我设法在Json数组中输出了'resort'列,但我也需要'country',以及'aantal'。不知道该怎么做。有人可以帮助我吗?
if ($numrows < 1 && strlen($sq) > 3)
{
$sql = "SELECT resort, country, COUNT(image) AS aantal FROM sx_cam
LEFT JOIN sv_orte ON sv_cam.res_id = sv_orte.res_id
WHERE sound=soundex('$sq') and (status < 1) GROUP BY resort order by aantal desc";
$result2 = mysql_query($sql) or die(mysql_error());
$numrows = mysql_num_rows($result2);
$suggest = 2;
}
$items = array();
while($row = mysql_fetch_assoc($result2)){
$items[$row['resort']] = $row['resort'];
}
foreach ($items as $key=>$value) {
echo strtolower($key)."|$value\n";
}
答案 0 :(得分:1)
你正在以错误的方式构建数组。一旦你掌握了数组,就像拨打json_encode
我不完全确定你希望你的json看起来如何,但这样的事情应该让你开始
$items = array();
while($row = mysql_fetch_assoc($result2)){
//first we build an 'object' of the current result
$item['country'] = $row['country'];
$item['resort'] = $row['resort'];
//now push it on the array of results
$items[] = $item;
}
echo json_encode($items);
一旦上述代码正常工作,您可以调整PHP数组以更改JSON的结构以满足您的需求。