我正在尝试将Json与我的数据库一起使用使用php5,但遭受了一个奇怪的结果。 这个数据库有四个字段 - 'id','Title','Thread','date',但jason的结果如下所示。PHP中的Json编码问题
[
{
"0": "1",
"id": "1",
"1": "Title 1",
"Title": "Title 1",
"2": "Thread 1",
"Thread": "Thread 1",
"3": "2011-10-19",
"date": "2011-10-19"
},
{
"0": "2",
"id": "2",
"1": "Title 2",
"Title": "Title 2",
"2": "Thread 2",
"Thread": "Thread 2",
"3": "2011-10-03",
"date": "2011-10-03"
}
]
您可以看到结果中有重复的信息。他们从哪里来?? 我会附上我写的代码... Jason & PHP大师,请赐教:'(.. 先谢谢了..我会尽力在我等待你的帮助时再解决它....
private static function queryAndFetch($tableName)
{
$query = "SELECT id, Title, Thread, date From $tableName";
$result = mysqli_query(self::$link, $query);
if(!($result))
{
echo "Error";
exit;
}
// $posts = mysqli_fetch_assoc(self::$result); - Working
self::$fetchedResult = array();
while($row = mysqli_fetch_array($result))
{
self::$fetchedResult[] = $row;
}
}
private static function encode()
{
//print_r(self::$fetchedResult);
//if($format == 'json') {
header('Content-type: application/json');
echo json_encode(self::$fetchedResult);
//}
//echo "hi".json_last_error();
}
}
哇...我真的很愚蠢。我甚至有一个正确的代码正确那里,并评论出来......这一切都是从我的误解和ASSOC阵列。谢谢你们的快速响应!!!! – Raccoon
您也可以指定要'mysqli_fetch_array()'返回的内容。你也可以使用'mysqli_fetch_array(MYSQLI_ASSOC)'或'mysql_fetch_array(MYSQLI_NUM)' –