Question

我正在做这个ajax调用

<script>
    function reserveSeat(showID) {  
        $.ajax({        
        url: 'reserve_seat.php',
        type: 'post',
        data: { "showID": showID}        
        }).done(function(data) {                    
            var booked_seats = JSON.parse( data1, data2, data3 ); //get multiple values 
            console.log(booked_seats);                     
        });
    };
    </script>

在我的reserve_seat.php中我想传递多个echo

$query = "SELECT * FROM `seats` WHERE show_id='" . $_POST['showID'] ."'";
    $result = mysql_query($query);

    if($result){
        $booked_seats = array();
    while($row = mysql_fetch_array($result)){
        array_push ($booked_seats, array($row['id'], $row['row_no'], $row['col_no']));   
    }
    echo json_encode($booked_seats, var2, var3); //echo multiple variable        
    } else {
        echo mysql_error();
    }

我想要的是在上面的代码中评论。我怎么能这样做？

Answer 1

将回声线更改为：

echo json_encode(array("booked_seats" => $booked_seats, "var2" => $var2, "var3" => $var3);

在你的ajax中

function(data) {                    
        var arr = JSON.parse( data );
        var booked_seats = arr["booked_seats"];
        console.log(booked_seats);                     
  }

Answer 2

为什么不打印编码的JSON输出？

更不用说json_encode()需要其他海报所说的数组

json_encode() PHP Manual

Answer 3

在侧面使用数组

json_encode(array($booked_seats, var2, var3)).

PHP在ajax调用中成功传递多个参数

3 个答案: