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Php <;选择>;使用JS函数正确下拉输出my MYSQL数据_Php_Javascript - Fatal编程技术网

Php <;选择>;使用JS函数正确下拉输出my MYSQL数据

Php <;选择>;使用JS函数正确下拉输出my MYSQL数据,php,javascript,Php,Javascript,我有一段JS/PHP代码,它从数据库中进行选择,使用表的ID做一些事情,并在下拉列表中输出数据 我的JS当前正确地追加了myfile.php?cat=X&projectd=Y,并且我正确地获取了变量,但是当我选择第二个下拉列表时,它不会输出名称,它只将ID追加到标题和变量 function reload(form) { var val=form.cat.options[form.cat.options.selectedIndex].value; var val2

我有一段JS/PHP代码,它从数据库中进行选择,使用表的ID做一些事情,并在下拉列表中输出数据

我的JS当前正确地追加了
myfile.php?cat=X&projectd=Y
,并且我正确地获取了变量,但是当我选择第二个下拉列表时,它不会输出名称,它只将ID追加到标题和变量

     function reload(form)
    {
    var val=form.cat.options[form.cat.options.selectedIndex].value;
    var val2=form.cat2.options[form.cat2.options.selectedIndex].value;
    self.location='add-recharge.php?cat=' + val + '&projectID=' + val2;
    }
PHP

<?

            @$cat=$_GET['cat'];
            if(strlen($cat) > 0 and !is_numeric($cat)){  
            echo "Data Error";
            exit;
            }

            //MYSQL stuff


            echo "<select name='cat' onchange=\"reload(this.form)\"><option value=''>Select a client</option>";

                while($noticia2 = mysql_fetch_array($quer2)) { 
                 if($noticia2['clientID']==@$cat){echo "<option selected value='$noticia2[clientID]'>$noticia2[clientName]</option>"."<BR>";}
                    else{echo  "<option value='$noticia2[clientID]'>$noticia2[clientName]</option>";}
                }
                echo "</select>";
                echo "&nbsp;<span class='req'><img src='/images/essentialInput.png' alt='*' title='*' border='0' /></span>";
                echo "&nbsp; &nbsp;<span class='label'>Project</span>" . "&nbsp;";
                echo "<select name='cat2' onchange=\"reload(this.form)\"><option value=''>Select a project</option>";

                while($noticia = mysql_fetch_array($quer)) { 
                    echo  "<option value='$noticia[projectID]'>$noticia[projectName]</option>";
                }
                 echo "</select>";
                 echo $intClientID;
                 //echo "<br /><br /><br /><br />";   
       ?>
尝试使用:

var val=form.cat.value

var val2=form.cat2.value

编辑

抱歉,我误解了这个问题

在PHP中需要某种形式的if/else逻辑,类似于客户端select上使用的逻辑,但依赖于projectID。比如:

if( $_GET['projectID'] == $noticia[projectID] ){
    echo "<option selected value='$noticia[projectID]'>$noticia[projectName]</option>";
} else {
    echo "<option value='$noticia[projectID]'>$noticia[projectName]</option>";
}
if($\u GET['projectID]==$noticia[projectID]){
echo“$noticia[projectName]”;
}否则{
echo“$noticia[projectName]”;
}

仍将projectID添加到变量中,但将其更改回“选择…”