doq70020 2014-11-29 11:06
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PHP在ajax调用中成功传递多个参数

I am doing this ajax call

<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>

and in my reserve_seat.php I want to pass multiple 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();
    }

What I want is commented in the above code. How can I do this?

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3条回答 默认 最新

  • dpnru86024 2014-11-29 11:10
    关注

    Change your echo line to :

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

    And in your ajax

    function(data) {                    
            var arr = JSON.parse( data );
            var booked_seats = arr["booked_seats"];
            console.log(booked_seats);                     
      }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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