Question

我在html中设计了一个注册表单。并尝试使用此PHP代码在mysql数据库中输入。但我无法在数据库中添加记录。 php文件的代码如下所示：

注意：数据库名称为rku_database_main

<?php
        $con = mysqli_connect("localhost","root","","rku_database_main");
        if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
              $us_student_id= $_POST['student_id'];
        $us_user_id= $_POST['user_id'];
        $us_name= $_POST['name'];
        $us_permanent_address= $_POST['permanent_address'];
        $us_local_address= $_POST['local_address'];
        $us_birthdate= $_POST['birthdate'];
        $us_gender= $_POST['gender'];
        $us_email_address= $_POST['email_address'];
        $us_student_cell_number= $_POST['student_cell_number'];
        $us_parents_cell_number= $_POST['parents_cell_number'];     
        $us_password1 = $_POST['password1'];
        $us_password2 = $_POST['password2'];
        mysqli_query($con,"INSERT INTO student_profile(student_id,user_id,name,permanent_address,local_address,password,birthdate,gender,email_address,student_cell_number,parents_cell_number,password1,password2) VALUES ('".$us_student_id."','".$us_user_id."','".$us_name."','".$us_permanent_address."','".$us_local_address."','".$us_birthdate."','".$us_gender."','".$us_email_address."','".$us_student_cell_number."','".$us_parents_cell_number."','".$us_password1."','".$us_password2.")");
        mysqli_close($con);


        if (!mysqli_query($con,$sql))
      {
      die('Error: redcord not added try again' . mysqli_error($con));
      }

      echo "1 record added";
    ?>`

这是我为数据库中的make条目编写的代码。表名是student_profile。字段如代码所示。

我无法将记录输入到数据库表中。输出中的错误显示为“未定义的索引”。

Answer 1

尝试以下代码。

mysqli_query($con,"INSERT INTO `student_profile` (`student_id`,`user_id`,`name`,`permanent_address`,`local_address`,`password`,`birthdate`,`gender`,`email_address`,`student_cell_number`,`parents_cell_number`,`password1`,`password2`) VALUES ('$us_student_id','$us_user_id','$us_name','$us_permanent_address','$us_local_address','$us_birthdate','$us_gender','$us_email_address','$us_student_cell_number','$us_parents_cell_number','$us_password1','$us_password2')");

Answer 2

以这种方式编写查询有助于调试和查找查询错误，正如Fred-ii所提到的，你错过了引用

$qry = "INSERT INTO 
            `student_profile`
            (
                    `student_id`,
                    `user_id`,
                    `name`,
                    `permanent_address`,
                    `local_address`,
                    `password`,
                    `birthdate`,
                    `gender`,
                    `email_address`,
                    `student_cell_number`,
                    `parents_cell_number`,
                    `password1`,
                    `password2`
            ) 
            VALUES 
            (
                    '".$us_student_id."',
                    '".$us_user_id."',
                    '".$us_name."',
                    '".$us_permanent_address."',
                    '".$us_local_address."',
                    '".$us_birthdate."',
                    '".$us_gender."',
                    '".$us_email_address."',
                    '".$us_student_cell_number."',
                    '".$us_parents_cell_number."',
                    '".$us_password1."',
                    '".$us_password2."'
            )" ;

mysqli_query($con,$qry);

Answer 3

首先创建connection.php页面并使用下面的代码。

<?php

$con = mysql_connect("localhost","root","") or die("can not connect");

$db = mysql_select_db("rku_database_main");

?>

现在在页面中添加包含上述代码的连接文件...使用下面的代码添加连接文件。

include('connection.php');

包含此文件后，在下一行使用下面的代码。

<?php
$us_student_id= $_POST['student_id'];
    $us_user_id= $_POST['user_id'];
    $us_name= $_POST['name'];
    $us_permanent_address= $_POST['permanent_address'];
    $us_local_address= $_POST['local_address'];
    $us_birthdate= $_POST['birthdate'];
    $us_gender= $_POST['gender'];
    $us_email_address= $_POST['email_address'];
    $us_student_cell_number= $_POST['student_cell_number'];
    $us_parents_cell_number= $_POST['parents_cell_number'];     
    $us_password1 = $_POST['password1'];
    $us_password2 = $_POST['password2'];

$sql1="INSERT INTO `student_profile` (`student_id`,`user_id`,`name`,`permanent_address`,`local_address`,`password`,`birthdate`,
`gender`,`email_address`,`student_cell_number`,`parents_cell_number`,`password1`,`password2`) 
VALUES ('$us_student_id','$us_user_id','$us_name','$us_permanent_address','$us_local_address','$us_birthdate','$us_gender','$us_email_address','$us_student_cell_number','$us_parents_cell_number','$us_password1','$us_password2')";
$query1=mysql_query($sql1) or die("query failed1".mysql_error());

Answer 4

如果您的数据库表名称和行名称正确，那么这应该有效：

mysqli_query($con,"INSERT INTO `student_profile` (`student_id`,`user_id`,`name`,`permanent_address`,`local_address`,`password`,`birthdate`,`gender`,`email_address`,`student_cell_number`,`parents_cell_number`,`password1`,`password2`) VALUES ('".$us_student_id."','".$us_user_id."','".$us_name."','".$us_permanent_address."','".$us_local_address."','".$us_birthdate."','".$us_gender."','".$us_email_address."','".$us_student_cell_number."','".$us_parents_cell_number."','".$us_password1."','".$us_password2."')");

只需在表格和行名称周围添加'标记，并且您的上一个值已丢失'。

可以肯定的是，您正在从表单中获取值，POST方法。

Answer 5

这可能是由于HTML元素的名称，请检查HTML元素名称和$ _POST ['元素名称']是否相同。

Answer 6

试试这个：

<?php 
 mysqli_query($con,"INSERT INTO student_profile(student_id,user_id,name,permanent_address,local_address,password,birthdate,gender,email_address,student_cell_number,parents_cell_number,password1,password2) VALUES ('$us_student_id','$us_user_id','$us_name','$us_permanent_address','$us_local_address','$us_birthdate','$us_gender','$us_email_address','$us_student_cell_number','$us_parents_cell_number','$us_password1','$us_password2')");

Answer 7

嗨未定义索引意味着数组中缺少某些内容。这意味着，无论你想从$ _POST数组得到什么都不可用。

执行print_r（$ _ POST）并检查是否有正确的数组元素调用。

不能使用PHP代码在mysql数据库中添加记录

7 个答案: