php输出json

我正试图让我从php/mysql中使用jquery请求json输出变得容易。现在我正在使用下面的。有人能推荐一个更好的方法吗

/php？username=bob

<?php
    $str = $_SERVER['QUERY_STRING'];
    if($str != ''){
        if(preg_match("/username/",$str)){
            parse_str($str);
            $json = json_encode(checkUserName($username));
            echo $json;
        }
    }
    function checkUserName($v){
        $db = new DB();
        $db->connectDB();

        $findUsername = mysql_query("SELECT COUNT(*) FROM user WHERE username = '$v'");
        $countUser = mysql_fetch_row($findUsername);
        if($countUser[0] < 1){
            return array('username' => 'false');
        }else{
            return array('username' => 'true');
        }

        $db->disconnectDB();
    }
?>

通过使用

$\u GET

superglobal，可以更轻松地编写：

if (isset($_GET['username'])) {
    echo json_encode(checkUserName($_GET['username']));
}

内部

检查用户名（）

：

您应该正确地转义

$v

：

$sql = sprintf("SELECT COUNT(*) FROM user WHERE username = '%s'", mysql_real_escape_string($v));
$findUsername = mysql_query($sql);

更好的是，学习PDO/mysqli并使用准备好的语句

$db->disconnectDB();

---

除非您使用的是持久连接，否则您不需要这些语句。如果您这样做了，您应该首先将返回值保留在变量中，并且只在断开连接后返回。

我不知道您的DB类是什么，但这看起来更漂亮

<?php

function checkUserName($v){
    $db = new DB();
    $db->connectDB();

    $findUsername = mysql_query("SELECT COUNT(*) FROM user WHERE username = '$v'");
    $countUser = mysql_fetch_row($findUsername);
    $db->disconnectDB(); // no code after "return" will do effect

    return ($countUser[0] != 0); // returning a BOOL true is better than a string "true"
}

// use addslashes to prevent sql injection, and use isset to handle $_GET variables.
$username = isset($_GET['username']) ? addslashes($_GET['username']) : '';

// the above line is equal to:
//  if(isset($_GET['username'])){
//      $username = addslashes($_GET['username']);
//  }else{
//      $username = '';
//  }

echo json_encode(checkUserName($username));

?>

---

如果需要修复，只需将checkUsername函数替换为以下函数：

function checkUserName($v){
  $db = new DB();
  $db->connectDB();  

  $findUsername = mysql_query("SELECT username FROM user WHERE username = '$v' LIMIT 1");

  if(mysql_num_rows($findUsername))
    return array('username' => mysql_result($findUsername,0));
  else
    return array('username' => 'false');

}

或者更简单的方法：

if(isset($_GET['username'])){

     $db = new DB();
     $db->connectDB();

     $query = mysql_query(sprintf("SELECT username FROM user 
                            WHERE username='%s'",
                            mysql_real_escape_string($_GET['username'])
                         );

     if(mysql_num_rows($query))
        $json = array('username'=>mysql_result($query,0));
     else
        $json = array('username'=>false);

     header('content-type:application/json');
     echo json_encode($json);
  }

---

顺便说一下，如果您想在

jquery

中处理json数据，您可以这样做

$.ajax({
            type:"POST",
            data:'username=bob',
            url: "do.php",
            success: function(jsonData){
                var jsonArray = eval('(' + jsonData + ')');

                if(jsonArray.username == 'true'){
                    // some action here
                }else if((jsonArray.username == 'false')){
                    // someother action hera
                }



            }
        },"json");

---

$db->disconnectDB（）不可访问。
DB
类在哪里？请修复SQL注入漏洞-请参阅（最好使用mysqli_*或pdo，而不是过时的mysql_*函数）。我使用$.getJSON（）关闭jquery
$db->disconnectDB();

<?php

function checkUserName($v){
    $db = new DB();
    $db->connectDB();

    $findUsername = mysql_query("SELECT COUNT(*) FROM user WHERE username = '$v'");
    $countUser = mysql_fetch_row($findUsername);
    $db->disconnectDB(); // no code after "return" will do effect

    return ($countUser[0] != 0); // returning a BOOL true is better than a string "true"
}

// use addslashes to prevent sql injection, and use isset to handle $_GET variables.
$username = isset($_GET['username']) ? addslashes($_GET['username']) : '';

// the above line is equal to:
//  if(isset($_GET['username'])){
//      $username = addslashes($_GET['username']);
//  }else{
//      $username = '';
//  }

echo json_encode(checkUserName($username));

?>

function checkUserName($v){
  $db = new DB();
  $db->connectDB();  

  $findUsername = mysql_query("SELECT username FROM user WHERE username = '$v' LIMIT 1");

  if(mysql_num_rows($findUsername))
    return array('username' => mysql_result($findUsername,0));
  else
    return array('username' => 'false');

}

if(isset($_GET['username'])){

     $db = new DB();
     $db->connectDB();

     $query = mysql_query(sprintf("SELECT username FROM user 
                            WHERE username='%s'",
                            mysql_real_escape_string($_GET['username'])
                         );

     if(mysql_num_rows($query))
        $json = array('username'=>mysql_result($query,0));
     else
        $json = array('username'=>false);

     header('content-type:application/json');
     echo json_encode($json);
  }

$.ajax({
            type:"POST",
            data:'username=bob',
            url: "do.php",
            success: function(jsonData){
                var jsonArray = eval('(' + jsonData + ')');

                if(jsonArray.username == 'true'){
                    // some action here
                }else if((jsonArray.username == 'false')){
                    // someother action hera
                }



            }
        },"json");