来自php类报告的mysql错误报告%0

Question

我试图通过一个类中的函数连接到MySql。同样的事情可以做得更容易，但这也是一次学习的经历。代码全部按预期工作，但是如果MySql查询失败，mysqli_error()返回空白，mysqli_errno()返回0。我在手动将数据输入mysql时发现了错误，并且它在db中是一个太短的列，但是我不明白为什么mysqli_error()和mysql_errno()都没有报告这个错误。提前感谢您的任何帮助。

<?php

class FrmAddProduct {

    protected function getDbConnection(){
        return new mysqli("localhost", "root", "****", "test");
    }

    /**
    *all variables are declares and assigned here
    *
    **/
    function commitSignUp (){
        $commitSignUp = "INSERT INTO Login ( `logTitle`,`logFirstName`, `logLastName`,`logLandLine`,) VALUE (\"$this->title\", \"$this->firstName\",\"$this->lastName\", \"$this->landLine\",)";
        if ($this->getDbConnection()->query($commitSignUp)){
            echo "The new product is added.";
            return 1;
        } else {
            echo "Mysql reported this error description: ".mysqli_error($this->getDbConnection()."<br>";
                echo "Mysql reported this error number: ".mysqli_errno($this->getDbConnection());
                return 0;
        }

Answer 1

不管你在这里做什么都是完全错误的做法。您正在两次或三次创建连接实例。并且您期望在那里写入错误。

以这种方式更改您的代码。

protected function getDbConnection(){
    return mysqli_connect("localhost", "root", "****", "test");
}

也这样做

function commitSignUp (){
      $commitSignUp = "INSERT INTO
                           Login (
                                     `logTitle`,
                                     `logFirstName`,
                                     `logLastName`,
                                     `logLandLine`,
                                   )
                            VALUE (
                                      \"$this->title\",
                                      \"$this->firstName\",
                                      \"$this->lastName\",
                                      \"$this->landLine\",
                       )";

            $conn = $this->getDbConnection();

            if ($conn->query($commitSignUp)) {
                echo "The new product is added.";
                return 1;
            } else {
                echo "Mysql reported this error description: ".mysqli_error($conn."<br>";
                echo "Mysql reported this error number: ".mysqli_errno($conn);
                return 0;
            }