unserialize() value insert with serialize() of database.? - php

Question

i insert in database values (array) with use function serialize(), how can echo they with unserialize() in tag <ul><li>...?
i have in database this: a:6:{i:0;s:15:"Coffee";i:1;s:14:"Satellite";i:2;s:11:"Game Notes";i:3;s:14:"Internet";i:4;s:10:"Pool";i:5;s:0:"";}
LIKE:
Coffee
Game Notes
Internet
Pool

Answer 1

You need to use unserialize(), as you said, with a foreach() loop, like this:
$arr = unserialize($dbString);
echo "<ul>";
foreach($arr as $key => $val)
{
echo "<li>$val</li>";
}
echo "</ul>";
This will echo a list containing value because foreach() iterates through the unserialize()d array, as you have specified in your question.
The $key => $part is the icing on the cake for foreach(); if you want the get the array key, simply reference $key. If you want the data for that key, use $val.
If you want to echo just one element (your example is Internet), just don't use a loop and reference it by key (integer):
$arr = unserialize($dbString);
echo $arr[2];
This echos the third element in the array on it's own.

Answer 2

Original question
Unserialize the data and loop over it printing it out. Notice that I have also added in checks to see that we are getting back an array and that it contains something before looping over it.
$data = unserialize($row['like']);
if(is_array($data) and count($data)) {
echo '<ul>';
foreach($like as $value) {
echo '<li>' . $value . '</li>';
}
echo '</ul>';
}
Internet
I think this is what you are asking for. To only echo out the value if its Internet.
$data = unserialize($row['like']);
if(is_array($data) and count($data)) {
echo '<ul>';
foreach($like as $value) {
if('Internet' != $value) {
continue;
}
echo '<li>' . $value . '</li>';
}
echo '</ul>';
}
JSON
If you can I would stop using serialize and use json_encode instead. It is easier to encode and decode in more programming languages and it is easier to edit by humans should you need to update the DB directly.