Mysql array data not display table - php

Question

I have use MySQL Table get values on table ,that values used another table .i got array values
first table query:
$ordersdetail = "SELECT * FROM `order_item` WHERE `Order_ID`='".$orderid ."'";
$customerorder = mysql_query($ordersdetail );
$i=0;
while($rows = mysql_fetch_array($customerorder ))
{
$orderproduct= $rows['Order_product_ID'];
$orderids= $rows['Order_ID'];
$productId[$i] = $rows['Product_ID'];
$sizeId[$i] = $rows['Size_ID'];
$colorId[$i] = $rows['Color_ID'];
$quantiy = $rows['Order_Quantity'][$i];
$price = $rows['Unit_Price'][$i];
$subTotal = $rows['Sub_Total'];
$customerccode = $rows['Record_Status'];
$customerphone = $rows['Created_Time'];
$i++;
}
table column productid array values get used another table:
given:
for($i=0;$i<count($productId);$i++)
{
$product = "SELECT * FROM `product` WHERE `Product_ID`='".$productId[$i]."'";
$products = mysql_query($product);
while($rows = mysql_fetch_array($products))
{
$productnames = $rows['Product_Name'];
}
}
color id used :
for($i=0;$i<count($colorId);$i++)
{
$color = "SELECT * FROM `color` WHERE `Color_ID`='".$colorId[$i] ."'";
$cname = mysql_query($color);
while($rows = mysql_fetch_array($cname))
{
$colorname = $rows['Color_Name'];
}
}
same way used the values table
my questions that values displayed table continuely
table structure
$message .= "<table width='100%' border='1' cellspacing='0' cellpadding='1'>
<thead>
<tr style='background: #eee;'>
<th><strong>Quantity</strong> </th>
<th><strong>ProductName</strong> </th>
<th><strong>Size</strong> </th>
<th><strong>Style</strong> </th>
<th><strong>Price</strong> </th>
<th><strong>Total</strong> </th>
</tr>
";
for($i=0;$i<count($productId);$i++)
{
$message .=" <tr style='color: #c40000;'>
<th>" .$quantiy. "</th>
<th>" .$productnames. "</th>
<th>" .$sizename. "</th>
<th>" .$colorname. "</th>
<th>" . $price. "</th>
<th>" . $subTotal. "</th>
</tr>";
}
this format is correct more values not displayed only first values display multiple time

Answer 1

This is a hard one to answer because there there is a lot of issues going on....
First off....unless this is a private/internal application, you cant place variables directly into your query...look into a proper database class, I can recommend : http://www.imavex.com/php-pdo-wrapper-class/
Secondly, your are mixing your php code and html too much...I'm not going to go into this too much but in general you want to keep this stuff seperate...research MVC programming patterns.
Now to answer your question...The reason you are seeing the same data repeat is because you you are using only one variable that is getting overwritten each time...ie. $quantiy (which is spelled wrong fyi)...so to hack your approach, assign it to an array and then use an $i counter on it like you are doing with $productID
A better approach would be to do this stuff in the loop and avoid assign the vars in the first place...I'm assuming you have assign stuff to the message var...if you can skip that and echo it out directly that would probably be better.
$ordersdetail = "SELECT * FROM `order_item` WHERE `Order_ID`='".$orderid ."'";
$customerorder = mysql_query($ordersdetail );
$i=0;
while($rows = mysql_fetch_array($customerorder ))
$product = "SELECT * FROM `product` WHERE `Product_ID`='".$row[$i]."'";
$products = mysql_query($product);
$color = "SELECT * FROM `color` WHERE `Color_ID`='".$colorId[$i] ."'";
$cname = mysql_query($color);
Now you would create your table from the $rows data directly and when you goto the color/size stuff you could look through in there.
}
Now I'm not 100% sure it will work for you but you could probably do something similar to above with a single query using JOINS
SELECT * FROM `order_item` LEFT JOIN product ON order_item.productID = product.productID LEFT JOIN color ON order_item.colorID = color.colorID WHERE order_item.`Order_ID`='".$orderid ."'
Please pardon the syntax/naming conventions...the above command will not run properly on your system, it is only meant to give you an idea on what to do.