Question

我想要一个代码来获取数据库行我有table-＆gt;（site）和两个Columns-＆gt;（url，rank）

排名应该超过0

我运行此代码：

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";
$conn = new mysqli($servername, $username, $password, $dbname);
$queryt  = "SELECT url FROM site WHERE rank>0 ORDER By RAND() LIMIT 1";
$resultt = mysql_query($queryt) or die(mysql_error());
while($row = mysql_fetch_assoc($resultt)) {
echo $row['url'];
}

但是说（未选择数据库）

Answer 1

将连接放在参数中：

$resultt = mysqli_query($conn,$queryt) or die(mysql_error());

Answer 2

尝试使用像这样的mysqli连接

$con=mysqli_connect("localhost","my_user","my_password","my_db");
mysqli_select_db($con,"test"); // not compulsory

或者尝试告诉mysqli抛出异常

mysqli_report(MYSQLI_REPORT_STRICT);
try {
         $connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db') ;
    } catch (Exception $e ) {
         echo "Service unavailable";
         echo "message: " . $e->message;   // not in live code obviously...
         exit;

此外，您使用mysqli连接和mysql查询数据库，两者之间存在差异

希望它有所帮助！

Answer 3

之前我遇到过这个问题并通过在表名前添加数据库名称来解决它，所以

 select * from database.table

通过随机获取mysql db的值

3 个答案: