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php接收post数据并写入mysql问题

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Jun 23, 2016 pm 01:47 PM

求专家解答：php如何接收远程post过来的txt文本内容并同时写入mysql数据库呢？求完整代码！合适可加分！
txt文本内容字段格式：
学生id、学生名字、学生邮箱、学生地址

回复讨论(解决方案)

数据表：

CREATE TABLE `student` (  `id` int(10) unsigned NOT NULL auto_increment,  `xid` varchar(20) NOT NULL,  `name` varchar(20) NOT NULL,  `email` varchar(100) NOT NULL,  `address` varchar(100) NOT NULL,  PRIMARY KEY  (`id`)) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

postdemo.php

<?php// 连接数据库  $conn=@mysql_connect("数据库ip","帐号","密码")  or die(mysql_error());  @mysql_select_db('数据库名',$conn) or die(mysql_error());  $action = isset($_REQUEST['action'])? $_REQUEST['action'] : '';if($action=='add'){    $xid = isset($_POST['xid'])? mysql_escape_string($_POST['xid']) : '';    $name = isset($_POST['name'])? mysql_escape_string($_POST['name']) : '';    $email = isset($_POST['email'])? mysql_escape_string($_POST['email']) : '';    $address = isset($_POST['address'])? mysql_escape_string($_POST['address']) : '';    if($xid=='' || $name=='' || $email=='' || $address==''){        echo 'please input data';        exit();    }    $sqlstr = "insert into student(xid,name,email,address) values('".$xid."','".$name."','".$email."','".$address."')";    mysql_query($sqlstr) or die(mysql_error());    header('location:postdemo.php');}else{?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html> <head>  <meta http-equiv="content-type" content="text/html; charset=utf-8">  <title>demo</title> </head> <body>  <form name="form1" method="post" action="postdemo.php">  <p>学生id:<input type="text" name="xid"></p>  <p>学生名字:<input type="text" name="name"></p>  <p>学生邮箱:<input type="text" name="email"></p>  <p>学生地址:<input type="text" name="address"></p>  <p><input type="hidden" name="action" value="add"><input type="submit" name="b1" value="提交"></p>  </form> <?php    $sqlstr = "select * from student order by id";    $query = mysql_query($sqlstr) or die(mysql_error());    $result = array();    while($thread=mysql_fetch_assoc($query)){        $result[] = $thread;    }    if($result){        echo '<table>';        echo '<th>NO</th><th>学生id</th><th>学生名字</th><th>学生邮箱</th><th>学生地址</th>';        foreach($result as $row){            echo '<tr>';            echo '<td>'.$row['id'].'</td>';            echo '<td>'.$row['xid'].'</td>';            echo '<td>'.$row['name'].'</td>';            echo '<td>'.$row['email'].'</td>';            echo '<td>'.$row['address'].'</td>';            echo '</tr>';        }        echo '</table>';    } ?>   </body></html><?php}?>

谢谢专家解答，有个地方没有表述清楚，txt文本内容非表单提交，是由远程客户端直接post过来整个txt文件，例如：
post的文件名：good.txt
文件内容：
001
王军
wangjun@126.com
山东省青岛市滨海路2号

请问专家，后台的接收代码是不是有变化？
使用php怎么模拟post good.txt文件给后台？
先在此谢过啦

只要是使用post方式提交的数据,都能够使用$_POST['变量名']的方式接收.

<?php$content = file_get_contents('php://input');?>

模拟文件上传？

干嘛非要用文件

用正则匹配找出对应数据

你是要上?一?txt文件，然後服?器把txt的?容?入?????？

fdipzone专家好，我没有表述清楚，现重新梳理：txt文件非表单提交，是远程的客户端post过来一个txt文件，并写入mysql数据库。
例如：post的文件名：good.txt，文件内容
001
王军
wangjun@126.com
山东省青岛市滨海路2号

由于是远程客户端直接post good.txt文件，所以提交的数据没有变量名，而是直接的字符串，是不是不能用$_POST接收？

文件流吗？
下面的例子将完成php post文件流，接收文件流，分析文件流，写入数据库四个步骤
1.conn.php 用于连接数据库
2.list.php 用于显示数据库表记录
3.add.php 用于接收流文件并分析写入数据库
4.post.ph 用于发送流文件
5.good.txt 文件，用来发送用，内容是

001王军wangjun@126.com山东省青岛市滨海路2号

假设以上文件都放在根目录，可以用http://localhost/文件名访问到。

首先创建数据库

CREATE TABLE `student` (  `id` int(10) unsigned NOT NULL auto_increment,  `xid` varchar(20) NOT NULL,  `name` varchar(20) NOT NULL,  `email` varchar(100) NOT NULL,  `address` varchar(100) NOT NULL,  PRIMARY KEY  (`id`)) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

conn.php

<?php// 连接数据库  $conn=@mysql_connect("数据库host","帐号","密码")  or die(mysql_error());  @mysql_select_db('数据库名',$conn) or die(mysql_error());  ?>

list.php

<?phprequire "conn.php";?><!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html> <head>  <meta http-equiv="content-type" content="text/html; charset=utf-8">  <title>demo</title> </head>  <body>  <?php    $sqlstr = "select * from student order by id";    $query = mysql_query($sqlstr) or die(mysql_error());    $result = array();    while($thread=mysql_fetch_assoc($query)){        $result[] = $thread;    }     if($result){        echo '<table>';        echo '<th>NO</th><th>学生id</th><th>学生名字</th><th>学生邮箱</th><th>学生地址</th>';        foreach($result as $row){            echo '<tr>';            echo '<td>'.$row['id'].'</td>';            echo '<td>'.$row['xid'].'</td>';            echo '<td>'.$row['name'].'</td>';            echo '<td>'.$row['email'].'</td>';            echo '<td>'.$row['address'].'</td>';            echo '</tr>';        }        echo '</table>';    }  ?>    </body></html>

add.php

<?phprequire "conn.php";$tmpfile = 'tmp.txt'; // 临时文件，用于保存接收到的文件流$content = $GLOBALS['HTTP_RAW_POST_DATA'];if(empty($content)){      $content = file_get_contents('php://input');}file_put_contents($tmpfile, $content, true);  $file = file_get_contents($tmpfile);$data = explode(chr(10), $file); // 分解成数组$xid = mysql_escape_string($data[0]);$name = mysql_escape_string($data[1]);$email = mysql_escape_string($data[2]);$address = mysql_escape_string($data[3]);$sqlstr = "insert into student(xid,name,email,address) values('".$xid."','".$name."','".$email."','".$address."')";    mysql_query($sqlstr) or die(mysql_error());header('location:list.php');?>

post.php

<?php$data = file_get_contents('good.txt'); // 要发送的文件$opts = array(	'http' => array(		'method' => 'POST',		'header' => 'content-type:application/x-www-form-urlencoded',		'content' => $data	));$context = stream_context_create($opts);$url = 'http://localhost/add.php'; // 接收的urlfile_get_contents($url, false, $context);?>

执行步骤如下，访问http://localhost/post.php
然后访问http://localhost/list.php
如果list.php能显示good.txt文件的内容，表示成功，每运行一次post.php，数据表都会增加一条记录。

fdipzone 专家的回复灰常给力，我先试试，晚些时候给分。

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