Question

可能重复：
How to generate .json file with PHP?

我想用php创建一个像下面这样的json。它将返回一个字符串json作为结果sql查询的响应。我该怎么办？

{"Orders":[  
            {"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"},  
            {"DeliveryId":"DeliveryId","CustomerName":"CustomerName","PhoneNumber":"PhoneNumber","Address":"Address"}               
]
}

我的代码

<?php
mysql_connect("mysql12.000webhost.com","a4602996_longvan","longvan2012");
mysql_select_db("a4602996_lv"); 
$id=$_POST[user];
$sql=mysql_query("select * from testlongvan where Status = 'PACKED'" ); 

$json = array();
if(mysql_num_rows($sql)){
while($row=mysql_fetch_row($sql)){
$json['Orders'][]=$row;
}
}

//while($row=mysql_fetch_assoc($sql))
//$output[]=$row;
print(json_encode($json)); 
mysql_close(); 
?>

但是当我使用我的代码我收到结果时，不要我想要的结果：

{ “订单”：[ [“longvan”，“10/12/2012”，“Be34433jh”，“Long Van”，“115 Pham Viet Chanh，quan Binh Thanh”，“http://longvansolution.tk/image/sample.jpg”，“压缩”， “0909056788”]， [“takehi”，“24/12/2012”，“BF6464633”，“Vn-zoom”，“16 nguyen cuu van，quan binh thanh”，“http://longvansolution.tk/image/hoadon3.jpg”， “压缩”， “098897657”] ]} 你能帮助我吗！

Answer 1

您可以使用json_encode()功能来执行此操作。

您的JSON格式无效。使用:代替=

Answer 2

// SERVER SIDE
<?php
    //... create array
    //....Obviously you would use your own SQL link and query
    $animals = array();
    $result = @mysqli_query('YOUR DB LINK', 'YOUR QUERY');

    while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
        array_push($animals, array(
            'animal' => $row['animal'],
            'sound'  => $row['sound']
    ));
    }
?>

// CLIENT SIDE
<html>
<script>
    // RETURN FROM PHP PAGE ECHO VIA AJAX PERHAPS //
    var animals = <?php echo json_encode($animals) ?>;

    $.each(animals, function (i, elem) {
        console.log(elem.animal + " makes a " + elem.sound);
    });
</script>
</html>

如何通过PHP创建Json

2 个答案: