Question

我一直致力于一个用户选择他的userame和日期范围的项目，当点击提交时，会出现一个html表。现在用户必须在其中一个中填写一个输入字段（下拉列表）。列和单击提交后，输入数据应相应地存储在数据库中。我设法将表格（下拉）放在正确的位置，但我如何调用所选数据并将其保存到数据库中。

<?php
    if (isset($_POST['submitted'])) {
    include ('connect.php');
    $username = $_POST['user'];
    $from = $_POST['from'];
    $to = $_POST['to'];
    $query = "SELECT * FROM errorreport WHERE user = '$username' AND date BETWEEN '$from' AND '$to' ORDER BY date DESC";
    $result = mysqli_query($dbcon, $query) or die ('cant find data');
    $num_rows = mysqli_num_rows($result);
    echo "welcome '$username' \n '$num_rows' results found";
    echo "<table>";
    echo "<table border='1' width:'100%'>";
    echo "<tr><th>DATE</th><th>TASK_QUEUE</th><th>AUDIO</th><th>TRANSCRIBED</th><th>VERIFIED</th><th>OVERTURNED_FIELDS</th><th>TYPE OF ERROR</th><th>BLUESHIFT_LINK</th></tr>";
    echo "<form action='insert.php' method='post'>";
    while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    echo $row['OVERTURNED_FIELDS'];
    echo "</td><td>";
    echo "<select name = error_type>";
    echo "<option value= > </option>
    <option value=Convention Error>Convention Error</option>
    <option value=Listening Error>Listening Error</option>
    <option value=Referencing Error>Referencing Error</option>
    <option value=Spelling/Typos>Spelling/Typos</option>
    <option value=Metadata Error>Metadata Error</option>
    </select>";
    echo "</td><td>";
    echo "<a href=".$row['BLUESHIFT_LINK'].">Click Here </a>";
    echo "</td></tr>";
    }
    echo "</table>";
    echo "<input type=submit value=submit>";
    echo "</form>";
    }

    ?>

please find the photo for a look at the webpage. When i select inputs from each dropdown and click submit at the end of the table, the data is not being saved to the db.

I get this error message when i submit.

请在下面找到insert.php代码。

<?php
DEFINE ('DB_USER', 'root');
DEFINE ('DB_PSWD', '');
DEFINE ('DB_HOST', 'localhost');
DEFINE ('DB_NAME', 'testdb');

$dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME);
if(!$dbcon)
{
    echo 'not connected to server';
}
if(!mysqli_select_db($dbcon,'testdb'))
{
    echo 'database not selected';
}

$errortype = $_POST['error_type'];
$audio=$_POST['audio'];
$username=$_POST['name'];


$sql = "UPDATE errorreport SET error_type = '$errortype'  WHERE user=  '$username' and audio = '$audio'";
if(!mysqli_query($dbcon,$sql))
{
    echo 'not inserted';
}
else
{
    echo 'inserted';
}
header("refresh:100; url=select1.php");

?>
<html>
 <head>
 <title>Welcome to Error Report Portal </title>
 <style type="text/css">
 table {
     background-color: #FFFACD;
 }
 th {
     width: 150px;
     text-align: center;
 }
</style>
</head>
<body>
  <h1> Welcome to Error Report Portal! </h1>
  </body>
  </html>

Answer 1

这些通知是因为您创建了名为error_type的选择框，因此$ _POST [＆＃39; error_type＆＃39;];可以访问，但$ _POST [＆＃39;音频＆＃39;]和$ _POST [＆＃39;名称＆＃39;]没有带有这些名称的输入字段，所以只提醒通知

1.请为name属性创建两个带有这些名称和音频的输入字段 2.或使用这样的属性 $音频= @ $ _ POST [＆＃39;音频＆＃39;]; $ USERNAME = @ $ _ POST [＆＃39;名称＆＃39;];

3.请使用这样的查询进行良好的编程练习...

"UPDATE errorreport SET error_type = '".$errortype."'  WHERE user=  '".$username."' and audio = '".$audio."'"

Answer 2

创建两个隐藏的输入字段，将名称和audiolink作为控件的值进行回显。例如。

<input type="hidden" name="name" value="<?php echo $name;" />

如何从while循环下的dropdown中获取值，并使用php将数据存储在db中

2 个答案: