php - 如何从与php中的dropdownlist相关的数据库中获取值-6ren

php - 如何从与php中的dropdownlist相关的数据库中获取值

转载作者：行者123 更新时间：2023-11-30 00:27:22

请帮忙解决这个问题。$query1 和 $query 语句没有给出正确的 o/p。$query1 显示 o/p as-->资源 id #5*错误是--> undefined variable :第 30 行 C:\xampp\htdocs\Programs\hosp.php 中的 hname*

我的问题是:编写一个 PHP 脚本，该脚本接受医院名称并以表格格式打印有关在该医院就诊的医生的信息。

我的数据库是:医院(hid,hname)doctor(did,dname,hid)//这里hid是外键

我写的程序:-

<html>
<body>
<form method="post" action="<?php $PHP_SELF;?>">
Select Hospital Name:<br>
<?php
$con = mysql_connect('localhost', 'root');
if (!$con)
{
    die("Could not connect..".mysql_error());
}
mysql_select_db('employee');

$sql = "SELECT hname FROM hospital";
$result = mysql_query($sql);

echo "<select name='hname'>";
while ($row = mysql_fetch_array($result))
{
    echo "<option value='" . $row['hname'] ."'>" . $row['hname'] ."</option>";
}
echo "</select>";
?>
<br><input type="submit" name="submit" value="submit">
<?
if (isset($_POST['submit']))
{
    //echo "Selected value: $_POST[hname]";
    $hosp = $_POST['hname'];
    echo "Hosp Name: $hosp";
    $query1=mysql_query("select hid from hospital where hname='$hname'");
    //$query1=mysql_query("select hid from hospital where hname='{$_POST[hname]}' ");
    echo $query1;
    $query=mysql_query ("SELECT did from doctor where hid='$query1' "); 
    while( $row = mysql_fetch_array($query))
    {
        $did=$row['did'];
        echo $did;
    }
}
?>
</form>
</body>
</html>

最佳答案

if (isset($_POST['submit']))
{
    $hname = $_POST['hname'];

    //First query
    $query1 = "select hid from hospital where hname='".$hname."'";
    //echo $query1;
    $result=mysql_query($query1);    
    $row = mysql_fetch_assoc($result);
    $hid = $row['hid'];
    //echo $hid;

    //second query
    $query2 = "SELECT did from doctor where hid=".$hid;
    //echo $query2;
    $result2=mysql_query($query2);    
    $row2 = mysql_fetch_assoc($result2);
    $did = $row['did'];
    //echo $did;    
}