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php - 关于PHP-Column count does not match value count at row 1的错误

问题描述

我是 PHP 新手。我创建了一个登录表单，然后填写了所有输入字段。我单击“提交”，然后它说..“列数与第 1 行的值计数不匹配”。有人帮我解决这个问题。我已经有将近 1 周的时间了。这是我的代码。和我的数据库连接。

<?php
    $connection = mysqli_connect("localhost","root","","db_violation");
    function DBCon(){
        try{
            $connection = new PDO("mysql:host=localhost;port=3306;dbname=db_violation;","root","");
            return $connection;
        }catch(PDOException $e){echo "Connection failed: ".$e->getMessage();}
    }function DBDiscon(){return $connection = null;}
?>

<?php
        include '../../PROCESS/dbconnect/dbconnect.php';

        $datauserssystems = json_decode(file_get_contents("php://input"));

            $user_firstname = mysqli_real_escape_string($connection, $datauserssystems->user_fname);
            $user_middlename = mysqli_real_escape_string($connection, $datauserssystems->user_mname);
            $user_lastname = mysqli_real_escape_string($connection, $datauserssystems->user_lname);
            $user_address = mysqli_real_escape_string($connection, $datauserssystems->user_address);
            $user_positionss = mysqli_real_escape_string($connection, $datauserssystems->user_positions);
            $user_age = mysqli_real_escape_string($connection, $datauserssystems->user_age);
            $user_username = mysqli_real_escape_string($connection, $datauserssystems->user_uname);
            $user_userpass = mysqli_real_escape_string($connection, $datauserssystems->user_pwd);

            $tableRows = DBCon()->prepare("SELECT COUNT (*) AS row FROM sb_account");
            $tableRows->execute();
            $count = $tableRows->fetch(); 
            $count = ($count["row"]+1);
            $gain ="SbAdmin";
            $encrpt = md5($gain.$user_userpass);
            $security = DBCon()->prepare("INSERT INTO sb_account_security VALUES('".$count."','".$encrpt."','".$user_userpass."')")->execute();

            $sbuserssssquery = mysqli_query($connection, "SELECT * FROM sb_account WHERE sb_username='$user_username'");
            $sbuserssssarray = mysqli_fetch_array($sbuserssssquery);
            $sbuserssss = $sbuserssssarray['sb_username'];

            if($sbuserssss == $user_username) {
                echo "sbtwice";
                die();
            }
            else {
            mysqli_query($connection, "INSERT INTO sb_account (sb_firstname,sb_middlename,sb_lastname,sb_fullname,sb_address,sb_position,sb_age,sb_username,sb_password)  VALUES ('$user_firstname','$user_middlename','$user_lastname','$user_address','$user_positionss','$user_age','$user_username','$encrpt')");
            if(mysqli_affected_rows($connection) > 0) {
                echo "AddUserSystem";
                die();
            }
            else {
                echo mysqli_error($connection);
                die();
            }
            }
    ?>

标签： php

解决方案

您的 mysql 插入查询有问题：

INSERT INTO sb_account 
            (sb_firstname, 
             sb_middlename, 
             sb_lastname, 
             sb_fullname, 
             sb_address, 
             sb_position, 
             sb_age, 
             sb_username, 
             sb_password) 
VALUES      ('$user_firstname', 
             '$user_middlename', 
             '$user_lastname', 
             '$user_address', 
             '$user_positionss', 
             '$user_age', 
             '$user_username', 
             '$encrpt');

这里有 9 列，但提供了 8 个值。您没有为sb_fullname列传递值。

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