0

这是我使用OO风格的代码:

$query = "SELECT * FROM articles LIMIT 1";
$result=mysqli_query($db, $query) or die("Could Not execute query.");
$row = $result->fetch_assoc()
echo $row;

使用程序的代码:

$query = "SELECT * FROM articles LIMIT 1";
$result=mysqli_query($db, $query) or die("Couldn't execute query.");
$row = mysqli_fetch_row($result);
echo $row;

我已经尝试了各种不同的方法来让它工作,并且查询似乎永远不会执行。

我是mysqlimysql 的新手,正在努力解决一些差异。如果这个问题有明显的答案,我们深表歉意。

非常感谢任何帮助。

4

2 回答 2

0

尝试:

$query  = "SELECT * FROM articles LIMIT 1";
$mysqli = mysqli_connect("example.com", "user", "password", "database");
$res = mysqli_query($mysqli, $query);
$row = mysqli_fetch_assoc($res);

或者:

$mysqli = new mysqli("example.com", "user", "password", "database");
$res = $mysqli->query($query);
$row = $res->fetch_assoc();
于 2013-01-29T22:08:13.660 回答
0

假设你的

 $db = mysqli_connect("localhost", "user", "password", "database");

那么你可以拥有

$query  = "SELECT * FROM articles LIMIT 1";
$res = $db->query($query);
$row = $res->fetch_assoc();
于 2015-10-10T11:30:21.103 回答