不在mysql、php中更新_Php_Html_Mysql_Forms

不在mysql、php中更新

php html mysql forms

不在mysql、php中更新,php,html,mysql,forms,Php,Html,Mysql,Forms,好的，我正在创建一个考勤系统，我想标记一个学生是否出席，这是我的代码 <?php if (isset($_POST['submit'])) { $present = $_POST['present']; } $test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_att

好的，我正在创建一个考勤系统，我想标记一个学生是否出席，这是我的代码

  <?php
if (isset($_POST['submit'])) {

$present = $_POST['present'];


}
$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3)  or die(mysql_error());


echo "</br><table border='1' align='center'><tr> <th><strong>Student ID</strong></th> <th><strong>First Name </strong></th> <th><strong>Last Name</strong></th> <th><strong>Present</strong></th> </tr> ";

while($rows=mysql_fetch_array($result)){
    echo"<form name='Biology_lecture11.php' method='post'>";
    echo "<tr><td width='100' align='center'>" .$rows['student_id'].
"</td><td width='120' align='center'>" .$rows['fname'].
"</td><td width='120' align='center'>" .$rows['lname'].
"</td><td><input type='text' name='present' value=" .$rows['present'] . ">";

}
echo "</table>";
?>
 <input type='submit' name='Submit' value='Submit'  >
   </form>

  <?php 


   $sql="UPDATE course_attendance SET present='$present' WHERE course_id='101' AND week_id='2' ";
  $result=mysql_query($sql);

  if($result){
      echo "Successfully logged the attendance";
  }
  else {
      echo"ERROR";

  }
  ?>

问题是，它不会更新数据库中的present字段，任何人都知道有什么问题

您在表标记和while循环中使用了一个表单，这将不起作用，下面是正确的代码

<?php
if (isset($_POST['submit'])) {

    $present = $_POST['present'];

    $sql="UPDATE course_attendance SET present='$present' WHERE course_id='101' AND week_id='2' ";
    $result=mysql_query($sql);

    if($result) {
        echo "Successfully logged the attendance";
    }
    else {
      echo"ERROR";
    }

}

?>

<form name='Biology_lecture11.php' method='post'>
<table border="1" align="center">
<tr>
    <th><strong>Student ID</strong></th>
    <th><strong>First Name </strong></th>
    <th><strong>Last Name</strong></th>
    <th><strong>Present</strong></th>
</tr>

<?php

$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3)  or die(mysql_error());

while($rows=mysql_fetch_array($result)) {

echo "<tr><td width='100' align='center'>" .$rows['student_id']."</td>
      <td width='120' align='center'>" .$rows['fname']."</td>
      <td width='120' align='center'>" .$rows['lname']."</td>
      <td><input type='text' name='present' value=" .$rows['present']."></td></tr>";
}
echo "</table>";
?>

<input type='submit' name='Submit' value='Submit'  >
</form>

我看到的一个错误是，你这样说：

echo"<form name='Biology_lecture11.php' method='post'>";

在你的while循环中。所以它不止一次被推出。尝试在循环之前的行中写入该部分。

我看到了几个问题：

1：您的更新代码在每次加载页面时都在运行。将更新块移到if-isset$_POST['submit']{}块中。 2：当你打印学生时，你为每个学生创建一个名为present的输入。如果要填写此字段并提交数据，则只会将最后一个字段添加到数据库中。 3：您没有更新特定的学生。我会将输入字段更改为复选框，并将其命名为present[$rows[student_id]]。然后，在处理页面后，循环遍历$\u POST['present']的键/值。并更新其中的所有学生

foreach (array_keys($_POST['present']) as $student_id) {
    if (is_numeric($student_id)) {
        $sql="UPDATE course_attendance SET present='true' WHERE course_id='101' AND week_id='2' and student_id='$student_id'";
    }
}

如果考勤表中没有自动填写学生，则必须修改更新。如果每个学生都不在，你必须运行一个查询，看看他们是否存在。如果他们不插入行。如果有，则更新该行。

4：在打开表之前将开始标记移动到student循环外部。

需要考虑两件事：第一，表单元素复制。正如上面的评论所说，去掉这条线

echo"<form name='Biology_lecture11.php' method='post'>";

