Question

include_once("php/db_connects.php");

$tbl_status = "CREATE TABLE IF NOT EXISTS status ( 
                id INT(11) NOT NULL AUTO_INCREMENT,
                osid INT(11) NOT NULL,
                account_name VARCHAR(16) NOT NULL,
                author VARCHAR(16) NOT NULL,
                type ENUM('a','b','c') NOT NULL,
                data TEXT NOT NULL,
                postdate DATETIME NOT NULL,
                PRIMARY KEY (id) 
                )"; 
$query = mysqli_query($tbl_status, $db_connects); 
if ($query === TRUE) {
    echo "<h3>status table created </h3>"; 
} else {
    echo "<h3>status table NOT created </h3>"; 
}

我一直收到mysqli错误。我很确定我在我的php中没有使用mysql。

Answer 1

更改此

$query = mysqli_query($tbl_status, $db_connects);

到

$query = mysqli_query($db_connects, $tbl_status);

示例 mysqli_query（mysqli $ link，string $ query）

Answer 2

mysqli_query的签名是：

mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )

假设$db_connects是存储mysqli_connect结果的变量，您需要翻转参数以适应：

mysqli_query($db_connects, $tbl_status);

mysqli_query documentation

警告：mysqli_query（）期望参数1为mysqli错误

2 个答案: