一、编程思路：
1. 使用XAJAX。
2. 从mysql数据库的nation表中获得各省的名称字段“province”.
3. 将该字段数组元素分别作为下拉列表的选项.（即列表项分别为：北京，天津，河南，河北……等）
4. 点击按钮“换吧”，之后，将下拉列表显示在
二、问题：
运行后，点击“换吧”按钮，下拉列表没有如期显示在中。什么也没显示。这是怎么回事？问题出在哪里？请高人指点。
三、以下是我的代码：
<?php // A部分：从数据库中取得province字段值include_once("../Connections/fycon.php");mysql_select_db($database_fycon,$fycon); $sqlone="select * from nation where province<>''";mysql_query("set names gb2312",$fycon);$rstone=mysql_query($sqlone,$fycon) or die(mysql_error());// A部分：结束。// ↓ 取得结果集的行数。$rows=mysql_num_rows($rstone);// ↓ 构建空字串，准备存放javascript代码$theOptions="";// B部分：用数据库结果集构建下拉列表框// 思路：将所有代码以字串形式放进变量$theOpotions中，// 将来接用XMLHttpResponse返回该字串，并显示在<div id="yes"></div>中。$theOptions.= "<form name=\"ok\" id=\"ok\">";$theOptions.= "<select>";while($rows){ $getvalue=mysql_fetch_assoc($rstone); //echo "<option value=\"".$getvalue['province']."\">".$getvalue['province']."</option>"; $theOptions.= "<option value=\"".$getvalue['province']."\">".$getvalue['province']."</option>"; $rows=$rows-1;}$theOpotions.= "</select>";$theOpotions.= "</form>";// B部分：结束?>// 下面是xajax部分<?php include_once("../xajax/xajax.inc.php");$myXajax= new xajax();$myFunct=$myXajax->register(XAJAX_FUNCTION,"myFunc");$myXajax->processRequest();function myFunc($arg){ $theOptions=iconv("gb2312","utf-8",$theOptions); //转码，将utf-8转为gb2312 $objResponse= new xajaxResponse(); $objResponse->assign('yes','innerHTML',$theOptions); return $objResponse;}// xajax部分结束?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312" /><title>?</title></head><?php $myXajax->printJavascript("../xajax");?><body ><br/><br/><div id="yes" style="width:897px; height:600px; background-color:#F00; margin-top:0px; margin-right:auto; margin-bottom:0px; margin-left:auto"></div><form><input type="button" value="换吧" onclick="xajax_myFunc('1');"/></form></body></html>
回复讨论(解决方案)
function myFunc($arg){
$theOptions=iconv("gb2312","utf-8",$theOptions); //转码，将utf-8转为gb2312
$objResponse= new xajaxResponse();
$objResponse->assign('yes','innerHTML',$theOptiones);
return $objResponse;
}
iconv("gb2312","utf-8",$theOptions) 是将 $theOptions 从 gb2312 转到 utf-8，确认是要这样吗（与你的注释是相反的）
$theOptions 不是传入的，也没有赋值
你的程序中并没有中文字符串，所以可 mysql_query("set names utf8",$fycon); 让数据库完成字符集转换
function myFunc($arg){
$theOptions=iconv("gb2312","utf-8",$theOptions); //转码，将utf-8转为gb2312
$objResponse= new xajaxResponse();
$objResponse->assign('yes','innerHTML',$theOptiones);
return $objResponse;
}
iconv("gb2312","utf-8",$theOptions) 是将 $theOptions 从 gb2312 转到 utf-8，确认是要这样吗（与你的注释是相反的）
$theOptions 不是传入的，也没有赋值
你的程序中并没有中文字符串，所以可 mysql_query("set names utf8",$fycon); 让数据库完成字符集转换
