PHP ajax 操作之后不要刷新页面更换图片

Question

ajaxphpjavascriptfunction

Answer 1

从你的ajax代码来看 只需要改变ajax的返回值,不再是200 100这样的代码,而是需要更换的新的图片地址然后用jquery定位到要改变的img标签将其src属性直接替换为新的图片地址即可.

Answer 2

function xingxing(id,name){
       $.ajax({
          type: "POST",
          url: "<?php echo base_url('test/fangfa'); ?>",
          processData: "false",
          data: "n=" + Math.random() + "&fn=" + escape(id) + "&starflag=" + escape(name) + "",
          success: function(msg){
            var msg = $.trim(msg);
            $("img").attr("src",msg);
            }
        }
   });
}

Answer 3

@xuzuning 
版主 去那里了，出来  出来 

Answer 4

 <?php if($value['starflag'] == 1){ ?>
        <a href="javascript:void(0);" onclick="xingxing('<?php echo $value['id']."','".$value['name']; ?>');">
          <img src="/themes/default/images/star02.gif" im="/themes/default/images/star02.gif" id="img_a"/> 
        </a>
        <?php 
          }
        else{
        ?>
        <a href="javascript:void(0);" onclick="xingxing('<?php echo $value['id']."','".$value['name']; ?>');">
          <img src="/themes/default/images/star01.gif"  im="/themes/default/images/star01.gif" id="img_b" /> 
        </a>
        <?php } ?>    
        </a>function xingxing(id,name){
       $.ajax({
          type: "POST",
          url: "<?php echo base_url('test/fangfa'); ?>",
          processData: "false",
          data: "n=" + Math.random() + "&fn=" + escape(id) + "&starflag=" + escape(name) + "",
          success: function(msg){
            if(msg =='1'){
              $("#img_b").attr("src", "/themes/default/images/star01.gif"); 
            }else{
              $("#img_a").attr("src", "/themes/default/images/star02.gif"); 
            }
             
        }
   });
}可恶 就是不变图片