将数据插入我的数据库不起作用 - Inserting data into my database isn't working

Inserting data into my database isn't working

我有一个很好的下拉列表，它正在从表团队（FK）填充。唯一不起作用的是将数据添加到匹配项中。我不断收到以下错误：

- team_home not set 
- team_away not set 
- Notice: Undefined index: team_home in vvo/insertmatch.php on line 28
- Notice: Undefined index: team_away in vvo/insertmatch.php on line 28
- Error: You have an error in your SQL syntax; check the manual that corresponds to your    MySQL server version for the right syntax to use near 'matches (team_home, team_away) VALUES ('','')' at line 1

谁能告诉我是什么导致了这些错误？

请参阅下面的代码，我知道它容易受到 sql 注入的攻击，但我只想让它工作。

**addmatch.php**
<?php
$con = mysql_connect("db.xx.nl","md190851db210288","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("md190851db210288", $con);
?>
<form action="insertmatch.php" method="GET">
<select name="team_home">
<?php
$sql = "SELECT team_id, team_name FROM teams ".
"ORDER BY team_name";
$rs = mysql_query($sql);
while($row = mysql_fetch_array($rs))
{
  echo "<option value='"".$row['team_id']."'">".$row['team_name']."</option>'n  ";
}
?>
</select>
<select name="team_away">
<?php
$sql = "SELECT team_id, team_name FROM teams ".
"ORDER BY team_name";
$rs = mysql_query($sql);
while($row = mysql_fetch_array($rs))
{
  echo "<option value='"".$row['team_id']."'">".$row['team_name']."</option>'n  ";
}
?>
</select>
<input type="submit" />

**insertmatch.php**
<?php
  error_reporting(E_ALL);
  ini_set("display_errors", 1);
$con = mysql_connect("db.xx.nl","md190851db210288","xxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (isset($_POST['team_home'])) { 
    echo $_POST['team_home']; 
} else { 
    echo 'team_home not set <br>'; 
}
if (isset($_POST['team_away'])) { 
    echo $_POST['team_away']; 
} else { 
    echo 'team_away not set <br>'; 
}  
mysql_select_db("md190851db210288", $con);
$sql="INSERT INTO matches (team_home, team_away)
VALUES
('$_POST[team_home]','$_POST[team_away]')";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
echo $team_home;
?>

试试这个

$sql=sprintf("INSERT INTO matches (team_home, team_away)VALUES('%s','%s')",mysql_real_escape_string($_POST['team_home']),mysql_real_escape_string($_POST['team_away']));

你的PHP无效：

$_POST[team_home]

里面的双引号是错误的。

用 {} 括起来，用单引号括起来team_home

{$_POST['team_home']}

其他领域也是如此。作为建议，请注意SQL注入，并在查询中使用$ _POST之前

清理数据

只是想指出你在这里做什么

$sql="INSERT INTO matches (team_home, team_away)
VALUES
('$_POST[team_home]','$_POST[team_away]')";

是非常危险的。不应直接使用任何全局变量。您需要筛选并验证它们，否则您将向应用程序开放 SQL 注入攻击。