希望有帮助

这应该适合你。这将为每个学生分配一个唯一的现值，然后在回发时检查该值，如果设置了该值，则该值将被清除并用于更新出勤中的学生记录

我还将PHP中的echo'd HTML提取为HTML，并将表单移到表外，这可能会在某些浏览器中导致问题

<?php
// Update present values
if (isset($_POST['submit'])) 
{
    // Get a list of student ids to check
    $idsResult = mysql_query("SELECT student_id from students");

    while($idRow = mysql_fetch_array($idsResult))
    {
       // if the textbox for this student is set 
       if(isset($_POST['present'.$idRow['student_id']]) && !empty($_POST['present'.$idRow['student_id']]))
       {
          // Clean the user input, then escape and update the database
          $cleanedPresent = htmlspecialchars(strip_tags($_POST['present'.$idRow['student_id']]));
          $sql = "UPDATE course_attendance SET present='".mysql_real_escape_string($present)."' WHERE course_id='101' AND week_id='2' AND student_id=".$idRow['student_id'];
          $result = mysql_query($sql);

          if($result){
            echo "Successfully logged the attendance for ID ".$idRow['student_id'];
          }
          else {
            echo "ERROR updating on ID ".$idRow['student_id'];
          }
       }
    }
}

$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3)  or die(mysql_error());
?>
<form name='Biology_lecture11.php' method='post'>
</br>
<table border='1' align='center'>
  <tr>
    <th><strong>Student ID</strong></th>
    <th><strong>First Name </strong></th>
    <th><strong>Last Name</strong></th>
    <th><strong>Present</strong></th>
  </tr>
<?php
while($rows=mysql_fetch_array($result)){
  echo "<tr><td width='100' align='center'>" .$rows['student_id'].
  "</td><td width='120' align='center'>" .$rows['fname'].
  "</td><td width='120' align='center'>" .$rows['lname'].
  "</td><td><input type='text' name='present".$rows['student_id']."' value=" .$rows['present'] . ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>

更好的替代方法：如果当前值可以设置为简单的0/1或真/假，那么为每个学生使用复选框会更容易。在回发中，您可以通过选中每个指示在场学生的复选框来检索一个值数组，然后在一个查询中更新数据库表。这也可以防止恶意文本输入

备选代码：

<?php
// Update present values
if (isset($_POST['submit'])) 
{
    // Get a list of student ids to check
    $idsResult = mysql_query("SELECT student_id from students");

    $presentIds = array();
    $absentIds = array();
    while($idRow = mysql_fetch_array($idsResult))
    {
       // If the student's checkbox is checked, add it to the presentIds array.
       if(isset($_POST['present'.$idRow['student_id']]))
       {
         $presentIds[] = $idRow['student_id'];
       }
       else
       {
         $absentIds[] = $idRow['student_id'];
       }
    }

      // Convert array to string for query
      $idsAsString = implode(",", $presentIds);

      // You can set present to whatever you want. I used 1. 
      $sql = "UPDATE course_attendance SET present='1' WHERE course_id='101' AND week_id='2' AND student_id IN (".$idsAsString.")";
      $result = mysql_query($sql);

      if($result){
        echo "Successfully logged the attendance for IDs ".$idsAsString;
      }
      else {
        echo "ERROR updating on IDs ".$idsAsString;
      }


      // OPTIONAL: Mark absent students as '0' or whatever other value you want
      $absentIdsAsString = implode(",", $absentIds);
      // You can set present to whatever you want. I used 1. 
      $absentQuery = "UPDATE course_attendance SET present='0' WHERE course_id='101' AND week_id='2' AND student_id IN (".$absentIdsAsString.")";
      $absentResult = mysql_query($absentQuery);

      if($absentResult){
        echo "Successfully logged absence for IDs ".$absentIdsAsString;
      }
      else {
        echo "ERROR updating absence on IDs ".$absentIdsAsString;
      }

}

$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3)  or die(mysql_error());
?>
<form name='Biology_lecture11.php' method='post'>
</br>
<table border='1' align='center'>
  <tr>
    <th><strong>Student ID</strong></th>
    <th><strong>First Name </strong></th>
    <th><strong>Last Name</strong></th>
    <th><strong>Present</strong></th>
  </tr>
<?php
while($rows=mysql_fetch_array($result)){
  echo "<tr><td width='100' align='center'>" .$rows['student_id'].
  "</td><td width='120' align='center'>" .$rows['fname'].
  "</td><td width='120' align='center'>" .$rows['lname'].
  "</td><td><input type='checkbox' name='present".$rows['student_id']."' ";

