doushou3814 2012-10-17 07:53 采纳率: 100%
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从数据库加载值以根据先前下拉列表中的选择创建下拉菜单

My problem is that I try to load a drop down based on selection from previous drop down. I'm loading the values from database. but when I use $_get['emirate'] I'm not getting the value. It says the value is not set.

$data = mysql_query("SELECT * FROM emirate") 
        or die(mysql_error()); 

    Print "<table border cellpadding=3>"; 
    print"<tr><th><form method=get action='index.php'><select id=EMIRATE size=1><OPTION value=all>all</option>";
    while($info = mysql_fetch_array( $data )) 
    { 
        Print "<option value=". $info['em_name'] .">".$info['em_name']."</option>"; 
    }
    print"</select></form></th>";
    if(isset($_GET['EMIRATE'])){
        $name=$_GET['EMIRATE'];
        echo $name;

        $data = mysql_query("SELECT a_name FROM areas where em_id=(select em_id from emirate where em_name=\"".$name."\")") 
        or die(mysql_error());
        Print "<th><select name=AREAS size=1>";
        while($info = mysql_fetch_array( $data ))
        {
            Print "<option >".$info['a_name']."</option>";
        }
        print"</select></th>";
    }
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3条回答 默认 最新

  • dongqiaozhe5070 2012-11-22 06:05
    关注

    I found out the problem.

    I haven't put a submit button and hence my form wasn't getting submitted. And hence I wasn't getting any value from $_GET.

    http://php.net/manual/en/reserved.variables.get.php

    Now I get the results smoothly.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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