我有一个JS/PHP代码段,从数据库中选择,做一些事情与表的ID,并在下拉菜单中输出数据。
我的JS目前追加myfile.php?cat=X&projectID=Y
正确,我正确抓取变量,但当我选择第二个下拉菜单时,它不输出名称,它只追加ID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 " <span class='req'><img src='/images/essentialInput.png' alt='*' title='*' border='0' /></span>";
echo " <span class='label'>Project</span>" . " ";
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 />";
?>
我如何得到它更新它与projectName,而不是重新加载与'选择项目' ?
尝试使用:
var val=form.cat.value;
var val2=form.cat2.value;
在PHP中需要某种形式的if/else逻辑,类似于在Client select中使用的逻辑,但是依赖于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>";
}