来自我的Php JSON的额外输出
我学习Php/SQL/JSON已经有24个多小时了,学习效果非常好。我已经做了一个数据库,现在有php页面向数据库添加数据 我创建了一个php页面来返回一个JSON对象。它确实这样做了,但它似乎也会返回一些文本来自我的Php JSON的额外输出,php,json,Php,Json,我学习Php/SQL/JSON已经有24个多小时了,学习效果非常好。我已经做了一个数据库,现在有php页面向数据库添加数据 我创建了一个php页面来返回一个JSON对象。它确实这样做了,但它似乎也会返回一些文本 [{"id":"5","udid":"4564645","name":"LastName","score":"999999.00","date":"2011-04-14 18:10:33"},{"id":"4","udid":"912345678901234567890123456759
[{"id":"5","udid":"4564645","name":"LastName","score":"999999.00","date":"2011-04-14 18:10:33"},{"id":"4","udid":"9123456789012345678901234567590123456789","name":"sdfdsf","score":"111110.13","date":"2011-04-14 18:10:01"},{"id":"3","udid":"0123456789012345678901234567890123456789","name":"derktreb","score":"710.13","date":"2011-04-14 18:09:12"},{"id":"1","udid":"0123456789012345678901234567890123456789","name":"brandontreb","score":"210.13","date":"2011-04-14 11:40:05"},{"id":"2","udid":"0123456789012345678901234567890123456789","name":"brandontreb","score":"210.13","date":"2011-04-14 18:08:35"}]
Name Score
[编辑]“名称”“分数”不再出现。是调用此php文件的旧版本导致它出现。下面的代码似乎工作正常。有什么问题吗
如果你看到我犯的任何错误,请向我指出
PHP代码:
<?php
// get_scores.php
/** MySQL database name */
define('DB_NAME', 'b_Chat');
/** MySQL database username */
define('DB_USER', 'b_App');
/** MySQL database password */
define('DB_PASSWORD', 'testtesttest');
/** MySQL hostname */
define('DB_HOST', $_ENV{DATABASE_SERVER});
$table = "highscores";
// Initialization
$conn = mysql_connect(DB_HOST,DB_USER,DB_PASSWORD);
mysql_select_db(DB_NAME, $conn);
// Error checking
if(!$conn) {
die('Could not connect ' . mysql_error());
}
$type = isset($_GET['type']) ? $_GET['type'] : "global";
$offset = isset($_GET['offset']) ? $_GET['offset'] : "0";
$count = isset($_GET['count']) ? $_GET['count'] : "10";
$sort = isset($_GET['sort']) ? $_GET['sort'] : "score DESC";
// Localize the GET variables
$udid = isset($_GET['udid']) ? $_GET['udid'] : "";
$name = isset($_GET['name']) ? $_GET['name'] : "";
// Protect against sql injections
$type = mysql_real_escape_string($type);
$offset = mysql_real_escape_string($offset);
$count = mysql_real_escape_string($count);
$sort = mysql_real_escape_string($sort);
$udid = mysql_real_escape_string($udid);
$name = mysql_real_escape_string($name);
// Build the sql query
$sql = "SELECT * FROM $table WHERE ";
switch($type) {
case "global":
$sql .= "1 ";
break;
case "device":
$sql .= "udid = '$udid' ";
break;
case "name":
$sql .= "name = '$name' ";
break;
}
$sql .= "ORDER BY $sort ";
$sql .= "LIMIT $offset,$count ";
$result = mysql_query($sql,$conn);
if(!$result) {
die("Error retrieving scores " . mysql_error());
}
//echo $result;
$rows = array();
while($row = mysql_fetch_assoc($result)) {
$rows[] = $row;
}
echo json_encode($rows);
mysql_free_result($result);
mysql_close($conn);
?>
这段代码之后似乎有其他内容正在执行
尝试添加exit();在mysql_close()之后,看看会发生什么。如果您得到了预期的结果,那么您的请求肯定也在执行其他代码 一次优化
而不是这个
// Protect against sql injections
$type = mysql_real_escape_string($type);
$offset = mysql_real_escape_string($offset);
$count = mysql_real_escape_string($count);
$sort = mysql_real_escape_string($sort);
$udid = mysql_real_escape_string($udid);
$name = mysql_real_escape_string($name);
try shorthand ( first handle mysql injection with `$_GET`)
$get_data = array_map('mysql_real_escape_string',$_GET);
注意:$\u GET
不应是多维数组
还要更改顺序
mysql_free_result($result);
mysql_close($conn);
echo json_encode($rows);
奇怪的这些单词在JSON编码的结果之外。这些必须来自其他地方。这是完整的密码吗?你百分之百肯定吗?这在include-somewehre中没有使用?如果不回显json会发生什么。这些工件还保留着吗?@diEcho JSON outputServer下面的字符串“Name Score”不允许我用我上传的新文件覆盖旧的php文件,因此输出来自以前的文件。我一删除它并上传上面显示的代码,它就正常工作了。Php代码中的其他内容看起来还好吗?有什么应该做的更好的吗?请参阅上面的命令ToonMariner,谢谢-代码