2013-07-29 72 views
1

我想补充一点,比如说某种标题。我的PHP返回是这样的:在PHP中添加标题

[{"Grad":"Beograd","Predmet":"matematika"},{"Grad":"Novi_Sad","Predmet":"matematika"},{"Grad":"Beograd","Predmet":"matematika"}] 

而且我想获得这样的

{"lista"[{"Grad":"Beograd","Predmet":"matematika"},{"Grad":"Novi_Sad","Predmet":"matematika"},{"Grad":"Beograd","Predmet":"matematika"}]} 

这是我的PHP

$con = mysqli_connect($host, $user, $pwd, $db); 

if(mysqli_connect_errno($con)) { 
    die("Failed to connect to MySQL: " . mysqli_connect_error()); 
} 


$sql = "SELECT Grad, Predmet FROM lista"; 
$result = mysqli_query($con, $sql); 


$rows = array(); 

while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 


     $rows[] = $row; 

} 

mysqli_close($con); 

echo json_encode($rows); 

回答

3
$rows = (object) array('lista' => $rows); 
echo json_encode($rows); 
+0

Tnx很多。我会在15分钟内接受你的回答。 –

+0

没问题,很高兴我可以帮助:) –

0

事情是这样的:$行['lista'] [] = $ row?

+0

这不会被编码为JSON对象,就像嵌套数组一样。 –