Array

laanes

I thought I would give it a try to get a grasp of PHP basics and started like this:

$getAddresses = mysql_query("SELECT address1 FROM customers WHERE id BETWEEN 1 AND 60");

if (!$getAddresses) {
    die('Invalid query: ' . mysql_error());
}

while ($row = mysql_fetch_array($getAddresses, MYSQL_ASSOC)) {

	echo $row;

Question:

Why does this give me an output:

ArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArrayArray

And when i try this:

<?php 

$getAllCustomers = mysql_query("SELECT address1 FROM customers WHERE id BETWEEN 1 AND 60");

if (!$getAllCustomers) {
    die('Invalid query: ' . mysql_error());
}

$array = array($getAllCustomers);

foreach ($array as $results) {

echo "$results\n";

}

?>

i get an output:

Resource id #2

bradgrafelman

The first code snippet outputs the word "Array" because you are trying to echo an array itself. An array is a collection of items, so telling PHP to directly output it doesn't make much sense. You might want to see the manual on what an [man]array[/man] is (and/or find some basic tutorials). The array that is being returned is an associative array with the column name as the index, thus $row['address1'] will contain the data from column address1 on each iteration of the loop.

The second code snippet doesn't make much sense either. You're taking the MySQL result resource $getAllCustomers (e.g. what [man]mysql_query/man returns when you execute a valid SELECT query) and then storing it in an array:

$array = array($getAllCustomers);

Thus $array is now an array with a single element - the MySQL result resource.

laanes

thus $row['address1'] will contain the data from column address1 on each iteration of the loop.

Thanks mate!

I did change it a bit now and I am facing the next question, why only one of the list items is is showing? I can see the list bullet of the second item but nothing coming after it.

Code:

//get first address in the list

$getfirstAddress = mysql_query("SELECT address1 FROM customers WHERE id LIKE 1");

if (!$getfirstAddress) {
    die('Invalid query: ' . mysql_error());
}

//get first product in the list

$getfirstProdGroup = mysql_query("SELECT GPname FROM PGroups WHERE PG LIKE 01");

if (!$getfirstProdGroup) {
    die("Error in query  " . mysql_error());

}

?>

<html>	
<head>

<body>
<ul class="ListOne">
<li><?php while ($row = mysql_fetch_array($getfirstAddress, MYSQL_ASSOC)) { echo $row['address1']; } ?> </li>
<li><?php while ($row = mysql_fetch_array($getfirstProdGroup, MYSQL_ASSOC)) { echo $row['GPname']; } ?> </li>
</ul>
</body>
</head>

</html

dagon

<?php 
while ($row = mysql_fetch_array($getfirstAddress, MYSQL_ASSOC)) 
{ 
echo '<li>.'$row['address1'] .'</li>';
echo '<li>.'$row['GPname'] .'</li>';
}

laanes

dagon;10962918 wrote:

<?php 
while ($row = mysql_fetch_array($getfirstAddress, MYSQL_ASSOC)) 
{ 
echo '<li>.'$row['address1'] .'</li>';
echo '<li>.'$row['GPname'] .'</li>';
}

Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in absolutepath/array.php on line 44

Line 44:

echo '<li>.'$row['address1'] .'</li>';

dagon

typo on my part, you should be able to see the error.

laanes

I tried something like this:

while ($row = mysql_fetch_array($getfirstAddress, MYSQL_ASSOC))
{
echo '<ul class="prodInfo">';
echo '<li>' . $row['address1'] . '</li>';
echo '<li>' . $row['GPname'] . '</li>';
echo '</ul>'; 
}

This still gives me only the first list item.

One thing that confuses me with this method - if i only iterate through $getfirstAddress, how come that I can echo GPname?

dagon

my bad, i should at looked at your code more carefully, the first error is with the query, you should use a join so that you get all the information in the one query rather than having to use two.

What joins the PGroups and Customers table?

laanes

dagon;10962946 wrote:
my bad, i should at looked at your code more carefully, the first error is with the query, you should use a join so that you get all the information in the one query rather than having to use two.

What joins the PGroups and Customers table?

Nothing is joining those two tables at the moment.

PGroups is joined only with products:

INNER JOIN PGroups ON (products.PGroup = PGroups.PG)

dagon

how do you know that id 1 in customers matches PG 01 in PGroups ?

laanes

dagon;10962952 wrote:
how do you know that id 1 in customers matches PG 01 in PGroups ?

I am not trying to match them. My purpose is to practice using php and html together. So I am pulling random data out of the database and try to put it in a list styled with CSS to see how it works.

There would be no practical reason to match customers with product groups anyway, or would there?

dagon

well for random data anything is as good as anything else.

laanes;10962955 wrote:
There would be no practical reason to match customers with product groups anyway, or would there?

how could any one possible know that based on the information supplied?

laanes

dagon;10962964 wrote:
well for random data anything is as good as anything else.

how could any one possible know that based on the information supplied?

True.

But I was only curious why the second list item does not display.

Thanks,
laanes

dagon

the value is not selected in the querry