How select rows from database mysql with foreach?

Question

For ex. adress page test.php?prid=4477535

Code page test.php

function query($query) {
    $database = 'test';
    $host = 'test';
    $username = 'test';
    $password = 'test';
    $link = mysql_connect($host,$username,$password);
    if (!$link) {
    die(mysql_error());
    }
    $db_selected = mysql_select_db($database);
    if (!$db_selected) {
    die(mysql_error());
    }
    $result = mysql_query($query);
    mysql_close($link);
    return $result;
    }


        $product_idn=$_GET['prid'];

        $select_image = query("SELECT * FROM products_images WHERE `product_idn`='$product_idn'") or die(mysql_error());
        foreach ($select_image as $row)
        {
        $select_image_array[]=$row->image;
        }

print_r ($select_image_array);

receives a request

SELECT * 
FROM products_images
WHERE  `product_idn` =  '4477535'

If make select from phpmyadmin i have 10 rows.

But if i use test.php?prid=4477535 i see empty page.

print_r ($select_image_array) not show array.

Tell me please why i see rows with phpmyadmin and not see rows with script?

Why are you opening a new connection each query?

Nathan
– Nathan

2012-11-25 01:32:13 +00:00
Commented Nov 25, 2012 at 1:32 — Nathan
– Nathan, Commented Nov 25, 2012 at 1:32

Felix Guo · Accepted Answer · 2012-11-25 01:32:53Z

3

Like the other said, you are prone to SQL injection since you don't serialize your input, but to fix your code, use this:

$select_image = query("SELECT * FROM products_images WHERE `product_idn`='$product_idn'") or die(mysql_error());

while($data = mysql_fetch_assoc($select_image))
{
    echo $data['image'];
}

answered Nov 25, 2012 at 1:32

Felix Guo

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Comments

Lior · Accepted Answer · 2012-11-25 01:27:27Z

0

You are doing it wrong. You have to fetch the resource (mysql_query returns a resource) into an array, and the keys of the array will be the names of the rows returned from your query.

$product_idn=$_GET['prid'];

$select_image = query("SELECT * FROM products_images WHERE `product_idn`='$product_idn'") or die(mysql_error());

    while($fetch=mysql_fetch_assoc($select_image))
    {
         echo $fetch['image'];
    }

    print_r ($select_image_array);

BTW, You have a security hole here - SQL Injection.

answered Nov 25, 2012 at 1:27

Lior

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Comments

Community · Accepted Answer · 2023-11-17 20:16:56Z

0

Test the following

$result = query("SELECT * FROM products_images WHERE `product_idn`='$product_idn'")
$select_image = mysql_fetch_assoc($result);
var_dump($select_image);

for more information look at https://www.php.net/mysql_query

edited Nov 17, 2023 at 20:16

CommunityBot

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answered Nov 25, 2012 at 1:28

martin-lundberg

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Comments

Bahadır İsmail Aydın · Accepted Answer · 2012-11-25 01:31:12Z

0

You just echo $row->image; Never initialize $select_image_array

answered Nov 25, 2012 at 1:31

Bahadır İsmail Aydın

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Comments

Popnoodles · Accepted Answer · 2012-11-25 01:34:07Z

0

print_r ($select_image_array); won't show anything because there is no $select_image_array defined. Did you mean print_r ($select_image);?

Is query() a function you've defined? If not and you don't have errors on you are likely to see nothing.

You also need to sanitize your SQL. Simplest method for now since it's an integer:

$product_idn=(int)$_GET['prid'];

answered Nov 25, 2012 at 1:34

Popnoodles

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