Question

所以我想为jquery分配一个url参数值，但它不会接受它..

  var product_id = '<?php echo $_GET["pID"]; ?>';

我的功能如下，包含文件准备就绪......刚发布相关内容

function sendOrderToServer() {
        var order = $(".sortable_table").sortable("serialize");
        var array_order = order.split("&");
        var product_id = '<?php echo $_GET["pID"]; ?>';
        alert(product_id);
        for (var i=0;i<array_order.length;i++){ 

            var id = array_order[i].split("=");
             id = id[1];
            var text_field_video = $("#products_video_sm_dynamic_"+ id).val();
            var text_field_video_caption = $("#products_video_sm_dynamic_"+ id +"_caption").val();
            var text_field_image = $("#products_image_sm_dynamic_"+ id).val();
            var text_field_image_caption = $("#products_image_sm_dynamic_"+ id +"_caption").val();
            var text_field_video_xl = $("#products_video_xl_dynamic_"+ id).val(); 
            var text_field_video_xl_caption = $("#products_video_xl_dynamic_"+ id +"_caption").val();
            var text_field_image_xl = $("#products_image_xl_dynamic_"+ id).val(); 

            if(text_field_video != undefined){

              var element = "products_video_sm_dynamic_" + id;
              var position = i + 1;
            }

            if(text_field_image != undefined){

              var element = "products_image_sm_dynamic_" + id ;
              var position = i + 1;
            }
            if(text_field_video_xl != undefined){

              var element2 = "products_video_xl_dynamic_" + id ;
              var position = i + 1;
            }
            if(text_field_image_xl != undefined){

              var element2 = "products_image_xl_dynamic_" + id ;
              var position = i + 1;
            }

            //alert(element);
            //alert("position is" + position);

            $.ajax({
            type:'POST',
             url :'includes/insert_database.php', 
            data:{ position : position, element1:element , element2:element2}, 
           success: function(result){

                 alert(result);

            }
          });


        }


    }

Answer 1

http://example.com/yourpage.php?pID=%27%3B+alert%28%27OMG+I+just+totally+hacked+your+site%21%27%29%3B+%27

请改为尝试：

var product_id = <?php echo json_encode($_GET['pID']); ?>

Javascript不会从PHP获取$ _GET值？

1 个答案: