PHP: insert query result into array elements - php

Question

How can I assign query result into array elements?
This is my code:
include('db.php');
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die("Database connection error");
mysql_select_db($dbname);
$query = "select * from test where value=20";
$result = mysql_query($query);
$vegetable_list = array('$rice', '$wheat', '$potato', '$pulses');
$i = 1;
while($row_result = mysql_fetch_row($result))
{
??????? = $row_result[$i];
$i++;
}
How can I assign the query result into the array? Let's say:
$rice = $row_result[1];
$wheat = $row_result[2];
$potato = $row_result[3];
How I can assign the values automatically?

Answer 1

Let try:
$vegetable_list= array('rice','wheat','potato','pulses'); //no $ symbol
while($row_result=mysql_fetch_row($result))
{
foreach($vegetable_list as $k => $v)
${$v} = $row_result[$k + 1]; //I think mysql_fetch_row should indexing from 0 -> n (not from 1)
}

Answer 2

Ok not sure but seems like you have a 150column x 20rows table that you want to convert into a two dimensional array. It is as simple as this:
$data = array( );
while( $row = mysql_fetch_assoc( $result ) )
{
// at this point, $row contains a single row as an associative array
// keys of this array consist of column names
// all you need to do is append $row to $data
$data[ ] = $row;
}
// $data is a two dimensional array
// $data[ 0 ] contains 1st row
// $data[ 1 ] contains 2nd row
// ...
// $data[ 0 ][ 'rice' ] contains rice column value for 1st row
// $data[ 0 ][ 'wheat' ] contains wheat column value for 1st row
// ...
// $data[ 1 ][ 'rice' ] contains rice column value for 2nd row
// $data[ 1 ][ 'wheat' ] contains wheat column value for 2nd row
// ...
// and so on
var_dump( $data );

Answer 3

Edited the code.
Here:
include('db.php');
$conn=mysql_connect($dbhost,$dbuser,$dbpass) or die("Database connection error");
mysql_select_db($dbname);
$query="select * from test where value=20";
$result=mysql_query($query);
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$vegetable_list= array( array () );
$rcols = mysql_query("SHOW COLUMNS FROM test");
if (!$rcols) {
echo 'Could not run query: ' . mysql_error();
exit;
}
if (mysql_num_rows($result) > 0) {
$i = 0;
$j = 0;
if (mysql_num_rows($result) == 0) {
echo "No rows found.";
} else {
while ($row = mysql_fetch_assoc($result)) {
while ($cols = mysql_fetch_assoc($rcols);) {
$vegetable_list[$i][$j] = $row[$cols['Field']];
$i++;
}
$j++;
}
}
} else {
//some error message
}

Answer 4

use switch case :
$i=1;
while($row_result=mysql_fetch_assoc($result))
{
switch($i) {
case 1 : $rice = $row_result[$i]; break;
case 2 : $wheat = $row_result[$i]; break;
}
$i++;
}
or you can do this :
$i=0;
$vegetable = array("rice", "wheat", "potato");
while($row_result=mysql_fetch_assoc($result))
{
$idx = 0;
foreach($row_result as $result){
${$vegetable[$i]}[$idx] = $result;
$idx++;
}
$i++;
}
then try to
var_dump($rice);
you should get array of your specific column.

Answer 5

The first thing is that, you dont need to make array element as variable.
You can simply write without dollar sign like : $rice as rice and so on.
$vegetable_list= array('rice', 'wheat', 'potato', 'pulses');
Now execute Select query to get the data from database
$query = "select * from test where value=20";
$result = mysql_query($query);
make while loop
while($row_result=mysql_fetch_row($result))
{
foreach($vegetable_list as $k => $v)
${$v} = $row_result[$k + 1];
from 0 -> n (not from 1)
}