Question

这是我的php sql代码

//fetch table rows from mysql db
$sql = "select * from tbl_sample";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));


//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
    $emparray[] = $row;
}

$info = json_encode($emparray);
echo $info;

//close the db connection
mysqli_close($connection);
?>

当我运行此代码时，我得到一个像这样的匿名json数组。

[{"id":"1","country":"india","domain":"MBA"},{"id":"2","country":"england","domain":"cricket"},{"id":"3","country":"pakistan","domain":"MTECH"},{"id":"4","country":"newzeland","domain":"bba"}]

有没有办法给这个数组命名，因为没有命名数组我不知道如何将这个json数据用于dust.js模板。如果没有建议我如何使用这些数据进行模板化。谢谢。

Answer 1

//fetch table rows from mysql db
$sql = "select * from tbl_sample";
$result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));


//create an array
$emparray = array();
while($row =mysqli_fetch_assoc($result))
{
    $emparray[] = $row;
}
$mydata['my_data'] = $emparray; 

$info = json_encode($mydata);
echo $info;

//close the db connection
mysqli_close($connection);

试试这个。您的数据将如下所示

{"my_data":[{"id":"1","country":"india","domain":"MBA"},{"id":"2","country":"england","domain":"cricket"},{"id":"3","country":"pakistan","domain":"MTECH"},{"id":"4","country":"newzeland","domain":"bba"}]}

Answer 2

构建阵列时，您需要使用您想要的任何唯一值来设置密钥。例如：

while($row =mysqli_fetch_assoc($result)) {
    $key = $row['domain'];
    $emparray[$key] = $row;
}

这将导致您的JSON使用domain值键入。

如何解决匿名数组？

2 个答案: