如何使用MySQL为现有值增加价值？

Question

hi dear friends yesterday I asked about a teachers rating system but did not get any idea please help me. i am making a teacher rating system where i want to rate every teacher by making text boxes in front of the name of each teacher .the number that the user will add in text box for a specific teacher is to be added to a current rate of teacher.but the code is not working properly.when i add values through text boxes it adds the value from the last text box to all rows.for example for the first teacher i want to add 2 ,for the second i want to add 4 and for the 3rd teacher i want to rate 3. so instead of 2 and 3 all the rating of all teachers are incremented by 3.how to solve the problem?please help i am giving code of 2 pages m.php and n.php

m.php

<form action="n.php" method="post" enctype="multipart/form-data">
<table width="642" height="215" border="10" align="left" cellspacing="0"  >
    <tr>
        <th class="style5">Teacher ID</th>
        <th width="90" class="style5">Teacher Name</th>
        <th width="127" class="style5">Teacher Registration</th>
        <th width="135" class="style5">Teacher Qualification</th>
        <th width="92" class="style5">Teacher Subject</th>
        <th width="92" class="style5">Action</th>
      </tr>
    <?php
    include 'conn.php';

    $sql = "SELECT * FROM teacher ";
    $result = $conn->query($sql);
    if ($result->num_rows > 0) {
        while($row = $result->fetch_array()){
        $id=$row['tid'];
        ?>
        <tr>
            <td height="50" align="center" class="style5"><?php echo $row['tid'];?></td>
            <td align="center" class="style5"><?php echo $row['tname'];?></td>
            <td align="center" class="style5"><?php echo $row['treg'];?></td>
            <td align="center" class="style5"><?php echo $row['qualification'];?></td>
            <td align="center" class="style5"><?php echo $row['subject'];?></td>
             <td align="center"> <input type="text" name="rating">
          </td>
        </tr>
        <?php
        }   
    }else{
        echo "<center><p><font size=10/> No Records</p></center>";
    }

    $conn->close();
    ?><tr><td colspan="6">
    <input type="submit" name="submit" value="Enter"></td></tr>
  </table>
</form>

n.php

<?php
mysql_connect("localhost","root","") or die ("couldnt connnect to server");
mysql_select_db("project")  or die ("couldnt connnect to database");


include 'conn.php';
if(isset($_POST['submit']))
{
    $sql = "SELECT * FROM teacher";
    $result = $conn->query($sql);
    if($result->num_rows > 0){
        while($row = $result->fetch_array()){
            $id=$row['tid'];
            $newvalue=$_POST['rating'];
            $sql_update="UPDATE teacher set hits =  hits +$newvalue where tid=".$id.""; 
            mysql_query($sql_update) or die(mysql_error());
            header("location:m.php");
        }
    }
}

Answer 1

Like this ... give rating name array with id from table and update the record with the right value and id ... but never do this way for username and password case ...

<form action="n.php" method="post" enctype="multipart/form-data">
<table width="642" height="215" border="10" align="left" cellspacing="0"  >
<tr>
    <th class="style5">Teacher ID</th>
    <th width="90" class="style5">Teacher Name</th>
    <th width="127" class="style5">Teacher Registration</th>
    <th width="135" class="style5">Teacher Qualification</th>
    <th width="92" class="style5">Teacher Subject</th>
    <th width="92" class="style5">Action</th>
</tr>
<?php
include 'conn.php';
$sql = "SELECT * FROM teacher ";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
    while($row = $result->fetch_array()){
    $id=$row['tid'];
    ?>
    <tr>
        <td height="50" align="center" class="style5"><?php echo $row['tid'];?></td>
        <td align="center" class="style5"><?php echo $row['tname'];?></td>
        <td align="center" class="style5"><?php echo $row['treg'];?></td>
        <td align="center" class="style5"><?php echo $row['qualification'];?></td>
        <td align="center" class="style5"><?php echo $row['subject'];?></td>
         <td align="center"> <input type="text" name="rating[<?php echo $id; ?>]">
      </td>
    </tr>
    <?php
    }   
}else{
    echo "<center><p><font size=10/> No Records</p></center>";
}

$conn->close();
?><tr><td colspan="6">
<input type="submit" name="submit" value="Enter"></td></tr>
</table>
</form>

---

<?php
mysql_connect("localhost","root","") or die ("couldnt connnect to server");
mysql_select_db("project")  or die ("couldnt connnect to database");
include 'conn.php';
if(isset($_POST['submit']))
{
$sql = "SELECT * FROM teacher";
$result = $conn->query($sql);
if($result->num_rows > 0){
    while($row = $result->fetch_array()){
        $id=$row['tid'];
        $newvalue=$_POST['rating'];
        if(isset($newvalue[$id])) {
            $new_temp_value = $newvalue[$id];
            $sql_update="UPDATE teacher set hits =  hits +$new_temp_value where tid=".$id."";
            mysql_query($sql_update) or die(mysql_error());
        }
    }
}

header("location:m.php");
}