感谢版主的回复和讲解。让我受益匪浅。而且受你的启发，终于找到问题所在，完善了代码，达到了自己想要的结果。
一、问题所在：
1. 用于返回的字符串变量：$theOptions并没有传入myFunc()函数。
2. 虽然看上去在myFunc()函数中也有一个$theOptions变量，但此变量非彼变量，函数内外的两个$theOptions变量并非同一个变量。【从内存角度来看，这两个变量代表不同的内存地址，根本就不是同一个变量】。
3. 换句话说：函数myFunc()内的$theOptions变量，相当于在函数内新定义的一个$theOptions变量，和函数外的那个同名变量没有一毛钱关系，就好像两个同名同姓的人一样，虽然同名同姓，但根本不是同一个人。
二、问题的解决：
1. 要想将函数外的$theOptions变量在函数内使用，则要让它成为一个全局变量，这样，就如同“同一个人，从函数外进入到了函数内” 。
2. 具体到代码，就是在myFunc()函数中加一行“global $theOptions;”。
3. 以前之所以没有显示“下拉列表框”，就是因为没有将函数外定义和赋值的$theOptions变量传入函数内，而是在函数内新定义了一个没有赋过值的$theOptions同名变量。既然这个新变量啥值也没有，怎么可能显示“下拉列表框”呢？
4. 另外，那句：$theOptions=iconv("gb2312","utf-8",$theOptions); 也是必要的，版主指出我的注释写错了。的确如此，注释将此句的作用写反了，应该是：将gb2312转码为utf-8。之所以说这句必不可少是因为，如果不进行转码，那么仍然会出现不显示下拉列表框的情况。
三、 总结：
1. 越是让人百思不得其解的错误，越是低级错误。
2. 有时，程序没有逻辑错误，但人脑出现了逻辑错误，程序逻辑≠人脑逻辑，就会出现“所得非所想”的结果。
3. 这段错误代码虽然披着xajax的外衣，但实际却是一个普通的低级语法错误，搞混了变量的作用域。
4. 网络诚不欺我，版主的认真解答，解决了我的问题，我也要认真总结以回报网络。下面贴出没有问题的代码。
<?php // A部分：从数据库中取得province字段值include_once("../Connections/fycon.php");mysql_select_db($database_fycon,$fycon); $sqlone="select * from nation where province<>''";mysql_query("set names gb2312",$fycon);$rstone=mysql_query($sqlone,$fycon) or die(mysql_error());// A部分：结束。// ↓ 取得结果集的行数。$rows=mysql_num_rows($rstone);// ↓ 构建空字串，准备存放javascript代码$theOptions="";// B部分：用数据库结果集构建下拉列表框// 思路：将所有代码以字串形式放进变量$theOptions中，// 将来接用XMLHttpResponse返回该字串，并显示在<div id="yes"></div>中。$theOptions.= "<form name=\"ok\" id=\"ok\">";$theOptions.= "<select>";while($rows){ $getvalue=mysql_fetch_assoc($rstone); //echo "<option value=\"".$getvalue['province']."\">".$getvalue['province']."</option>"; $theOptions.= "<option value=\"".$getvalue['province']."\">".$getvalue['province']."</option>"; $rows=$rows-1;}$theOpotions.= "</select>";$theOpotions.= "</form>";// B部分：结束?>// 下面是xajax部分<?php include_once("../xajax/xajax.inc.php");$myXajax= new xajax();$myFunct=$myXajax->register(XAJAX_FUNCTION,"myFunc");$myXajax->processRequest();function myFunc($arg){ global $theOptions; //与错误代码相比，就加了这么一句，解决了问题。 $theOptions=iconv("gb2312","utf-8",$theOptions); //转码，将gb2312转为 utf-8 $objResponse= new xajaxResponse(); $objResponse->assign('yes','innerHTML',$theOptions); return $objResponse;}// xajax部分结束?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312" /><title>?</title></head><?php $myXajax->printJavascript("../xajax");?><body ><br/><br/><div id="yes" style="width:897px; height:600px; background-color:#F00; margin-top:0px; margin-right:auto; margin-bottom:0px; margin-left:auto"></div><form><input type="button" value="换吧" onclick="xajax_myFunc('1');"/></form></body></html> 
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