我試圖讀取我的MySQL數據並將其作爲JSON。但我只想要一個JSON對象,它會嘗試給我一個JSONArray,但該數組是空的。數組總是從一個開始,爲零
這就是它給我的,頂部是JSON,底部是簡單的數組。
[
[],
{
"id": "1",
"description": null,
"copyright": "© Ian Ransley, flickr.com\/CC BY 2.0"
}
]
Array
(
[0] => Array
(
)
[1] => Array
(
[id] => 1
[description] =>
[copyright] => © Ian Ransley, flickr.com/CC BY 2.0
)
)
正如你可以看到,零指數爲空出於某種原因,和右我的對象之前開始有一個開閉括號([]),這使得我的Android應用認識到,作爲一個數組代替對象。
這裏是我的PHP代碼:
<?php
header('Content-type: text/plain');
$connection = mysqli_connect("localhost", "bananatime", "Yoyobanana", "bananatime")
or die("Error " . mysqli_error($connection));
if(isset($_GET['id'])){
$sql = "SELECT * FROM bananas WHERE id = '".$_GET['id']."' ORDER BY id";
}else{
$sql = "SELECT * FROM bananas ORDER BY id";
}
$result = mysqli_query($connection, $sql) or die("Error in Selecting " .mysqli_error($connection));
$array[] = array();
while($row = mysqli_fetch_array($result, MYSQL_ASSOC))
{
array_push($array, $row);
}
array_values($array);
echo json_encode($array, JSON_PRETTY_PRINT);
echo "\n\n";
print_r($array);
mysqli_close($connection);
?>
如果您不發佈您的代碼,我們應該如何知道您做錯了什麼? –
該死的,對不起,我會在第二秒編輯我的帖子, – TemporaryName