Question

这是我的代码：

$con = mysqli_connect($databasehost,$dbname,$dbpassword) or die(mysql_error());
mysqli_select_db($con, $dbname) or die(mysql_error());
mysqli_query($con,"SET CHARACTER SET utf8");
$query = file_get_contents("php://input");
$sth = mysqli_query($con,$query);

if (mysql_errno()) {
    header("HTTP/1.1 500 Internal Server Error");
    echo $query.'\n';
    echo mysql_error();
}
else
{
    $rows = array();
    while($row = mysql_fetch_assoc($sth)){
        $rows[] = $r;
    }
    print json_encode($rows);
}

问题是什么？

这是错误：

[19-Jan-2018 09:17:39 UTC] PHP Warning:  mysql_fetch_assoc() expects parameter 1 to be resource on line 22

Answer 1

您正在混合mysqli和mysql。

自版本5.5.0起，后者已弃用。

尝试使用mysqli_fetch_assoc($sth)代替mysql_fetch_assoc($sth)

另见http://php.net/manual/en/mysqli-result.fetch-assoc.php。

Answer 2

你的问题是你正在混合mysqli和mysql。

尝试使用mysqli_fetch_assoc($sth)代替mysql_fetch_assoc($sth)。

而且，在mysqli_connect("host", "user", "password", "database")中，您缺少用户名。

// (1) make sure $username is defined.
$con = mysqli_connect($databasehost,$username,$dbname,$dbpassword) or die(mysqli_error());

// check to use mysqli_, not mysql_
mysqli_select_db($con, $dbname) or die(mysqli_error());
mysqli_query($con,"SET CHARACTER SET utf8");
$query = file_get_contents("php://input");
$sth = mysqli_query($con,$query);

if (mysql_errno()) {
    header("HTTP/1.1 500 Internal Server Error");
    echo $query.'\n';
    echo mysql_error();
}
else
{
    $rows = array();

    // (2) Use mysqli_fetch_assoc instead of mysql_fetch_assoc.
    while($row = mysqli_fetch_assoc($sth)) {

        // (3) $r is not defined. Use $row.
        $rows[] = $row;
    }
    print json_encode($rows);
}

我的代码中存在一些问题

2 个答案: