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

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...');

其他回答

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"]);

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...');

用PHP发送GET或POST请求的更好方法如下:

<?php
    $r = new HttpRequest('http://example.com/form.php', HttpRequest::METH_POST);
    $r->setOptions(array('cookies' => array('lang' => 'de')));
    $r->addPostFields(array('user' => 'mike', 'pass' => 's3c|r3t'));

    try {
        echo $r->send()->getBody();
    } catch (HttpException $ex) {
        echo $ex;
    }
?>

代码摘自官方文档http://docs.php.net/manual/da/httprequest.send.php

[编辑]:请忽略,现在在php中不可用。

还有一个你可以用的

<?php
$fields = array(
    'name' => 'mike',
    'pass' => 'se_ret'
);
$files = array(
    array(
        'name' => 'uimg',
        'type' => 'image/jpeg',
        'file' => './profile.jpg',
    )
);

$response = http_post_fields("http://www.example.com/", $fields, $files);
?>

详情请按此处

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;