doumei1955
2013-03-19 07:46
浏览 57

Jquery Ajax用于从php mysql获取数据而不刷新页面 - 语法错误[关闭]

I need to get data from mysql database without refreshing the page using jquery ajax. I have a php script which is working fine. However, my JS seems to be having some problem. Here is the jquery script. Also, I am using multiple jquerys on same page like maximage and custom scrollbar.

var count = jQuery.noConflict();

count('#CountryName').on ('change',function(){ 
var Country = count('#CountryName').val();
count.ajax({ 
  type: 'GET' ,
  url: 'getcountry2.php', 
  data: 'q=', 
}).done(function( html ) { 
 count('#result').append(html); 
 or 

  count('#result').html(html); 
});
});

Here is my php

<link href="main2.css" rel="stylesheet" type="text/css" />
<link href="country.css" rel="stylesheet" type="text/css" />
<?php
$q=$_GET["q"];

$con = mysql_connect('localhost', 'root', 'password');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
$user = mysql_real_escape_string($user);
$pwd = mysql_real_escape_string($pwd);

mysql_select_db("mahc", $con);

$sql="SELECT * FROM texttesti WHERE Country = '".$q."'";

$result = mysql_query($sql);

echo "<div>";

    while($row = mysql_fetch_array($result))
      {

    echo "<blockquote>". $row['Review'] ."</blockquote>";

    echo "<p>" . $row['Name'] . " " . $row['ReasonForPanchakarma']."</p>";
    }
echo "</div>";

mysql_close($con);
?>

Here is my HTML

<p>Select the Fields below to see the testimonials of your choice:</p>
   <p>  <form name="Country">
 <select id="CountryName" onChange="showCountry(this.value)">
<option value="">Select a Country:</option>
<option value="USA">USA</option>
<option value="India">India</option>
<option value="Germany">Germany</option>
<option value="Russia">Russia</option>
</select>
</form>
</p>
<div id="result_data">
<script type="text/javascript" src="country2.js"></script>
</div>
</div>

图片转代码服务由CSDN问答提供 功能建议

我需要从mysql数据库获取数据而不使用jquery ajax刷新页面。 我有一个PHP脚本,工作正常。 但是,我的JS似乎遇到了一些问题。 这是jquery脚本。 此外,我在同一页面上使用多个jquerys,如maximage和自定义滚动条。

  var count = jQuery.noConflict(); 
 
count('#CountryName')。  on('change',function(){
var Country = count('#CountryName')。val(); 
count.ajax({
 type:'GET',
 url:'getcountry2.php',  
 data:'q =',
})。done(function(html){
 count('#result')。append(html); 
或
 
 count('#result')  .html(html); 
}); 
}); 
   
 
 

这是我的php

 &lt; link href =“main2.css”rel =“stylesheet”type =“text / css”/&gt; 
&lt; link href =“country.css”rel =“stylesheet”type =“text / css”/  &gt; 
&lt;?php 
 $ q = $ _ GET [“q”]; 
 
 $ con = mysql_connect('localhost','root','password'); 
if(!$ con)\  n {
 die('无法连接:'。mysql_error()); 
} 
 $ user = mysql_real_escape_string($ user); 
 $ pwd = mysql_real_escape_string($ pwd); 
 
mysql_select_db(“mahc  “,$ con); 
 
 $ sql =”SELECT * FROM texttesti WHERE Country ='“。$ q。”'“; 
 
 $ result = mysql_query($ sql); 
 
echo”&lt  ; div&gt;“; 
 
 whi  le($ row = mysql_fetch_array($ result))
 {
 
 echo“&lt; blockquote&gt;”。  $ row ['Review']。“&lt; / blockquote&gt;”; 
 
 echo“&lt; p&gt;”  。  $ row ['名称']。  “”。  $ row ['ReasonForPanchakarma']。“&lt; / p&gt;”; 
} 
echo“&lt; / div&gt;”; 
 
mysql_close($ con); 
?&gt; 
  <  / pre> 
 
 

这是我的HTML

 &lt; p&gt;选择下面的字段以查看您选择的推荐:&lt; / p&gt; \  n&lt; p&gt;  &lt; form name =“Country”&gt; 
&lt; select id =“CountryName”onChange =“showCountry(this.value)”&gt; 
&lt; option value =“”&gt;选择国家/地区:&lt; / option&gt  ; 
&lt; option value =“USA”&gt; USA&lt; / option&gt; 
&lt; option value =“India”&gt; India&lt; / option&gt; 
&lt; option value =“Germany”&gt; Germany&lt; / option&gt; \  n&lt; option value =“Russia”&gt; Russia&lt; / option&gt; 
&lt; / select&gt; 
&lt; / form&gt; 
&lt; / p&gt; 
&lt; div id =“result_data”&gt; 
&lt; script type =  “text / javascript”src =“country2.js”&gt;&lt; / script&gt; 
&lt; / div&gt; 
&lt; / div&gt; 
   
 
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3条回答 默认 最新

  • dongzhuan1185 2013-03-19 07:51
    已采纳

    Use simple ajax as:

      $.ajax({
       url:'getcountry2.php',
       datatype:"application/json",
       type:'get',
       data: 'q='+Country, 
       success:function(data){
          count('#result').append(html); 
       },
       error:function(){
          // code for error
       }
     });
    

    hopw it will help!

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  • douwaz34842 2013-03-19 07:49

    You didn't pass country value to PHP but you are getting the value in PHP. So it could be a problem. Try this,

    count.ajax({ 
      type: 'GET' ,
      url: 'getcountry2.php', 
      data: 'q='+Country, 
    
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  • dqd22496 2013-03-19 07:49

    Here you go..Country is missing

    var count = jQuery.noConflict();
    
        count('#CountryName').on ('change',function(){ 
        var Country = count('#CountryName').val();
        count.ajax({ 
          type: 'GET' ,
          url: 'getcountry2.php', 
          data: 'q='+Country, 
        }).done(function( html ) { 
         count('#result').append(html); 
         or 
    
          count('#result').html(html); 
        });
        });
    
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