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Home Backend Development PHP Tutorial php新人请求知道,关于post传递的问题

php新人请求知道,关于post传递的问题

Jun 23, 2016 pm 02:16 PM

PHP post传递



因为是刚学PHP,所以有很多不懂的地方,所以还请热心的朋友指导下。

回复讨论(解决方案)

不能直接在php标签中写文本内容,要显示要么echo,要么写到标签之外

text.php是什么东西...

后台代码修改为

<?phpecho "Welcome to my world!";echo "Welcome," . $_POST['name'];echo "You are " . $_POST['age'] . " old .";
Copy after login


字符串用双引号或者单引号包裹

后台代码修改为

<?phpecho "Welcome to my world!";echo "Welcome," . $_POST['name'];echo "You are " . $_POST['age'] . " years old .";
Copy after login


字符串用双引号或者单引号包裹
这样是对的, lz那个text.php是干什么的?

楼主说出你的疑问让我帮你解答一下。

4楼的方法是对的。
接收form过来的值,当然用$_POST。
而且在页面中,也是要用 例如:

<?=$_POST['a']?>
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一个不错的参考代码:

<?php/**  * 发送post请求  * @param string $url 请求地址  * @param array $post_data post键值对数据  * @return string  * @edit by www.jbxue.com */  function send_post($url, $post_data) {      $postdata = http_build_query($post_data);    $options = array(      'http' => array(        'method' => 'POST',        'header' => 'Content-type:application/x-www-form-urlencoded',        'content' => $postdata,        'timeout' => 15 * 60 // 超时时间(单位:s)      )    );    $context = stream_context_create($options);    $result = file_get_contents($url, false, $context);      return $result;  }    //使用方法  $post_data = array(    'username' => 'stclair2201',    'password' => 'handan'  );  send_post('http://www.qianyunlai.com', $post_data);          <?php  /**  * Socket版本  * 使用方法:  * $post_string = "app=socket&version=beta";  * request_by_socket('chajia8.com', '/restServer.php', $post_string);  */  function request_by_socket($remote_server,$remote_path,$post_string,$port = 80,$timeout = 30) {    $socket = fsockopen($remote_server, $port, $errno, $errstr, $timeout);    if (!$socket) die("$errstr($errno)");    fwrite($socket, "POST $remote_path HTTP/1.0");    fwrite($socket, "User-Agent: Socket Example");    fwrite($socket, "HOST: $remote_server");    fwrite($socket, "Content-type: application/x-www-form-urlencoded");    fwrite($socket, "Content-length: " . (strlen($post_string) + 8) . "");    fwrite($socket, "Accept:*/*");    fwrite($socket, "");    fwrite($socket, "mypost=$post_string");    fwrite($socket, "");    $header = "";    while ($str = trim(fgets($socket, 4096))) {      $header .= $str;    }      $data = "";    while (!feof($socket)) {      $data .= fgets($socket, 4096);    }      return $data;  }  ?>    <?php  /**   * Curl版本   * 使用方法:   * $post_string = "app=request&version=beta";   * request_by_curl('http://www.qianyunlai.com/restServer.php', $post_string);   */  function request_by_curl($remote_server, $post_string) {    $ch = curl_init();    curl_setopt($ch, CURLOPT_URL, $remote_server);    curl_setopt($ch, CURLOPT_POSTFIELDS, 'mypost=' . $post_string);    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);    curl_setopt($ch, CURLOPT_USERAGENT, "qianyunlai.com's CURL Example beta");    $data = curl_exec($ch);    curl_close($ch);      return $data;  }  ?>
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表单里面是input name='name'

你在$_POST里面也要小写 $_POST['name']

可以这样
echo "welcome".
?>
welcome

you are years old

可以这样

<?phpecho "welcome".?>welcome <?php echo $_POST["name"];?><br />you are <?php echo $_POST['age'];?> years old
Copy after login
Copy after login

可以这样

<?phpecho "welcome".?>welcome <?php echo $_POST["name"];?><br />you are <?php echo $_POST['age'];?> years old
Copy after login
Copy after login

11楼的做法是比较好的做法,就是在HTML中插入php代码

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