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      2. PHP json_encode() 在 while 循环中

        时间:2023-09-20
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                • 本文介绍了PHP json_encode() 在 while 循环中的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                  问题描述

                  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"]}

                  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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