  // NOTE: REPLACE 1 with whatever value you store in the database for being present. 
  // I used 1 since the update at the top of the code uses 0 and 1.
  if($rows['present']=='1')
  {
    echo "checked='checked' ";
  }
  // With a checkbox, you don't need to assign it a value.
  echo "value=" .$rows['present'];

  echo ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>

你连接到数据库了吗？是的，我连接到数据库了。看起来你正在更新所有学生的现值，因为你的唯一检查是课程id和周id。如果你的while循环包含多条记录，您将有多个元素实例，MySQL中的表是course_Attention:Attention_id，week_id FK，course_idFK，student_idFK，PresentYou真的给您的儿子起名叫Robert吗；升降台课程_出勤率；-？哦，是的，小鲍比桌子，我们叫他这行不通。while循环中仍然存在表单元素的重复。@Sumit感谢您写出它，但是它仍然没有更新每个特定学生的字段，这是一个奇怪的问题，因为代码在逻辑上似乎是这样的correct@JasjootMuhdar检查我编辑的答案，有遗漏不工作，我不知道为什么！这真的很烦人，我很感激你花时间在这个家伙身上@SumitBijvaniDo你认为如果我有复选框的话可能会更容易吗？我现在正在查看，谢谢你花时间来帮助我，我会让你知道它是否有效！这是行不通的。因为您已将表单标记放入表中，并在表标记后关闭它，所以我发现的第二个问题是，提交按钮名为Submit，并且在开始时您要求$U POST['Submit']。它们应该有相同的名字，注意大写字母和小写字母。我要试试这个out@Gareth对不起，这让人很痛苦，你知道如何用复选框而不是字母来编码吗textbox@JasjootMuhdar是的，我很快会把它添加到答案中。非常感谢您，我在几天后就拿到了我的最后一年项目演示，你会救我的命@JasjootMuhdar请查看我的备用代码。您可以使用当前用于将学生标记为数据库中存在的任何内容替换0和1。

<?php
// Update present values
if (isset($_POST['submit'])) 
{
    // Get a list of student ids to check
    $idsResult = mysql_query("SELECT student_id from students");

    $presentIds = array();
    $absentIds = array();
    while($idRow = mysql_fetch_array($idsResult))
    {
       // If the student's checkbox is checked, add it to the presentIds array.
       if(isset($_POST['present'.$idRow['student_id']]))
       {
         $presentIds[] = $idRow['student_id'];
       }
       else
       {
         $absentIds[] = $idRow['student_id'];
       }
    }

      // Convert array to string for query
      $idsAsString = implode(",", $presentIds);

      // You can set present to whatever you want. I used 1. 
      $sql = "UPDATE course_attendance SET present='1' WHERE course_id='101' AND week_id='2' AND student_id IN (".$idsAsString.")";
      $result = mysql_query($sql);

      if($result){
        echo "Successfully logged the attendance for IDs ".$idsAsString;
      }
      else {
        echo "ERROR updating on IDs ".$idsAsString;
      }


      // OPTIONAL: Mark absent students as '0' or whatever other value you want
      $absentIdsAsString = implode(",", $absentIds);
      // You can set present to whatever you want. I used 1. 
      $absentQuery = "UPDATE course_attendance SET present='0' WHERE course_id='101' AND week_id='2' AND student_id IN (".$absentIdsAsString.")";
      $absentResult = mysql_query($absentQuery);

      if($absentResult){
        echo "Successfully logged absence for IDs ".$absentIdsAsString;
      }
      else {
        echo "ERROR updating absence on IDs ".$absentIdsAsString;
      }

}

$test3= "SELECT * FROM course_attendance, students, courses, attendance WHERE course_attendance.course_id=courses.course_id AND course_attendance.week_id=attendance.week_number_id AND course_attendance.student_id= students.student_id AND courses.course_id='101' AND attendance.week_number_id='2' ";
$result = mysql_query($test3)  or die(mysql_error());
?>
<form name='Biology_lecture11.php' method='post'>
</br>
<table border='1' align='center'>
  <tr>
    <th><strong>Student ID</strong></th>
    <th><strong>First Name </strong></th>
    <th><strong>Last Name</strong></th>
    <th><strong>Present</strong></th>
  </tr>
<?php
while($rows=mysql_fetch_array($result)){
  echo "<tr><td width='100' align='center'>" .$rows['student_id'].
  "</td><td width='120' align='center'>" .$rows['fname'].
  "</td><td width='120' align='center'>" .$rows['lname'].
  "</td><td><input type='checkbox' name='present".$rows['student_id']."' ";

  // NOTE: REPLACE 1 with whatever value you store in the database for being present. 
  // I used 1 since the update at the top of the code uses 0 and 1.
  if($rows['present']=='1')
  {
    echo "checked='checked' ";
  }
  // With a checkbox, you don't need to assign it a value.
  echo "value=" .$rows['present'];

  echo ">";
}
?>
</table>
<input type='submit' name='Submit' value='Submit'>
</form>