我想在获取数据库结果的同时在while循环中使用json_encode().这是我的代码:
<?
$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);
$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");
while($row = sqlite_fetch_array($results)) {
$data = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));
sqlite_close($database);
?>
这个的输出是
{“response”:”lastUserID lastUser lastXPos lastYPos”}
我希望它是……
{“response”:[“1 Alex 10 12”, “2 Fred 27 59”, “3 Tom 47 19”]}
等等
所以我希望json_encode()函数将所有用户放入数组而不是最后一个.我该怎么做?谢谢
解决方法:
尝试:
<?
$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);
$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");
$data = array();
while($row = sqlite_fetch_array($results)) {
$data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));
sqlite_close($database);
?>