未定义的索引：

-2

我得到这个错误：注意：未定义的索引：性别\ form.php在第26.please help..how解决这个错误..？未定义的索引：

如何从形式员工性别男性或女性获得并插入MySQL数据库

**my html code**

<form action="form.php" method="post"> 
 
<table border="1"> 
 
    <tr> 
 
    <td align="center">Form Input Employees Data</td> 
 
    </tr> 
 
    <tr> 
 
    <td> 
 
     <table> 
 
     <form method="post" action="form.php"> 
 
     <tr> 
 
      <td>Name</td> 
 
      <td><input name="employe_name" type="text" id="employe_name" size="20"> 
 
      </td> 
 
     </tr> 
 
     <tr> 
 
      <td>Address</td> 
 
      <td><input name="employe_add" type="text" id="employe_add" size="40"> 
 
      </td> 
 
     </tr> 
 
     <tr> 
 
      <td height="23">Gender</td> 
 
\t \t <td>Male<input type="radio" name="gender"value="m" /> 
 
\t \t Female<input type="radio" name="gender"value="f"/> 
 
\t \t </td></tr> 
 
\t \t 
 
\t \t <td> 
 
      <td align="right"><input type="submit" 
 
      name="submit" value="Submit"> add</td> 
 
     </tr> 
 
     </table> 
 
     </td> 
 
    </tr> 
 
</table></form>

而且我的PHP代码...

<?php 

//Open a new connection to the MySQL server 
$mysqli = new mysqli('localhost','root','','employe'); 

//Output any connection error 
if ($mysqli->connect_error) { 
    die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error); 
}else{ 
    echo"connection good"; 
} 
if(isset($_POST['gender'])) 
{ 

    $gender=$_POST['gender']; 
    echo("your are select" .$gender); 


} 
//get values from form 
$employe_name=$_POST['employe_name']; 
$employe_add=$_POST['employe_add']; 
$gender=$_POST['gender']; 


//insert data into mysql 



$sql = "INSERT INTO employe_detail(employe_name,employe_add,gender) VALUES('$employe_name','$employe_add','$gender')"; 

print '<h3>Insert a record</h3>'; 



if($sql) 
{ 
    print 'Success! ID of last inserted record is : ' .$mysqli->insert_id .'<br />'; 
}else{ 
    die('Error : ('. $mysqli->errno .') '. $mysqli->error); 
} 

$mysqli->close(); 
?>

请帮我............

来源

2015-07-28 sasi kumar

嵌套表格，你也不能查询 –

从

if(isset($_POST['gender'])) 
{ 

    $gender=$_POST['gender']; 
    echo("your are select" .$gender); 


} 
//get values from form 
$employe_name=$_POST['employe_name']; 
$employe_add=$_POST['employe_add']; 
$gender=$_POST['gender'];

要

$gender = ""; 
if(isset($_POST['gender'])) 
{ 

    $gender=$_POST['gender']; 
    echo("your are select" .$gender); 


} 
//get values from form 
$employe_name=$_POST['employe_name']; 
$employe_add=$_POST['employe_add'];

性别将被分配更改您的代码或将保持空白

来源

2015-07-28 12:15:43

这不过是一个带-援助。 –

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