Question

我有两个页面，一个帖子值到另一个，然后它进行mysql查询并将数据返回到第一个。如果我使用jquery json post，而不是jquery ajax post，如何取回整个页面？（我想回到mysql查询结果）。谢谢。

<html>
<head>
<script type='text/javascript' src='jquery-1.4.2.min.js'></script>
<script type="text/javascript">

$(document).ready(function(){
    $("#submitbutton").click(function(){
      $.ajax({
         url: "m2.php", 
         dataType: "json",
           data: "number1="+$('#number1').val(), 
         success: function(json){
            //$("#result").html(json.number1);// original json data get, it can success.
            $("#result").load('m2.php');//jquery load only load an empty page...
         }
      });
    });
});
</script>
  </head>
  <body>
    <input name="number1" id="number1" value="today" type="text" />
    <input type="submit" value="Calc" id="submitbutton" name="submitbutton"/>
    <div id="result"></div>
  </body>
</html>


<?php
 $db = @mysql_connect("localhost","root","root") or die("can not connect Mysql Server");
 mysql_select_db("contact",$db); 
 $number1 = $_GET['number1'];
 $arr = array(); 
 $arr['number1'] = $number1;
 //echo json_encode($arr); // original json encode data, back to the first page 
 $searchword = json_encode($arr);
 $result = mysql_query("SELECT * FROM msg where title like '%$searchword%' Order By id DESC Limit 0,10");
 while($row = mysql_fetch_array($result)){
    echo $row['name'];// This is what I want get the data. 
 }
?>

Answer 1

试试这个：

$strName = '';
while($row = mysql_fetch_array($result)){
    $strName .= $row['name'];// This is what I want get the data. 
}
echo json_encode($strName);

在你的html页面中使用json解码的值。

jquery json post返回整页html

1 个答案: