我正在嘗試將Json與我的數據庫一起使用使用php5,但遭受了一個奇怪的結果。 這個數據庫有四個字段 - 'id','Title','Thread','date',但jason的結果如下所示。PHP中的Json編碼問題
[
{
"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"
}
]
您可以看到結果中有重複的信息。他們從哪裏來?? 我會附上我寫的代碼... Jason & PHP大師,請賜教:'(.. 先謝謝了..我會盡力在我等待你的幫助時再解決它....
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。我甚至有一個正確的代碼正確那裏,並評論出來......這一切都是從我的誤解和ASSOC陣列。謝謝你們的快速響應!!!! – Raccoon
您也可以指定要'mysqli_fetch_array()'返回的內容。你也可以使用'mysqli_fetch_array(MYSQLI_ASSOC)'或'mysql_fetch_array(MYSQLI_NUM)' –