嘗試使用php創建表時出錯

Question

我是php的新手，我正在嘗試使用mysql創建一個表。 當我運行我編寫的php代碼時，它說：

Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in /home/engelsmj/public_html/Table/CreateTable.php on line 16
Warning: mysqli_query() expects at least 2 parameters, 1 given in /home/engelsmj/public_html/Table/CreateTable.php on line 23
Notice: Undefined variable: sql in /home/engelsmj/public_html/Table/CreateTable.php on line 25
Warning: mysqli_query(): Empty query in /home/engelsmj/public_html/Table/CreateTable.php on line 25

創建表時出錯：

我相信它是連接到mysql，但實際創建表時出錯。 我已經多次完成了我的代碼，但我無法弄清楚它有什么問題以及為什么它不會創建我的表。 感謝您的幫助。

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

    $con=mysqli_connect("*******","******","*******");
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    mysqli_select_db("*****", $con);

    mysqli_query("CREATE TABLE Friends(
    idNumber INT NOT NULL AUTO_INCREMENT,  
    Name varchar(30),
    Phone varchar(30), 
    Age int,
    PRIMARY KEY (idNumber))");

    if (mysqli_query($con,$sql)) {
     echo "Friends table created successfully";
    } else {
      echo "Error creating table: " . mysqli_error($con);
    }
    mysqli_close($con);

?>

Answer 1

你必須切換這些值：

//Wrong 
mysqli_select_db("*****", $con);

//Right
mysqli_select_db($con, "*****");

至於查詢，您必須添加$ sql而不是mysqli_query，因為您稍后會在if語句中執行此操作：

$sql = "CREATE TABLE Friends(
idNumber INT NOT NULL AUTO_INCREMENT,  
Name varchar(30),
Phone varchar(30), 
Age int,
PRIMARY KEY (idNumber))";

Answer 2

重寫你的代碼：

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);

$con=mysqli_connect("*******","******","*******");
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

mysqli_select_db($con, "*****");

$sql = "CREATE TABLE Friends(
idNumber INT NOT NULL AUTO_INCREMENT,  
Name varchar(30),
Phone varchar(30), 
Age int,
PRIMARY KEY (idNumber))";

if (mysqli_query($con,$sql)) {
 echo "Friends table created successfully";
} else {
  echo "Error creating table: " . mysqli_error($con);
}
mysqli_close($con);
?>