Question

好的我已经构建了一个面向数据库的应用程序（Mysql），就是它从JSON php文件连接到数据库，我希望它有多个用户使用不同的UN和Pass。

我对Json文件没有经验，我发现了一些例子，但无法将它们组合在一起。以下是代码：

<?php
$link = mysql_pconnect("host", "db_user", "db_password") or die("Could not connect");
mysql_select_db("db_name") or die("Could not select database");
$arr = array();
$rs = mysql_query("SELECT * FROM news");
while($obj = mysql_fetch_object($rs)) {
    $arr[] = $obj;
}
echo json_encode($arr);
?>

这是一种方式，但它不是多用户。

<?php 
header('Content-type: application/json');
if($_POST) {
    if($_POST['username'] == '****' && $_POST['password'] == '*****') {
        echo '{"success":1}';
    } else {
        echo '{"success":0,"error_message":"Username and/or password is invalid."}';
    }
}else {    echo '{"success":0,"error_message":"Username and/or password is invalid."}';}
?>

这是多用户但不包含数组。

<?php
$host = "************"; //Your database host server
$db = "****"; //Your database name
$user = "****"; //Your database user
$pass = "****"; //Your password

$connection = mysql_connect($host, $user, $pass);
//Check to see if we can connect to the server
if(!$connection)
{
    die("Database server connection failed.");
}
else
{
    //Attempt to select the database
    $dbconnect = mysql_select_db($db, $connection);
    //Check to see if we could select the database
    if(!$dbconnect)
    {
        die("Unable to connect to the specified database!");
    }
    else
    {
        $query = "SELECT * FROM *****";
        $resultset = mysql_query($query, $connection);
        $records = array();
        //Loop through all our records and add them to our array
        while($r = mysql_fetch_assoc($resultset))
        {
            $records[] = $r;
        }
        //Output the data as JSON
        echo json_encode($records);
    }
}
?>

这是完美的，但不能与第二个混合使用。

知道怎么做吗？

Answer 1

我在应用程序中有多用户功能，我正在使用它，它适用于我：

我希望它对你有用，

$sql ="select * from login WHERE user_name='$s1' and password='$s2'";
$data = mysql_query($sql);
$count = mysql_num_rows($data);

if($count==1)
{
    $response['success']=1;
    $response['userdetail'] = mysql_fetch_assoc($data);

}
else
{
    $response['success']=0;
}


echo json_encode($response);

与JSON的数据库连接无法解决它

1 个答案: