任何人都可以幫助我解決這個問題。爲了顯示來自MySql數據庫的rezults,我必須選擇學校,課程,科目和考試。但列出所有課程或所有考試不是很實用,所以我想做另一個功能,當我在第一個選擇框中選擇一些學校時,它會在第二個選擇框中顯示我選擇的學校的課程。相關選擇輸入
我的代碼是:在abc.php得到學校的值或ID,讓你需要的標籤,並將結果返回到HTML 功能get_options的相關ID
<div id="allselects">
<form action="viewing.php" method="post" name="filt">
<div class="multsarrange">
<h1 class="choosetext" >Chose Schools</h1>
<select class="multipleselect" name="schools[]" size="8" multiple="multiple" id="shkll">
<?php
$sql = "SELECT * FROM schools ";
$scc=mysql_query($sql);
while ($db_f = mysql_fetch_assoc($scc)) {
$schcd=$db_f['schoolcode'];
$schc=$db_f['schoolname'];
echo "<option value=$schcd >$schc</option>";
}
?>
</select>
</div>
<div class="multsarrange" id="clasaajax">
<h1 class="choosetext" >Chose an Classes</h1>
<select class="multipleselect" name="classes[]" size="8" multiple="multiple" ">
<?php
$c = "SELECT * FROM classes ";
$cl=mysql_query($c);
while ($db_f = mysql_fetch_assoc($cl)) {
$clsc=$db_f['schoolID'];
$claid=$db_f['classID'];
$clay=$db_f['year'];
$clanm=$db_f['className'];
$name=schoolidton($clsc)." ".$clay." ".$clanm;
echo "<option value=$claid >$name</option>";
}
?>
</select>
</div>
<div class="multsarrange">
<h1 class="choosetext" >Chose Subjects</h1>
<select class="multipleselect" name="subjects[]" size="8" multiple="multiple">
<?php
$sb = "SELECT * FROM subjects ";
$sbi=mysql_query($sb);
while ($db_f = mysql_fetch_assoc($sbi)) {
$sbnm=$db_f['subjectName'];
$sbid=$db_f['subjectID'];
echo "<option value=$sbid >$sbnm</option>";
}
?>
</select>
</div>
<div class="multsarrange">
<h1 class="choosetext" >Chose Exams</h1>
<select class="multipleselect" name="exams[]" size="8" multiple="multiple">
<?php
$e = "SELECT * FROM exams ";
$ex=mysql_query($e);
while ($db_f = mysql_fetch_assoc($ex)) {
$id=$db_f['examID'];
$sub=$db_f['subjectID'];
$desc=$db_f['description'];
$year=$db_f['year'];
$data=$db_f['data'];
$exnam=subidton($sub)." - ".$year." - ".$desc." - ".$data;
echo "<option value=$id >$exnam</option>";
}
?>
</select>
</div>
<div id="longsubmit">
</br></br>
<input name="submit" type="submit" value="View" />
</div>
</form>
</div>
我已經完成了這段代碼,它對於一個值很有效 當我選擇一所學校時,該學校的課程列在了sexond選擇框中,但我無法修改它以使用多個選擇值。 這是我使用的代碼 – 2012-04-24 11:41:14
函數AjaxFunction(cat_id) var httpxml; 嘗試 {0},{0},{0},Firefox 8.0+,Safari httpxml = new XMLHttpRequest(); } 趕上(E) {// Internet Explorer的 \t \t嘗試 \t \t \t \t \t { \t \t \t \t httpxml =新的ActiveXObject(「MSXML2。XMLHTTP 「); \t \t \t \t} \t \t \t趕上(E) \t \t \t \t { \t \t \t嘗試 \t \t { \t \t httpxml =新的ActiveXObject(」 Microsoft.XMLHTTP「); \t \t} \t \t \t catch(e) \t \t { \t \t alert(「您的瀏覽器不支持AJAX!」); \t \t return false; \t \t} \t \t} }'' – 2012-04-24 11:41:39
功能stateck() { 如果(httpxml.readyState == 4){ 的document.getElementById( 「CLS」)的innerHTML = httpxml.responseText。 } } \t var url =「clasajax.php」; url = url +「?schools [] =」+ cat_id; url = url +「&sid =」+ Math.random(); httpxml.onreadystatechange = stateck; httpxml.open(「GET」,url,true); httpxml.send(null); } ' – 2012-04-24 11:42:53