首先我有这个:
$myArray = getDatabaseData($p1);
header('Content-Type: application/json');
echo json_encode($myArray );
这很好用,json数据的回声正常工作。
但是现在我想添加另一个数组数据从其他查询接收数据。像这样:
$myArray = getDatabaseData($p1); //returns 2 dimensions array
$myArray2 = getDatabaseData2($p2); //returns 2 dimensions array
$finalArray = array();
$finalArray['data1'] = $myArray;
$finalArray['data2'] = $myArray2;
header('Content-Type: application/json');
echo json_encode($finalArray);
并没有回应。
我发现如果我做echo json_encode($myArray);
它就会产生回声。
但是,如果我echo json_encode($myArray2);
它没有。
注意: getDatabaseData函数只对数据库进行一次查询。 getDatabaseData2执行4,然后将它们组合在一个数组中。
是因为我在我的getDatabaseData2
函数中组合了多个数据库查询吗?
以下是$myArray
和$myArray2
的 print_r :
myArray的:
Array
(
[values] => Array
(
[0] => 0
[1] => 2
[2] => 3
[3] => 2
[4] => 7
[5] => 17
[6] => 6
[7] => 5
[8] => 9
[9] => 0
)
[keys] => Array
(
[0] => G. M.
[1] => G. S.
[2] => Cruz.
[3] => At.
[4] => Rem. C.
[5] => Rem. S.
[6] => Fs.
[7] => Rec.
[8] => B. P.
[9] => V. F.
)
)
myArray2:
Array
(
[names] => Array
(
[77] => André
[78] => Daniel
[79] => Rúben
[80] => Ant�nio
[81] => João
[83] => João
)
[nums] => Array
(
[77] => 0
[78] => 2
[79] => 0
[80] => 0
[81] => 0
[83] => 6
)
[nums2] => Array
(
[77] => 0
[78] => 0
[79] => 4
[80] => 0
[81] => 3
[83] => 0
)
)
感谢您的帮助。
答案 0 :(得分:1)
我在这里试过但它的工作!
<!DOCTYPE html>
<html>
<head>
<title>My Title</title>
</head>
<body>
<?php
$arr = array(array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5),array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5));
$arr2 = array(array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5),array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5),array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5));
$finalArray = [];
$finalArray[] = $arr;
$finalArray[] = $arr2;
echo json_encode($finalArray); die();
?>
</body>
</html>