Question

我是PHP编程的新手。所以我不明白为什么我的代码不起作用（并且没有显示任何错误）我想在我的表sinhvien中插入新记录这是我的代码：

<?php
$con=new mysqli("localhost","root");
$con->select_db('learn_mysql');
$query="INSERT INTO sinhvien (ho,ten,tuoi) VALUES ('le','van cuong',26)";
$con->query($query);
$con->close();
?>

运行正常但没有记录插入我的表格。

Answer 1

您没有正确连接，请尝试以下操作：

<?php
  $servername = "localhost";
  $username = "root";
  $password = "YOURS";
  $dbname = "learn_mysql";

  $conn = new mysqli($servername, $username, $password, $dbname);
  if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error());  exit(); }

  $sql = "INSERT INTO `sinhvien`(`ho`, `ten`, `tuoi`) VALUES ('le','van coung','26')";
  if ($conn->query($sql) === TRUE) {    } else {echo "Error: " . $sql . "<br>" . $conn->error;}

  $conn->close();
?>

Answer 2

我建议你浏览一些php mysqli基础教程。另请尝试使用mysqli in a object oriented way with mysqli Prepared statements。

以下是我的两个例子

0）使用MYSQLI连接到数据库

<?php
$localhost='localhost';
    $username='root';
    $password='admin';
    $databaseName='example';

    $con = new mysqli($localhost,$username,$password,$databaseName);

    if(mysqli_connect_errno()){
        die("connection Failed: ".mysqli_connect_errno());
    }

1）MYSQLI OBJECT OREINTED WAY

//THIS IS ONE OF THE WAYS OF INSERTING DATA IN MYSQL USING MYSQLI
$sql="INSERT INTO sinhvien (ho,ten,tuoi) VALUES ('THIS','WAY ALSO WORKS',26)";
if($con->query($sql) === TRUE){
    echo "INSERTED INTO THE DATA BASE";
}

2）以MYSQLI准备语句为导向的MYSQLI对象

//THIS IS ONE OF THE WAYS OF INSERTING DATA IN MYSQL USING MYSQLI PREPARED STATEMENT WHICH IS MORE SECURE AGAINST MYSQL INJECTION
$query=$con->stmt_init(); //INITIATE
$a='THIS IS';
$b='MYSQLI PREPARED STATEMENT';
$c=30;
if($query->prepare("INSERT INTO sinhvien(ho,ten,tuoi)VALUES(?,?,?)")){
    $query->bind_param('ssi',$a,$b,$c);//BIND THE ? TO THES VARIABLES
    $query->execute();//NOW EXECUTE THEM
    echo "INSERTED INTO THE DATABASE WITH MYSQLI PREPARED STATEMENT";
}

?>

我无法使用mysqli_query（）将记录插入到我的表中

2 个答案: