Question

我是新手，我想更新数据库，当我提交表单更新记录时，我收到此错误

Notice: Undefined index: idno in /Library/WebServer/Documents/practice/employee/edit_employee.php on line 6 Call Stack: 0.0001 633952 1. {main}() /Library/WebServer/Documents/practice/employee/edit_employee.php:0

这是代码

<?php
require_once '../includes/configuration.php';

    if (!isset($_POST['enter']))
    {
        $employee_id_passport = $_GET['idno'];

        $sql_query = "SELECT * FROM employee_master WHERE employee_id_passport = '$employee_id_passport'";
        $result = mysql_query($sql_query, $connection);
        $row = mysql_fetch_assoc($result);


?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http:www.w3.org/TR/xhthml1/DTD/xhtml1-transitional.dtd">

<html xmls="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">

<head>
    <title> </title>

    <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
    <link rel="stylesheet" type "text/css" href="styles/global.css" />
</head>

<body>
    <form name="view_employee" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" />
        Employee ID/Passport: <input type="text" name="id_passport" disabled="disabled" value="<?php echo $row['employee_id_passport']; ?>" /> <br />
        First Name: <input type="text" name="first_name" value="<?php echo $row['first_name']; ?>" /> <br />
        Surname: <input type="text" name="surname" value="<?php echo $row['surname']; ?>" /> <br />
        Mobile Number: <input type="text" name="mobile_number" value="<?php echo $row['mobile_number']; ?>"/> <br />
        <input type="submit" value="Enter" name="submit" />
    </form>
</body>

<html>
<?php
    }
    else
    {
        $_POST['employee_id_passport'] = $employee_id_passport;
        $_POST['first_name'] = $first_name;
        $_POST['surname'] = $surname;
        $_POST['mobile_number'] = $mobile_number;

        $sql_query_update = "UPDATE employee_master SET first_name = '$first_name',  SET surname = '$surname', SET mobile_number = '$mobile_number', WHERE employee_id_passport = '$employee_id_passport'";

        $result = mysql_query($sql_query_update, $connection);
    }


?>

Answer 1

$ _GET超全局中未设置键idno。检查&indo=的查询字符串。

Answer 2

您的代码中存在一些错误。

永远不要相信用户在没有验证和清理数据的情况下输入任何内容。
始终对错误或意外情况进行测试并采取相应措施。
SQL命令中的WHERE子句之前没有逗号。

尝试以下方法：

<?php

require_once '../includes/configuration.php';

if( !isset( $_POST ) ){
 # No Update Form Submission
  if( isset( $_GET['idno'] ) ){
   # ID Number Set for Query
    $employee_id_passport = mysql_real_escape_string( $_GET['idno'] );
    $sql_query = "SELECT * FROM employee_master WHERE employee_id_passport = '$employee_id_passport'";
    if( !( $result = mysql_query( $sql_query , $connection ) ) ){
     # Query Failed
    }elseif( mysql_num_rows( $result )==0 || mysql_num_rows( $result )>1 ){
     # Query Succeeded, but No Rows Returned OR More than One Row Returned
    }else{
      $row = mysql_fetch_assoc( $result );
    }
  }else{
   # No ID Number sent for Query
  }

?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
  "http:www.w3.org/TR/xhthml1/DTD/xhtml1-transitional.dtd">
<html xmls="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>Employee Search</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<link rel="stylesheet" type "text/css" href="styles/global.css" />
</head>

<body>
<?php if( $row ){ ?>
    <form name="view_employee" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" />
        Employee ID/Passport: <input type="text" name="id_passport" disabled="disabled" value="<?php echo $row['employee_id_passport']; ?>" /> <br />
        First Name: <input type="text" name="first_name" value="<?php echo $row['first_name']; ?>" /> <br />
        Surname: <input type="text" name="surname" value="<?php echo $row['surname']; ?>" /> <br />
        Mobile Number: <input type="text" name="mobile_number" value="<?php echo $row['mobile_number']; ?>"/> <br />
        <input type="submit" value="Enter" name="submit" />
    </form>
<?php } ?>
</body>
<html>
<?php
} else {
 # Declare Error Holder
  $error = array();
 # Declare Field Holder
  $field = array();
 # Validate
  if( !isset( $_POST['employee_id_passport'] ) || $_POST['employee_id_passport']=='' )
    $error['employee_id_passport'] = 'No ID/Passport Set';
  elseif( !is_int( $_POST['employee_id_passport'] ) )
    $error['employee_id_passport'] = 'ID/Passport is not a Number';
  else
    $field['employee_id_passport'] = mysql_real_escape_string( $_POST['employee_id_passport'] );

  if( !isset( $_POST['first_name'] ) || $_POST['first_name']=='' )
    $error['first_name'] = 'No First Name Set';
  else
    $field['first_name'] = mysql_real_escape_string( $_POST['first_name'] );

  if( !isset( $_POST['surname'] ) || $_POST['surname']=='' )
    $error['surname'] = 'No First Name Set';
  else
    $field['surname'] = mysql_real_escape_string( $_POST['surname'] );

  if( !isset( $_POST['mobile_number'] ) || $_POST['mobile_number']=='' )
    $error['mobile_number'] = 'No First Name Set';
  else
    $field['mobile_number'] = mysql_real_escape_string( $_POST['mobile_number'] );

  if( !count( $error ) ){
   # Validation was Passed
    $sql_query_update = "UPDATE employee_master
                           SET first_name = '{$field['first_name']}',
                           SET surname = '{$field['surname']}',
                           SET mobile_number = '{$field['mobile_number']}'
                         WHERE employee_id_passport = '{$field['$employee_id_passport']}'";
    if( !( $result = mysql_query($sql_query_update, $connection) ) ){
     # Update Query Failed
    }else{
     # Update Query OK
    }
  }

?>

使用PHP更新MySQL数据库时出错

2 个答案: