-4
$.ajax({
        url:'ajax.php?action=wordFind&word='+arrString,
        cache: true,
        type: "GET",
        success:function(res){
              if(res=="find") { }
        }
    });
4

1 回答 1

0
$.ajax({
        url:'ajax.php?action=wordFind&word='+arrString,
        cache: true,
        async:false,
        dataType:html,//If the response is json replace it with "json"
        type: "GET",
        success:function(res){
           console.log(res);//To check you are getting any reponse
           if(res=="find")
               {
                //Do the stuff you want.
               }
            }
    });
于 2013-04-03T11:01:34.377 回答