PHP+MySQL实现下拉框展示数据库信息

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Release: 2016-06-13 12:10:31
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PHP+MySQL实现下拉框显示数据库信息

本帖最后由 kitttyfjy12 于 2012-05-20 17:18:03 编辑

  Classroom research



  

CRMS - Classroom research



  

    
      
        
         
      
      
  
  
      

      
        
      
    
Classroom ID
Course ID

        

  




1、我希望在第一个框输入课室号,第二框通过下拉框显示数据库中course2表的CID的内容。
这应该就是不对的,毫无头绪,不知道怎么实现。求详细代码!
2、怎样实现填写其中一个数据(即不用填两个)就可以查询数据?

------解决思路----------------------
仅供参考:
<br /><?php<br />/* Created on [2012-5-16] */<br />#查询标题信息<br />$sql="select * from table";<br />	$res=mysql_query($sql);<br />	if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());<br />	if(mysql_affected_rows() > 0){<br />		$titles = array();<br />		while($rows = mysql_fetch_array(MYSQL_ASSOC)){<br />			array_push($titles,$rows);<br />		}<br />	}<br />?><br /><br /><table border=1><br /><?php foreach($titles as $row_Recordset_task){ ?><br />	<tr><br />		<td><br />			<a href="javascript:void(0)" onclick="record(<?=$row_Recordset_task['TID']?>)" ><br />				<?=$row_Recordset_task['csa_title']?><br />			</a><br />		</td><br />	</tr><br /><?php } ?><br /></table><br /><div id="show"></div><br /><br /><br /><form name="frm"><br /><select name="s1" onChange="redirec(this.value)"><br /> <option selected>请选择</option><br /> <option value="1">内科</option><br /> <option value="2">内科</option><br /> <option value="3">内科</option><br /></select><br /><div id="s2"></div><br /></form><br /><script><br />//Ajax<br />var xmlHttp;<br /><br />	function createXMLHttpRequest() {<br />		if(window.XMLHttpRequest) {<br />			xmlHttp = new XMLHttpRequest();<br />		} else if (window.ActiveXObject) {<br />			xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");<br />		}<br />	}<br /><br />	function record(id){<br />		createXMLHttpRequest();<br />		url = "action.php?id="+id+"&ran="+Math.random();<br />		method = "GET";<br />		xmlHttp.open(method,url,true);<br />		xmlHttp.onreadystatechange = show;<br />		xmlHttp.send(null);<br />	}<br /><br />	function show(){<br />		if (xmlHttp.readyState == 4){<br />			if (xmlHttp.status == 200){<br />				var text = xmlHttp.responseText;<br />				document.getElementById("s2").innerHTML = text;<br />			}else {<br />				alert("response error code:"+xmlHttp.status);<br />			}<br />		}<br />	}<br /></script><br /><?php<br />#action.php<br />if(isset($_GET['id'])){<br />	$sql="select * from table where id=".$_GET['id'];<br />	$res=mysql_query($sql);<br />	if(!$res) die("SQL: {$sql} <br>Error:".mysql_error());<br />	if(mysql_affected_rows() > 0){<br />		$arrMenu=array();<br />		while($rows = mysql_fetch_array(MYSQL_ASSOC)){<br />			array_push($arrMenu,$rows);<br />		}<br />	}<br />	mysql_close();<br />	if(!empty($arrMenu)){<br />		echo "<select name='menu2'>";<br />		foreach($arrMenu as $item2){<br />			echo "<option value='{$item2['id']}'>{$item2['name']}</option>";<br />		}<br />		echo "</select>";<br />	}<br />}<br /><br />?><br /><br />
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------解决思路----------------------
就这么简单,不需要上面那么复杂!!伱直接复制过去就可以用了。。。。

//require_once('conn.php'); //伱最好写个连接数据库的文件 每次包含一下就行了! 而且要写在最上面

//最好把下面三行写在conn.php文件里 以后每次用时 向上面那样包含一下就OK了!!
$con = mysql_connect("localhost","root","***") or die("错误信息:".mysql_error()); //连接
$db = mysql_select_db("表course2所在的数据库名"); //这个要不写就取不着数据 但不会报错
mysql_query("set names gb2312");

?>




无标题文档






------解决思路----------------------
搞这么复杂?script标签在代码底部加也行啊,只要有:
<br /><script><br />//Ajax<br />var xmlHttp;<br /><br />    function createXMLHttpRequest() {<br />        if(window.XMLHttpRequest) {<br />            xmlHttp = new XMLHttpRequest();<br />        } else if (window.ActiveXObject) {<br />            xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");<br />        }<br />    }<br /><br />    function record(id){<br />        createXMLHttpRequest();<br />//指定目标地址及参数<br />        url = "action.php?id="+id+"&ran="+Math.random();<br />        method = "GET";<br />        xmlHttp.open(method,url,true);<br />        xmlHttp.onreadystatechange = show;<br />        xmlHttp.send(null);<br />    }<br /><br />    function show(){<br />        if (xmlHttp.readyState == 4){<br />            if (xmlHttp.status == 200){<br />//回调函数,返回的后端结果<br />                var text = xmlHttp.responseText;<br />                document.getElementById("s2").innerHTML = text;<br />            }else {<br />                alert("response error code:"+xmlHttp.status);<br />            }<br />        }<br />    }<br /></script><br />
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能运行就行
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