.url.com", ["myField1"=>"myValue1"], ["myFitstHeaderName"=>"myFirstHeaderValue"]);
其他回答
尝试PEAR的HTTP_Request2包来轻松地发送POST请求。或者,您可以使用PHP的curl函数或使用PHP流上下文。
HTTP_Request2还使模拟服务器成为可能,因此您可以轻松地对代码进行单元测试
你可以使用cURL:
<?php
//The url you wish to send the POST request to
$url = $file_name;
//The data you want to send via POST
$fields = [
'__VIEWSTATE ' => $state,
'__EVENTVALIDATION' => $valid,
'btnSubmit' => 'Submit'
];
//url-ify the data for the POST
$fields_string = http_build_query($fields);
//open connection
$ch = curl_init();
//set the url, number of POST vars, POST data
curl_setopt($ch,CURLOPT_URL, $url);
curl_setopt($ch,CURLOPT_POST, true);
curl_setopt($ch,CURLOPT_POSTFIELDS, $fields_string);
//So that curl_exec returns the contents of the cURL; rather than echoing it
curl_setopt($ch,CURLOPT_RETURNTRANSFER, true);
//execute post
$result = curl_exec($ch);
echo $result;
?>
上面的答案对我不起作用。这是第一个完美运行的解决方案:
$sPD = "name=Jacob&bench=150"; // The POST Data
$aHTTP = array(
'http' => // The wrapper to be used
array(
'method' => 'POST', // Request Method
// Request Headers Below
'header' => 'Content-type: application/x-www-form-urlencoded',
'content' => $sPD
)
);
$context = stream_context_create($aHTTP);
$contents = file_get_contents($sURL, false, $context);
echo $contents;
根据主要答案,以下是我使用的方法:
function do_post($url, $params) {
$options = array(
'http' => array(
'header' => "Content-type: application/x-www-form-urlencoded\r\n",
'method' => 'POST',
'content' => $params
)
);
$result = file_get_contents($url, false, stream_context_create($options));
}
使用示例:
do_post('https://www.google-analytics.com/collect', 'v=1&t=pageview&tid=UA-xxxxxxx-xx&cid=abcdef...');
我做了一个函数来请求一个使用JSON的帖子:
const FORMAT_CONTENT_LENGTH = 'Content-Length: %d';
const FORMAT_CONTENT_TYPE = 'Content-Type: %s';
const CONTENT_TYPE_JSON = 'application/json';
/**
* @description Make a HTTP-POST JSON call
* @param string $url
* @param array $params
* @return bool|string HTTP-Response body or an empty string if the request fails or is empty
*/
function HTTPJSONPost(string $url, array $params)
{
$content = json_encode($params);
$response = file_get_contents($url, false, // do not use_include_path
stream_context_create([
'http' => [
'method' => 'POST',
'header' => [ // header array does not need '\r\n'
sprintf(FORMAT_CONTENT_TYPE, CONTENT_TYPE_JSON),
sprintf(FORMAT_CONTENT_LENGTH, strlen($content)),
],
'content' => $content
]
])); // no maxlength/offset
if ($response === false) {
return json_encode(['error' => 'Failed to get contents...']);
}
return $response;
}