Question

我正在尝试为我的俱乐部创建一个表单，该表单从数据库中获取信息，以便从数据库中选择受训人员。然后另外从事件列表中选择一个并将INSERT返回到数据库中。它写入数据库确定并循环正确的次数但没有将$ trainee值传递给数据库我认为失败的是从

传递信息

print ' <input type="hidden" name="Trainee" value= ' . $trainee . ' />

到if(isset($_POST['formSubmit'])) loop.

中的$查询

有谁告诉我哪里出错了？代码如下：

<?php 

//Retrieve trainees of specified grade

$data = mysql_query('SELECT * FROM membership WHERE grade = "Trainee"  ') 
or die(mysql_error()); // select works

// Writes to database OK, including Trainee if manual value entered into form like done in instrucot
   $query = "INSERT INTO testtraining ( trainee_no, activity, instructor, entered_by, entered_by_date) VALUES ( '{$_POST['Trainee']}', '{$_POST['activity']}', '{$_POST['instructor']}',  '{$_POST['enteredBy']}', NOW())"; 

// Feedback and posting
    if(isset($_POST['formSubmit'])) 

{
    $aTrainee = $_POST['data'];
    $training = $_POST['activity'];

    if(empty($aTrainee)) 
        {
            echo("<p>You didn't select trainees.</p>\n");               
        }   else {
        $N = count($aTrainee);
        echo("<p>You selected $N trainee(s): ");

            for($i=0; $i < $N; $i++) // loop thru all selected checkbox adding 
                {
                    $trainee = $aTrainee[$i];
                    // Execute the query.
                    if (@mysql_query ($query))  { 
                        // lists OK on screen but does not pass to form for writing to database
                        print "<p>The $training added for  $trainee.</p>";  
                                    }
                }
        }                                   
}
// end of posting

// Start of form
// Creates list with checkbox, cycles through info from membership database and makes a     multi select checkbox list
 while($info = mysql_fetch_array( $data )) //repeat while there is still data from SELECT
{ 
?> 
<form action ="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post"  >  
<input id= "<?= $info['no'] ?>" type="checkbox" name="data[]" value="<?= $info['no'] ?>" />
<label for="<?= $info['no'] ?>"><?= $info['_no'] ?></label>
<br />
<? 
 }  

// Training Activities checkbox, Displays training activity to be selected from
print '<p><input type="radio" name="activity" value="Training1" /> Training1</p>';       //works
print '<p><input type="radio" name="activity" value="Training2" /> Training2</p>'; //works

print ' <input type="hidden" name="Trainee" value= ' . $trainee . ' />
<input type="hidden" name="instructor" value= anInstructor />   
<input type="hidden" name="enteredBy" value=' . ($_SESSION['username']) . ' /> 
<input type="submit" name="formSubmit" value="Add Training" />
</form>';

mysql_close(); // Close the database connection;
?>

Answer 1

您的查询不会从字符串中突破以插入变量。而是尝试：

$query = "INSERT INTO testtraining ( trainee_no, activity, instructor, entered_by, entered_by_date) VALUES ( '".$_POST['Trainee']."', '".$_POST['activity']."', '".$_POST['instructor']."','".$_POST['enteredBy']."', NOW())";

虽然我建议先将这些$ _POST变量添加到$ variables中并运行一些验证以确保它是干净的。 addslashes()是确保不会弹出SQL错误的良好开端。但这不是关于安全插入已清理的用户输入的讲座。

多条目表单不将信息传递给mysql数据库

1 个答案: