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2 回答 2

1

您在 ajax 函数中有错误..var 术语应该在 ajax 调用之外。数据应该是对象而不是字符串

尝试这个

  ...
  else{
   var term = $form.find( 'input[name="vrname"]' ).val();  //<---here declare it outside the ajax
    //or  
  var term =$("#vrname").val(); //<--not sure what $form is but i think this should work
   $.ajax({
    type: "POST",
    url: "nviewdetail.php",
    data: { vrname: term }, //<--here this should be object
     success:function(data){  //<--- callback function which is called whn ajax call succeed
         //do your stuff
     }
    });
  }
于 2013-09-05T05:35:27.803 回答
0

将以下代码用于 ajax:

function IsBlank_Post(){
    var x = $("#vrname").val();
    if (x==null || x==""){
      alert("Company name must be filled out");
      return false;
    } else {
        $.ajax({
        type: "POST",
        url: "nviewdetail.php",
        data: {'vrname': x},
        success: function(returnData){
            alert("Successfull");
            return true;
        }
        });
    }
}

在页面nviewdetail.php你可以通过

$vrname = $_REQUEST['vename'];
于 2013-09-05T05:35:20.897 回答