在mysql表中我有一组记录。我想让他们想要在json响应下显示。从数据库获取数据显示为json响应
"results":[
{
"timestamp":"2014-03-04 17:26:14",
"id":"440736785698521089",
"category":"sports",
"username":"chetan_bhagat",
"displayname":"Chetan Bhagat"
}
我从数据库中获取上面的值,即时间戳,id,类别,用户名。如何以类似上面的json响应形式显示结果?
UPDATE:
我以这种方式获取数据:
$con = mysqli_connect('127.0.0.1', 'root', '', 'mysql');
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
return;
}
$today = date("Ymd");
$result = mysqli_query($con,"SELECT url,img_url,sentiment,title,category from frrole_cateogry_article where category='".$category."' AND today <= '".$today."' AND title != '' AND img_url != '' order by url desc limit 3 ");
while ($row = @mysqli_fetch_array($result))
{
$url = $row['url'];
$img_url = $row['img_url'];
$screen_name = $row['screen_name'];
}
入住这一点 - http://stackoverflow.com/questions/6818441/sql-server-select-to-json-function – TechMaze
@TechMaze,这个问题是有关Microsoft SQL Server,而不是MySQL。 –
@BillKarwin:对不起,我正在使用MySql –