从下拉菜单中捕获用户所选值的数据库ID

Question

您好，我目前正在为我的学位项目构建一个粗略的原型，并且我在上周遇到了一些问题。基本上我有一个测试人员记录防病毒软件测试结果的系统。我已经填充了父表中记录的下拉列表，现在我想捕获用户选择的值和用户标识，并将其输入到链接表中以记录此信息。但我只想捕获该测试的主键testID而不是测试的名称，这是我卡住的地方。如果有人有任何想法，这将帮助我很多。

继承人的第一页

<?php 
//Template for all pages 
$pagename="Select File Associations"; 
session_start(); 
include ("mysqli_test.php"); 
// css style sheet for template 
echo "<link rel=stylesheet type=text/css href=stylesheet.css>"; 
// displays name in window tab 
echo "<title>".$pagename."</title>"; 

// use header file 

include("header.html"); 

// display the current date and time 
echo date ('l d F Y H:i:s'); 

echo "<p></p>"; 
// display page name on the page 
echo "<h2>".$pagename."</h2>"; 

$testsql = "SELECT testID, testName from tests"; 
$testresult = mysql_query($testsql); 

echo "<td>Select Test:</td>"; 
echo "<select name= testName>"; 
while ($row = mysql_fetch_array($testresult)) 
{ 
echo "<option value=\"". $row['testID']."\">". $row['testName']. "</option>"; 
} 
echo "</select>"; 



echo "<form method=post action= confirm_selection2.php>"; 
echo "<tr><td><input type=submit value='Submit'></td>"; 
echo "<td><input type=reset value='clear'></td></tr>"; 


// use footer file 
include("footer.html"); 
?> 

那么表单处理页面confirm_selections.php代码

<?php 
//Template for all pages 
$pagename="confirm_selection"; 
session_start(); 
include ("mysqli_test.php"); 

// css style sheet for template 
echo "<link rel=stylesheet type=text/css href=stylesheet.css>"; 
// displays name in window tab 
echo "<title>".$pagename."</title>"; 

// use header file 
include("header.html"); 

// display the current date and time 
echo date ('l d F Y H:i:s'); 


echo "<p></p>"; 
// display page name on the page 
echo "<h2>".$pagename."</h2>"; 

// retrieve POST data from the user selected test 
$testselection=$_POST['testName']; 

// check if the selections were made 

if(isset($testselection)){ 

$instestSQL = "INSERT INTO test_performance SET 
testID='$testselection' 
testerID='$_SESSION['testID']); 
$exeintestSQL=mysql_query($instestSQL) OR die(mysql_error()); 
} 

// use footer file 
include("footer.html"); 

?> 

Answer 1

的选择元件的形式以外，将其更改为：

echo "<form method=post action= confirm_selection2.php>"; 
echo "<select name= testName>"; 
while ($row = mysql_fetch_array($testresult)) 
{ 
echo "<option value=\"". $row['testID']."\">". $row['testName']. "</option>"; 
} 
echo "</select>"; 
echo "<tr><td><input type=submit value='Submit'></td>"; 
echo "<td><input type=reset value='clear'></td></tr>"; 
echo "</form>"; 

所有表单元素应放置在<form></form>元素内。