[RESOLVED] Need a little help...

[dassyka]

I'm new with php and mysql and i have a little problem.

Here's my code:

<?php

function db_result_to_array ($result)
{
$res_array = array();

for ($count=0; $row=$result -> fetch_assoc(); $count++)
$res_array[$count] = $row;
return $res_array;
}

function display_categories($cat_array)
{
if(!is_array($cat_array))
{
echo 'no categorie available<br/>';
return;
}
echo '<ul>/n';
foreach ($cat_array as $row)
{
echo '<li>'$cat_array'</li>/n';
echo '</ul>/n';
}
}

function get_categories()
{
$query="SELECT cat_name, id_cat FROM categorie";
$result = @$cxn ->query($query);
if (!$result)
return false;
$num_cats = @$result -> num_rows;
if (num_cats ==0)
return false;
$result = db_result_to_array ($result);
return $result;
}

echo '<p> Here categorie<p/>';

$cat_array = get_categories();
display_categories($cat_array);

?>

And here what i have as a result:

Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in C:\Program Files\wamp\www\index.php on line 42

Can someone help me to fin what is my error and how to correct it?

Thank you very much

[laserlight]

The problem is probably on this line:

echo '<li>'$cat_array'</li>/n';

It should be:

echo '<li>' . $cat_array . "</li>\n";

Note that I used double quotes for the last string since \n is only interpreted as a special character in double quote delimited strings. Oh, and of course I changed /n to \n. You need to do the same elsewhere (e.g., on the next line).

Oh, and please post your PHP code in php bbcode tags.

[dassyka]

Thank you!
But now, i don't have error message but i can not see a list of my categories and i really don't know what to do 😕

Does somebody can continue to help me a little bit?

[igorlopez]

Maybe this works:

echo '<li>' . $row . "</li>\n";

Just one tip, Avoid using the error suppression operator, @, during development.

[dassyka]

thank you to you all but i think it's to much complicate for me! i'll try an other solution!!