新手问个函数的有关问题
新手问个函数的问题
function del_news($nid)
{
$selnews_str = "select n_class,n_image,n_pname,n_path from news where n_id=".$nid;
$selnews_result = mysql_query($selnews_str,$con);
$selnews_row = mysql_fetch_array($selnews_result);
$class = $selnews_row["n_class"];
if($selnews_row)
{
echo("有内容");
}
else
{
echo("无内容");
}
echo($nid);
}
if($_GET["nid"] != null || $_GET["nid"] != "")
{
del_news($_GET["nid"]);
}
以上是代码,如果把del_news里的代码放在外面的话就显示有内容,放在函数里调用的话就无内容,参数是正确的。这是什么原因呢?
------解决方案--------------------
- PHP code
function del_news($nid) { } $selnews_str = "select n_class,n_image,n_pname,n_path from news where n_id=".$nid; $selnews_result = mysql_query($selnews_str,$con); $selnews_row = mysql_fetch_array($selnews_result); $class = $selnews_row["n_class"]; if($selnews_row) { echo("有内容"); } else { echo("无内容"); } echo($nid); if($_GET["nid"] != null || $_GET["nid"] != "") { del_news($_GET["nid"]); } <br><font color="#e78608">------解决方案--------------------</font><br>错误信息是什么?<br><br>猜测:<br>可能是函数作用域的问题。 你的数据库连接句柄是全局的。在函数里面操作数据被认为是没有连接数据库的。 <br><font color="#e78608">------解决方案--------------------</font><br>$selnews_result = mysql_query($selnews_str,$con);<br>这里的$conn是局部变量<br>没有声明Global。<br>可能是这个原因造成的。<br>把数据库连接句柄加入函数内部试下。<br> <br><font color="#e78608">------解决方案--------------------</font><br>$selnews_result = mysql_query($selnews_str); <br><font color="#e78608">------解决方案--------------------</font><br>执行顺序! 你放在函数里边 怎么调用啊! <br><font color="#e78608">------解决方案--------------------</font><br>你那种可以解决。 还有版主那样也行。 <div class="clear"> </div>

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