How to fetch next row from database on each button click in PHP

Question

Whenever I click the button, the page got refreshed and lost the previous values in variables. When I use mt_rand function, the fetched rows are sometimes duplicated. What to do to get next row on each button click?

Fetch random row is not necessary, at least I want to get the next row in order.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<?php

//Database initialization
$StudName ="";
$Name="";
$Task="";
$connection=mysqli_connect("localhost","APB","12345678","apb");

$number=1;
if(isset( $_POST["button1"]))
{
    $number = mt_rand(1,10);
    $s="SELECT * FROM student_list WHERE NUM='$number'";
    $r=  mysqli_query($connection,$s);
    $row = mysqli_fetch_array($r);

    $StudName = $row["NAME"];
    $Name="";
}

if(isset($_POST["Submit5"]))
{
    $numb = mt_rand(1,4);

    $s="SELECT * FROM computer WHERE rn='$numb'";
    $r=  mysqli_query($connection,$s);
    $row=mysqli_fetch_array($r);

    $Task = $row["task"];
    $StudName ="";
    $Name ="";
}
?>

<html xmlns="http://www.w3.org/1999/xhtml">
    <body>
        <form id="form" action="" name="form1" method="POST" >
            <h1>MPTC FRESHER'S DAY CELEBRATION - 2017</h1>
            <h5>This time the lucky one is ...</h5>

            <p>
                <input type="hidden" name="process" value="1";>
                <input onclick="<?php $number=$number+1;?>" name="button1"   type="submit" id="button1" value="GENERATE" />
                <br/>
                <input id="t1" name="textfield" type="text"  value="<?php echo $StudName; ?>" />
            </p>

            <p>
                <h5>Be Ready !! </h5>

                <input type="submit" value="TASK" name="Submit5" value="COMPUTER" />
                <br/>
                <input name="textfield3" type="text" value="<?php echo $Task; ?>" />
            </p>
        </form>
    </body>
</html>

Kempeth · Accepted Answer · 2017-07-26 08:53:43Z

2

First of all that onclick="<?php $number=$number+1;?>" does nothing. You cannot use a js event handler to execute php code.

What you can do, and you're actually very close to the solution, is this:

        <p>
            <input type="hidden" name="process" value="<?php echo $number;?>";>
            <input name="button1" type="submit" id="button1" value="GENERATE" />
            <br/>
            <input id="t1" name="textfield" type="text"  value="<?php echo $StudName; ?>" />
        </p>

and change the php part like this:

$number= $_POST['process'];
if(isset( $_POST["button1"]))
{
    $number = $number + 1;
}
$s="SELECT * FROM student_list WHERE NUM='$number'";
$r=  mysqli_query($connection,$s);
$row = mysqli_fetch_array($r);

$StudName = $row["NAME"];
$Name="";

answered Jul 26, 2017 at 8:53

Kempeth

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Athar Iqbal · Accepted Answer · 2017-07-26 08:02:10Z

0

Use AJAX and append response in variable and show.

answered Jul 26, 2017 at 8:02

Athar Iqbal

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Gillian Lo Wong · Accepted Answer · 2017-07-26 11:05:28Z

0

To get the previous or next row (number) in the database you can do this query:

SELECT * FROM student_list WHERE num = (SELECT min(num) FROM student_list WHERE num > '$number') LIMIT 1;

And for the previous its the same but use smaller than (<);

Another tip I would like to give you is to not inject variables directly into the query. But if you have to, at least if you know the value is going to be a number you can cast it to int by doing $number = (int) $number; Other functions you could look into are is_numeric, and $number > 0, etc.

But rather avoid placing these variables directly into the query. Look into prepared statements, PDO works out of the box and is much safer than using mysqli_*.

answered Jul 26, 2017 at 11:05

Gillian Lo Wong

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