将值从一个.php传递到另一个.php

时间:2015-07-16 00:59:12

标签: php database

我想将值从一个表单发送到将其发布到数据库的php文件,然后发送到发布相同值的另一个php文件,我该如何使用php执行此操作?

这是我的php

<?php
$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("database_name", $con);

$sql="INSERT INTO players (first_name, last_name, company, phone, email, zip, street, city, state, country, reasons, notes, callback)
VALUE
('$_POST[first_name]','$_POST[last_name]','$_POST[company]','$_POST[phone]','$_POST[email]','$_POST[zip]','$_POST[street]','$_POST[city]','$_POST[state]','$_POST[country]','$_POST[reasons]','$_POST[notes]','$_POST[callback]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con)
?>

感谢您的帮助。 (我很抱歉,如果这是重复的,我没有找到任何相关内容)

1 个答案:

答案 0 :(得分:2)

在submit.php文件中

<?php
    $con = mysql_connect("localhost","user","password");
    if (!$con)
      {
      die('Could not connect: ' . mysql_error());
      }

    mysql_select_db("database_name", $con);

    $sql="INSERT INTO players (first_name, last_name, company, phone, email, zip, street, city, state, country, reasons, notes, callback)
    VALUE
    ('$_POST[first_name]','$_POST[last_name]','$_POST[company]','$_POST[phone]','$_POST[email]','$_POST[zip]','$_POST[street]','$_POST[city]','$_POST[state]','$_POST[country]','$_POST[reasons]','$_POST[notes]','$_POST[callback]')";

    if (!mysql_query($sql,$con))
      {
      die('Error: ' . mysql_error());
      }
    //Add more
    $id = mysql_insert_id();
    mysql_close($con);
header('Location: review.php?id=' . $id);
    ?>

然后从review.php文件中获取数据

<?php
  $id = $_GET['id'];
  //SELECT * FROM tbl ... WHERE id = $id
  //Your code
?>