Question

我使用以下代码创建一个带有提交按钮的表单，并在提交后以相同的形式显示信息..我写的代码是

<html>
<head>

<script type="text/javascript" src="jquery.js"></script></script>
<script type="text/javascript">
function get(){
$.post('data.php', { name: form.name.value },
function(output) {
    $('#age').html(output).show();
    $('#number').html(output).show();

}) ;

 }

</script>
</head>
<body> 
<p>
 <form name="form">
   name:
 <input type ="text" name="name"><input type ="button" value ="Get" onclick="get();">
 </form>
<div id="age"></div>
</p>

</body>

 </html>

文件2

<?php

Connect to DB
$name = mysql_real_escape_string($_POST['name']);



if ($name==NULL)
echo "please enter an name!";
else
{   
$age= mysql_query("SELECT age FROM parentid WHERE id ='$name'");
$age_num_rows = mysql_num_rows($age);   
  if ($age_num_rows==0)
  echo "id does not exist";

 else
 {
  $age = mysql_result($age, 0);
  echo "$name's title is $age'";
 }
$number= mysql_query("SELECT number FROM parentid WHERE id ='$name'");
$number_num_rows = mysql_num_rows($number);   
  if ($name_num_rows==0)
  echo "name does not exist";

 else
 {
  $number = mysql_result($number, 0);
  echo "$name's number is $number'";
 }
 }
 ?>

我得到的输出是

12882's title is xxxx'12882's reportno is Resource id #4'

这里12882的标题是xxxx是数据库中的内容 12882的reportno是资源ID＃4不是数据库中的东西，我不知道它从哪里生成结果

Answer 1

它是一个mysql资源，避免使用mysql_result。使用类似的东西

$row = mysql_fetch_row($age);
echo $name . '\'s title is ' . $row[0];

使用jquery的Ajax帮助（使用php和mysql）

1 个答案: