MySQL Fetch Array Each Row Function

Question

This is the first time I'm using this website, I have recently started working on a project where the webpage will display two table, one of the table will gather all the information from a SQL Database and place them in the table along with a button next to each rows allowing you to insert that record to another to the other table, so far, this is the code I have written:

//Include the Following Files
include 'connect.php';

//Select All Price Plans
$query = "SELECT * FROM pricing";
$result = mysql_query($query);

//Print all Plans in Table 
while ($pricing = mysql_fetch_array($result)) 
{
    echo "<table border=1>";
    echo "<tr>";
    echo "<td>" .$pricing['planID'].  "</td>";
    echo "<td>" .$pricing['planTitle']. "</td>";
    echo "<td>" .$pricing['planDescription']. "</td>";
    echo "<td>" .$pricing['planPricing']. "</td>";
    echo "<td>" . **<a href="'. $link .'">BUTTON HERE</a>** . "</td>";
    echo "</tr>";
    echo "<table>";
}  


//Assign $link to Case     
$link = '?run=planA'; 

//Pre-defined Functions
function planA()
{ 
$query = "INSERT into restaurant_plan (`planID`, `planTitle`, `planDescription`, `planPricing`) SELECT * FROM pricing WHERE planID='2' ";
$result = mysql_query($query);
} 

//Setting Case to Run Each Functions
if (isset($_GET['run'])) $linkchoice=$_GET['run']; 
else $linkchoice=''; 

switch($linkchoice)
{       
    case 'planA': 
    planA(); 
    break; 

    case 'planB': 
    planB(); 
    break;              
}

Could anyone suggest any guide or possibly an example how I could assign a function to each rows? Thanks a lot!

It may not help answer your question, but you should stop using mysql_* functions. They're being deprecated. Instead use PDO (supported as of PHP 5.1) or mysqli (supported as of PHP 4.1). If you're not sure which one to use, read this article. — Matt
– Matt, Commented Aug 20, 2012 at 14:22
Well, this is the script I have so far, but I'm confused how I can get it to assign a function based on each rows, I did try a counter but that didn't worked out. Thanks — Beeraj Jhummun
– Beeraj Jhummun, Commented Aug 20, 2012 at 14:25

Yoones Mehdian · Accepted Answer · 2012-08-20 15:05:05Z

your code prints one table for each record in table "pricing", use this code instead :

//Select All Price Plans
$mysqli = new mysqli("hostname", "username", "pass", "dbname");

$query = "SELECT * FROM pricing";

$result = $mysqli->query($query);

//Print all Plans in Table    

echo "Available Price Plans";          
echo "<table border=1>";
while ( $pricing = $result->fetch_assoc() ) {
    echo "<tr>";
    echo "<td>" .$pricing['planID'].  "</td>";
    echo "<td>" .$pricing['planTitle']. "</td>";
    echo "<td>" .$pricing['planDescription']. "</td>";
    echo "<td>" .$pricing['planPricing']. "</td>";
    echo "<td>" .'<a href="'. $link .'"><img style="border:none;" src="'. $icon .'" /></a>'. "</td>";
    //print a button as you mentioned
    echo "<td>"."<form action='#' method='get'><input type='hidden' name='id' value='".$pricing['planID']."'/><input type='submit' value='copy'/></form></td>";
    echo "</tr>";
}
echo "</table>";

and your function :

function planA()
{ 
    // get selected plan info
    $query = "SELECT * FROM pricing";
    $result = $mysqli->query($query);
    $row = $res->fetch_assoc();

    //copy info to table 'restaurant_plan'
    $query = "INSERT into restaurant_plan (".$_GET["id"].", ".$row["planTitle"].", ".$row["planDescription"].", ".$row["planPricing"].")";
    $result = $mysqli->query($query);

}

Matt · Accepted Answer · 2012-08-20 14:27:21Z

Not an answer, but if you want your code to be cleaner & more readable/maintainable, you should use HTML with PHP inserted, rather than echoing HTML via PHP:

<?php
//Select All Price Plans
$query = "SELECT * FROM pricing";
$result = mysql_query($query);

//Print all Plans in Table     
?>
Available Price Plans
<?php while ($pricing = mysql_fetch_array($result)) : ?>
    <table border=1>
        <tr>
            <td><?= $pricing['planID'] ?></td>
            <td><?= $pricing['planTitle'] ?></td>
            <td><?= $pricing['planDescription'] ?></td>
            <td><?= $pricing['planPricing'] ?></td>
            <td><a href="<?= $link ?>"><img style="border:none;" src="<?= $icon ?>" /></a></td>
        </tr>
    <table>
<?php endforeach; ?>

Also, as I mentioned in the comments above, you should stop using mysql_* functions. They're being deprecated. Instead use PDO (supported as of PHP 5.1) or mysqli (supported as of PHP 4.1). If you're not sure which one to use, read this article.

Conrad Lotz · Accepted Answer · 2012-08-20 14:41:28Z

0

Change your table output to the following:

<table border="1">
<?php
while ($pricing = mysql_fetch_array($result)) { 
?>
       <tr>
       <td><?php echo $pricing['planID']; ?></td>
       <td><?php echo $pricing['planTitle']; ?></td>
       <td><?php echo $pricing['planDescription']; ?></td>
       <td><?php echo $pricing['planPricing']; ?></td>
       <td><a href='<?php echo $link; ?>'><img style="border:none;" src='<?php echo $icon; ?>' /></a></td> 
       </tr>
   <?php     
   }  
   ?>
<table>

Double check you database connection as well to ensure data can be retrieved.

answered Aug 20, 2012 at 14:41

Conrad Lotz

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