Array into mysql row - php

Question

Hello i have a dynamic form that can change field names upon user selection. i would like for the form result to be inserted into one mysql row.. i create an array as follows:
<?php
include_once 'dbconnect.php';
if (isset($_POST['item_name'])) {
$table = $_POST['ItemTypeSelect'];
foreach ($_POST as $key => $variable) {
$chesckColumn = mysql_query("SELECT `$key` from $table");
if (!$chesckColumn) {
echo "column ".$key." does not exist <br/>";
}else{
$results = $variable;
$columnName = $key;
$array[] = array($columnName => $results);
}
}
mysql_query("INSERT INTO $table (`$columnName`) VALUES ('$results') ")or die(mysql_error());
}
?>
my array result are
(
[0] => Array
(
[Server_id] =>
)
[1] => Array
(
[Server_IP_Address] => 12345
)
[2] => Array
(
[Server_IP] => 12345
)
[3] => Array
(
[Server_Name] => dad
)
)
How can i have the array be inserted in one table row and reference the $results and the field value and $columnName as the field value.

Answer 1

You could use a few php built in functions:
function put_quotes(&$item){
$item = "'$item'";
}
$columns = implode(",",array_walk(array_keys($_POST),'put_quotes'));
$values = implode(",",array_walk(array_values($_POST),'put_quotes'));
mysql_query("INSERT INTO $table (columns) VALUES ($values) ")or die(mysql_error());
Where :
array_walk executes a function over all array, here put quotes over the item.
array_keys gets the keys from an array
array_values gets the values from an array
implode joins an array to a string with a separator

Answer 2

Use PDO because mysql_* functions are depreceated.
Connect to MySQL in PDO :
$db="database";
$host="127.0.0.1";
$usr="username";
$pass="password";
$dbh=new PDO("mysql:dbname=$db;host=$host", $usr, $pass);
First convert the array to JSON and save it on the table :
$json=json_encode($array);
$sql=$dbh->prepare("INSERT INTO ? (?) VALUES (?)");
$sql->execute(array($table,$columnName,$json));
If you want to retrieve the array back, decode the JSON data got from table:
$sql=$dbh->prepare("SELECT ? FROM ?");
$sql->execute(array($columnName,$table));
while($r=$sql->fetch()){
$array=json_decode($r[$columnName],true);
print_r($array);
}