Php 我可以从一个页面抛出多个json吗?
我想从这个页面抛出多个json,以便获取一个json文件,用java编程显示我的移动应用程序的输出。下面是我的代码,它显示了表“news”中正确的json,但是,我想抛出数据库中其他可用对象的json。可能吗?Php 我可以从一个页面抛出多个json吗?,php,mysql,sql,json,Php,Mysql,Sql,Json,我想从这个页面抛出多个json,以便获取一个json文件,用java编程显示我的移动应用程序的输出。下面是我的代码,它显示了表“news”中正确的json,但是,我想抛出数据库中其他可用对象的json。可能吗? $dblink = new mysqli($dbhost, $dbuser, $dbpass, $dbname); mysqli_query($dblink, 'SET NAMES utf8'); //Check connection was successf
$dblink = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
mysqli_query($dblink, 'SET NAMES utf8');
//Check connection was successful
if ($dblink->connect_errno) {
printf("Failed to connect to database");
exit();
}
$result = $dblink->query("SELECT * FROM news ORDER BY id DESC");
$dbdata = array();
while ( $row = $result->fetch_assoc()) {
$dbdata[]=$row;
}
echo json_encode($dbdata);
?>
这取决于您如何组合它们,可以简单到: 合并:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
不同的键:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
MySQL加入:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
这取决于您如何组合它们,可以简单到: 合并:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
不同的键:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
MySQL加入:
$dbdata = array_merge($dbdata1, $dbdata2);
echo json_encode(dbdata);
$dbdata = array(
'table_1' => $dbdata1,
'table_2' => $dbdata2
);
echo json_encode($dbdata);
SELECT n.*, m.* FROM news n LEFT JOIN news_meta m ON m.news_id = n.news_id ORDER BY n.id DESC;
/* $dbdata will then contain your two tables worth of data: */
echo json_encode($dbdata);
合并到另一个数组中并将其作为json发送。是的,您可以发送、合并到数组中并使用json_encode发送。您所说的抛出是什么意思?假设你不是这个意思?合并到另一个数组中并以json的形式发送。是的,你可以发送,合并到数组中并使用json编码发送。你所说的抛出是什么意思?假设你不是这个意思?是的,它给出了两个表的数组,但我如何知道哪个数组来自表1,哪个数组来自表2。如果你想让JSON清楚地显示它来自哪个表,我的答案会告诉你“不同的键”方法。是的,它给出了这两个表的数组,但我如何知道哪个数组来自表1,哪个数组来自表2。如果您想让JSON清楚地显示它来自哪个表,我的答案会告诉您“不同的键”方法。