多个单选按钮评级相同

Question

I tried to search yet i unable to find out what i am looking for.Please help me.I am working on a rating form which using radio button star for rating,I am using while loop of mysqli_fetch_assoc to pull the date from database.As shown below

<?php
    $sql = "SELECT * FROM orderdetail,orderuser,food WHERE orderuser.CustomerID=orderdetail.CustomerID AND orderuser.OrderID=orderdetail.OrderID AND orderdetail.FoodID=food.FoodID AND orderuser.confirmstatus=1";
    $result = $conn->query($sql) or die($conn->error);


    while($colum = mysqli_fetch_assoc($result))
    {
?>
<img src="../php/<?php echo $colum['Foodphoto'];?>" style="height:150px;width:150px">
<br>
<?php echo $colum['FoodName'];?>
<br>
<div class="stars">
    <form action="">
        <input class="star star-5" id="rate-5[]" type="radio" name="rate[]" />
        <label class="star star-5" for="rate-5[]"></label>
        <input class="star star-4" id="rate-4[]" type="radio" name="rate[]" />
        <label class="star star-4" for="rate-4[]"></label>
        <input class="star star-3" id="rate-3[]" type="radio" name="rate[]" />
        <label class="star star-3" for="rate-3[]"></label>
        <input class="star star-2" id="rate-2[]" type="radio" name="rate[]" />
        <label class="star star-2" for="rate-2[]"></label>
        <input class="star star-1" id="rate-1[]" type="radio" name="rate[]" />
        <label class="star star-1" for="rate-1[]"></label>
    </form>
</div>
<hr>
<?php
    }
?>

but the rating for the second result is not working,i find out that is due to the problem of name but i tried to edit it by adding the array still not working. And how could i store them into mysql using php?those tutorial i found only using ajax or jquery i need the ratings submit once the user submitted the form.

Answer 1

The problem I think is that you cannot relate the form being submitted with the record so any update will not work as expected. One of the simplest ways to accomplis this would be a hidden field with the ID of the record - the ID can then be used in the update statement - ie: ( pseudo sql update )

update orderdetail set rating=$rate where ID=$id

So, add a hidden field per form like this perhaps

    while($colum = mysqli_fetch_assoc($result)){
?>

<img src='../php/<?php echo $colum['Foodphoto'];?>' style='height:150px;width:150px'>
<br>
<?php echo $colum['FoodName'];?>
<br>

<div class='stars'>
  <form action=''>
    <input class='star star-5' type='radio' id='rate-5' name='rate' />
    <label class='star star-5' for='rate-5'></label>

    <input class='star star-4' type='radio' id='rate-4' name='rate' />
    <label class='star star-4' for='rate-4'></label>

    <input class='star star-3' type='radio' id='rate-3' name='rate' />
    <label class='star star-3' for='rate-3'></label>

    <input class='star star-2' type='radio' id='rate-2' name='rate' />
    <label class='star star-2' for='rate-2'></label>

    <input class='star star-1' type='radio' id='rate-1' name='rate' />
    <label class='star star-1' for='rate-1'></label>
    <!--
        HIDDEN FIELD WITH ID
    -->
    <input type='hidden' name='id' value='<?php echo $colum['id'];?>' />
  </form>
</div>
<hr>