duanchao1002 2018-04-27 02:19
浏览 25
已采纳

php编程错误[重复]

<?php
if (isset($_POST['value'])) {
    $StudentFirstName = $_POST['FirstName'];
    $StudentLastName = $_POST['LastName'];
    $OrgID = $_POST['OrganizationId'];
    $AmountRaised = $_POST['AmountRaised'];
    $Captain = $_POST['Captain'];
}

$server = "mssql.up.ist.psu.edu";
$connectionInfo = array(
    'Database' => 'pjb5422',
    'UID' => 'sqlpjb5422',
    'PWD' => 'vIPY5De2',
    'Encrypt' => '0',
    'CharacterSet' => 'UTF-8');
$connection = sqlsrv_connect($server, $connectionInfo);

if (!($connection)) {
    echo "Connection could not be established.";
    die(print_r(sqlsrv_errors(), true));
} else {

    $query = "INSERT INTO STUDENT VALUES (?,?, ?, ?, ?, ?)"; //**BUILDING AN INSERT STATEMENT

    $var = array(rand(40, 100000000), $StudentFirstName, $StudentLastName, $OrgId, $AmountRaised, $Captain);

    $sendIt = sqlsrv_query($connection, $query, $var); //submit the query

    echo "<br>Thanks for Registering!!<br><br>Go <a href='MiniThon.html'>Home</a>";
}

?>

I get the following errors and it never submits to my SQL data base

Notice: Undefined variable: StudentFirstName in \UP.IST.LOCAL\WEBSITES\pjb5422\AddDancer1.php on line 50

Notice: Undefined variable: StudentLastName in \UP.IST.LOCAL\WEBSITES\pjb5422\AddDancer1.php on line 50

Notice: Undefined variable: OrgId in \UP.IST.LOCAL\WEBSITES\pjb5422\AddDancer1.php on line 50

Notice: Undefined variable: AmountRaised in \UP.IST.LOCAL\WEBSITES\pjb5422\AddDancer1.php on line 50

Notice: Undefined variable: Captain in \UP.IST.LOCAL\WEBSITES\pjb5422\AddDancer1.php on line 50

</div>
  • 写回答

3条回答 默认 最新

  • douken0530 2018-04-27 02:51
    关注

    change this line

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

    to check other values that are required in your form since value doesn't have anything in it causing your declared variables to be null.

    I also suggest that before executing your insert query you should first check if

    $StudentFirstName

    and other variables are not empty before you execute the insert query.

    else {
        if(!empty($StudentFirstName) && !empty($StudentLastName)){
             $query = "INSERT INTO STUDENT VALUES (?,?, ?, ?, ?, ?)"; //**BUILDING AN INSERT STATEMENT
    
             $var = array(rand(40,100000000),$StudentFirstName, $StudentLastName, $OrgId, $AmountRaised, $Captain);
    
             $sendIt = sqlsrv_query($connection, $query, $var); //submit the query
    
             echo "<br>Thanks for Registering!!<br><br>Go <a href='MiniThon.html'>Home</a>"; 
        }else{
          echo "Failed to insert record empty variables!";
    
        }
    
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 公交车和无人机协同运输
  • ¥15 stm32代码移植没反应
  • ¥15 matlab基于pde算法图像修复,为什么只能对示例图像有效
  • ¥100 连续两帧图像高速减法
  • ¥15 组策略中的计算机配置策略无法下发
  • ¥15 如何绘制动力学系统的相图
  • ¥15 对接wps接口实现获取元数据
  • ¥20 给自己本科IT专业毕业的妹m找个实习工作
  • ¥15 用友U8:向一个无法连接的网络尝试了一个套接字操作,如何解决?
  • ¥30 我的代码按理说完成了模型的搭建、训练、验证测试等工作(标签-网络|关键词-变化检测)