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up vote 0 down vote favorite

The following code shows the name of the customer in a line with all the items he/she has purchased in lines below the name line.

As you will see, in the items purchased line I also print the item's price which I get from the getPrice function.

And here's the problem, the first query, the customers query, retrieves several rows of data but as soon as I call the getPrice function I get just the first row.

If instead of calling the function I give price just a value, everything goes right.

Taking this into account, I have changed the getPrice function query, thinking it could be wrong but the problem persists .. the while loop prints just one row.

Any ideas on what is going wrong? What can I try? I’m stuck :(

Thanks !!

<?php

$query1  = "SELECT * FROM Customers"; //this retrieves several rows of data
$result1 = mysql_query($query1) or die(mysql_error());

$flag = -999;
while($row = mysql_fetch_array($result1))
{

    if ($row["id"] != $flag) 
    {

        $content .="<span>".$row["lastName"].", ".$row["Name"]." </span><br />";
        $flag = $row["id"];
    }

    $content .=("Item purchased: " .$row["item"]."Price: ".getPrice($row["ItemID"])."<br />");


    $content .="</br></br>";
}                


function getPrice ($itemID)
{

    $query2  = "SELECT Price FROM Items WHERE ItemID= '".$itemID."'";
    $result2 = mysql_query($query2) or die(mysql_error());

    if($row2 = mysql_fetch_array($result2))
    {
        return $row2[0];
    } 
}

echo $content;
?>
share|improve this question
    
Please post the table structure and some sample records, preferably in form of CREATE TABLE ... and INSERT INTO ... sql statements. –  VolkerK Jun 26 '13 at 10:55

2 Answers 2

up vote 0 down vote accepted

use while instead of if in getPrice function

function getPrice ($itemID)
{
//Changed this line
$query2="SELECT Price FROM Items WHERE ItemID= '".$itemID."'";
$result2 = mysql_query($query2) or die(mysql_error());

while($row2 = mysql_fetch_array($result2))
{
    return $row2[0];
} 
}
share|improve this answer
    
thanks but does not work .. –  user523129 Jun 26 '13 at 11:17
    
i think you should check your database.... –  Ammar Hayder Khan Jun 26 '13 at 11:31
up vote 0 down vote 
function getPrice ($itemID)
{
    //Changed this line
    $query2="SELECT Price FROM Items WHERE ItemID= '".$itemID."'";//;was missing
    $result2 = mysql_query($query2) or die(mysql_error());

    if($row2 = mysql_fetch_array($result2))
    {
        return $row2[0];
    } 
}
share|improve this answer
    
My error when copying the code, in my script is correct .. thanks –  user523129 Jun 26 '13 at 10:41

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