[英]fetch a foreign key data from mysql database into select dropdownlist using php
我想創建一個選擇下拉列表,該列表從表“ teamtable”中檢索數據並將其顯示在用戶輸入其選擇的頁面上,並且該選擇的相應ID在其他數據庫“用戶”中提交,其中該列是外鍵。
表格及其內容
團隊表 -
idTeam(INT)(PK)-1,2,3
teamName(VARCHAR)-團隊1,團隊2,團隊3
用戶 -
球隊(INT)(FK)
<html> <head> <script type="text/javascript"> function validateForm() { var f=document.forms["reg"]["team"].value; if ((f==null || f=="")) { alert("All Field must be filled out"); return false; } } </script> <form name="reg" action="user_exec.php" onsubmit="return validateForm()" method="post"> <table width="274" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td colspan="2"> <div align="center"> <?php $remarks=$_GET['remarks']; if ($remarks==null and $remarks=="") { echo 'Register a new user'; } if ($remarks=='success') { echo 'Registration Success'; } ?> </div></td> </tr> <tr> <td><div align="right">Team:</div></td> <td> <?php $mysqli_hostname = "localhost"; $mysqli_user = "root"; $mysqli_password = "my_pass"; $mysqli_database = "my_db"; $prefix = ""; $bd = mysqli_connect($mysqli_hostname, $mysqli_user, $mysqli_password) or die("Could not connect database"); mysqli_select_db($mysqli_database, $bd) or die("Could not select database"); $sql = "SELECT idTeam,teamName FROM teamtable "; $result = mysqli_query($sql); echo "<select name='team'>"; while ($row=mysqli_fetch_array($result)) { echo "<option value='" . $row['idTeam'] ."'>" . $row['teamName'] ."</option>"; } echo "</select>"; ?> </td> </tr> <tr> <td><div align="right"></div></td> <td><input name="submit" type="submit" value="Submit" /></td> </tr> </table> </form> </head> </html>
<?php include('connection.php'); $sql = "SELECT idTeam, teamName FROM team"; $result = $conn->query($sql); echo "<select name='team'>"; if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "<option value='" . $row['idTeam'] ."'>" . $row['teamName'] ."</option>"; } echo "</select>"; } else { echo "0 results"; } $conn->close(); ?>
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