2011-07-12 80 views
0

到PHP我怎么可以在这个代码发送rowNumber变量dataSource PHP文件?发送变量与AJAX

function getData(dataSource, divID,rowNumber) 
{ 

if(XMLHttpRequestObject) 
{ 
    var obj = document.getElementById(divID); 
    XMLHttpRequestObject.open("GET", dataSource); 
    XMLHttpRequestObject.onreadystatechange = function() 
     { 
      if (XMLHttpRequestObject.readyState == 4 && 
       XMLHttpRequestObject.status == 200) 
      { 
       obj.value = XMLHttpRequestObject.responseText; 
      } 
     } 

     XMLHttpRequestObject.send(null); 
    } 
} 

PHP文件(数据源):

<?php 
    //mysql connection 
    $result = mysql_query('CALL view_polls(`rowNumber`);'); 
    $row=mysql_fetch_array($result); 
    echo $row['title']; 
?> 
+5

采取jQuery框架来看看......它使AJAX确实容易:) –

回答

2
function getData(dataSource, divID,rowNumber) 
{ 

if(XMLHttpRequestObject) 
{ 
    var obj = document.getElementById(divID); 
    XMLHttpRequestObject.open("GET", dataSource + "?rownumber=" + rowNumber); 
    XMLHttpRequestObject.onreadystatechange = function() 
     { 
      if (XMLHttpRequestObject.readyState == 4 && 
       XMLHttpRequestObject.status == 200) 
      { 
       obj.value = XMLHttpRequestObject.responseText; 
      } 
     } 

     XMLHttpRequestObject.send(null); 
    } 
} 

P HP:

<?php 
    if (isset($_GET['rownumber']) && is_numeric($_GET['rownumber'])) { 
     $rowNumber = $_GET['rownumber']; 
     //mysql connection 
     $result = mysql_query('CALL view_polls(`' . $rowNumber . '`);'); 
     $row=mysql_fetch_array($result); 
     echo $row['title']; 
    } 
    else { 
     echo "Error"; 
    } 
?> 
+0

谢谢@Jules非常 –

5
在JavaScript

XMLHttpRequestObject.open("GET", dataSource + '?rowNumber=' + rowNumber); 

在PHP中:

$result = mysql_query('CALL view_polls(`' . $_GET['rowNumber'] . '`);'); 
+3

不要忘了消毒查询并确保该查询字符串ROWNUMBER确实存在:'如果(isset($ _GET ['rowNumber'])&& is_numeric($ _ GET ['rowNumber'])){...}'! – Leonard

+0

谢谢@Sascha厨房非常 –