PHP json_encode() 在 while 循环中

2021-12-26 00:00:00 json arrays while-loop php sqlite

我试图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:

I am trying to use json_encode() in a while loop while getting database results. Here is my code:

<?

$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);

?>

这个输出是

{"response":"lastUserID lastUser lastXPos lastYPos"}

{"response":"lastUserID lastUser lastXPos lastYPos"}

我希望它是...

{"response":["1 Alex 10 12", "2 Fred 27 59", "3 Tom 47 19"]}

{"response":["1 Alex 10 12", "2 Fred 27 59", "3 Tom 47 19"]}

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

So I want the json_encode() function to put ALL users into the array rather than the last one. How would I do this? Thanks

推荐答案

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);

?>

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