Inputting multiple values in database from HTML form using PHP

Question

I'm trying to add multiple values into a database using PHP from an HTML. However, I can't just refer to the name attribute of the HTML form because each field in the form is generated by a PHP script. I've tried Googling around, but since I don't exactly know what I'm looking for, my search has been futile.

Here's the bit of code that I use to generate the HTML form:

<form action="input_points.php" method="post">
<?php
    while($row = mysql_fetch_array($result)) {
        echo $row['Name'] . ' <input type="text" name="userpoints">';
    }
?>
<button type="submit" name="add_points">Add Points </button>
</form>

I don't know what names are currently in the directory so I need this piece of php to determine what names are in the database. Afterwards, I want to have a bunch of boxes for people to input points (hence the form). I'm having trouble figuring out how to link the particular text box with the user.

For example, if I have a text box for Bob, how would I link up the input text field that contains the number of points Bob earns with Bob's entry in the database?

I know you can do this with regular form fields:

$userpoints = $_POST['userpoints'];

UPDATE members SET points = $userpoints where $user = "Bob";

But since I have multiple users, how do I link up the correct database entry with the right user? Also, how would I determine which boxes are empty and which boxes are updated with a value?

Answer 1

The update you're trying to do is not safe unless you treat values to prevent SQL injection... but if you really want it, instead of mysql_fetch_array() , try using mysql_fetch_assoc() .

Using mysql_fetch_assoc() you can extract the keys (database field names) with array_keys() . The keys will be your the name property of your form fields and the values of will be the fields' values.

Hope it helps.

Answer 2

You can use an array to store all the data that you need and add a hidden field that contains the missing data:

<form action="input_points.php" method="post">
<?php
    for($i=0; $row = mysql_fetch_array($result); $i++ ) {
        echo ' <input type="hidden" name="user[0]['name'] value ='". $row['name'] ."'">';
        echo $row['Name'] . ' <input type="text" name="user[$i]['points'] ">';
    }
?>
<button type="submit" name="add_points">Add Points </button>
</form>

Answer 3

If you want to update multiple filed then are using array

Please changes some code

<form action="input_points.php" method="post">
<?php
    $userCount=mysql_num_rows($result);
    echo '<input type="hidden" name="userCount" value="' .$userCount. '">';
    while($row = mysql_fetch_array($result)) {
        echo '<input type="hidden" name="userid[]" value="' .$row['id']. '">'; //Give the uniq id
        echo $row['Name'] . ' <input type="text" name="userpoints[]">';
    }
?>
<button type="submit" name="add_points">Add Points </button>
</form>

PHP Code -

$userCount = $_POST['userCount'];
   for($i=1; $i=$userCount; $i++){
        $userpoints = $_POST['userpoints'];
        $userid = $_POST['userid'];
        //UPDATE members SET points = $userpoints where $user = $userid;
        //YOUR CODE HERE
   }

Answer 4

当你点击提交userpoints 问题只包含最后一个值前面的所有值都将被覆盖解决方案的名称=“userpoints”每次都必须是不同的，为什么不是你在数据库中定义它，然后把它拿来就像你取$行[“名称”]？