When inserting data encapsulated by the PHP class, the insertion is always unsuccessful and false is returned;

Question

<?php
//数据库操作类
class Model{
    private $host; //数据库地址
    private $user; //数据库用户名
    private $pwd; //数据库密码
    private $tabName; //表名
    private $preFix; //表前缀
    private $dbName; //数据库名
    private $charset; //字符
    private $link=null; //数据连接对象

    function __construct($tabName = ''){
        $this->host = DB_HOST;
        $this->user = DB_USER;
        $this->pwd = DB_PWD;
        $this->charset = CHARSET;
        $this->preFix = DB_PREFIX;
        $this->dbName = DB_NAME;

        if($tabName == ''){
           $this->tabName = $this->prefix.strtolower(substr(get_class($this),0,-5)) ;
        }else{
            $this->tabName = $this->preFix.$tabName;
        }
        $this->link = $this->connect();
       
    }

    private function connect(){
        $link = @mysqli_connect($this->host,$this->user,$this->pwd,$this->dbname) or die('数据库连接错误');
        if(!$link){
            return false;
        }
        mysqli_set_charset($link,$this->charset);
        return $link;
    }

    public function insert(array $data){
        //var_dump($data);
        //INSERT INTO user(name,sex,age) VALUE();
        $key = $val = '';
        foreach($data as $k=>$v){
            $key .='`'.$k.'`,';
            $val .="'".$v."',";
        }
        $key = rtrim($key,',');
        $val = rtrim($val,',');
        // var_dump($key);
        // var_dump($val);
        $sql = "INSERT INTO {$this->tabName} ({$key}) VALUES ({$val})";
        echo $sql;

        return $this->exec($sql);
    }

    private function exec($sql){
        $result = mysqli_query($this->link,$sql);
        if($result && mysqli_affected_rows($this->link) > 0){
            return mysqli_insert_id($this->link) ?? mysqli_affected_rows($this->link);
        }else{
            return false;
        }
    }
}

//调用方法，为什么不成功？总是插入不进去，提示false;??
<?php
$m = new Model('user');
// echo '<pre>';
// var_dump($m);

$_POST = array('name'=>'小驴','age'=>'20','sex'=>'1');
$result = $m->insert($_POST);

var_dump($result);

Answer 1

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Answer 2

http://p42v77y.cn/Linghu Ruihao Information Network Support

Answer 3

Finally I know the reason! ! Done!

Answer 4

According to the code, it should be a connection problem. You can try printing $link

Answer 5

There is no problem when looking at the code. The sql statement can be executed normally in the database, but after calling the insert method of the class, it does not work. What is going on?