Echoing MySQL query in PHP

Question

I have been trying for the past while to echo out the result of a a MySQL query in PHP. I can't seem to get it to work so I am a bit lost. I know for a fact that the query works as I have done it in PHPMYADMIN and it is working fine, however whenever I load the webpage, nothing is outputted. For the purposes of this question I have generalized the things in the query as I obviously don't want anyone accessing my database. This is what I have tried:

<?php
    $connect = mysql_connect('host', 'user', 'password', 'dbname');
    if (!$connect) {
        die('Could not connect: ' . mysql_error());
    }
    if (!mysql_select_db('dbname')) {
        die('Could not select database: ' . mysql_error());
    }
    $result = mysql_query('SELECT SUM(row name) FROM `table`');
    if (!$result) {
        die('Could not query:' . mysql_error());
    }
    echo mysql_result($result);

    mysql_close($connect);
?>

I look forward to your replies.

MYSQL_* is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. — Krish R
– Krish R, Commented Dec 25, 2013 at 10:04

xdazz · Accepted Answer · 2013-12-25 10:04:43Z

3

mysql_result need the second parameter (The row number from the result that's being retrieved).

echo mysql_result($result, 0);

answered Dec 25, 2013 at 10:04

xdazz

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Comments

Titi master · Accepted Answer · 2013-12-25 10:09:17Z

2

you have to change

echo mysql_result($result); by

echo mysql_result($result,0);

or you can use

print_r(mysql_fetch_assoc($result));

answered Dec 25, 2013 at 10:09

Titi master

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Atanu Saha · Accepted Answer · 2013-12-25 10:30:24Z

1

Try this

echo mysql_result($result,0);

For more information, please give a look on http://www.php.net/mysql_result

or you can use mysql_fetch_row instead of mysql_result.

$row = mysql_fetch_row($result);
echo $row[0];

answered Dec 25, 2013 at 10:30

Atanu Saha

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Baby Groot · Accepted Answer · 2013-12-25 12:29:52Z

1

You may use this following format: Give your Sql command like this:

 $result = mysql_query('SELECT SUM(row name) as new_column_name FROM `table`');

then use this loop

while($fetched_values = mysql_fetch_array($result)){
    echo $fetched_values['new_column'];
}

edited Dec 25, 2013 at 12:29

Baby Groot

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answered Dec 25, 2013 at 11:57

Ajitha Ms

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M I · Accepted Answer · 2013-12-25 10:15:33Z

0

In your mysql_connect you have provided 4 parameters. it accepts 3 params

host, username and password. dbname is given inside mysql_select_db

as you can see from official docs. So remove the 4th param from mysql_connect function.

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');

http://php.net/function.mysql-connect

$db_selected = mysql_select_db('foo', $link);

http://www.php.net/manual/en/function.mysql-select-db.php

but these mysql_* function will be deprecated use mysqli instead.

answered Dec 25, 2013 at 10:15

M I

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Adarsh Nahar · Accepted Answer · 2013-12-25 10:53:00Z

0

pass a row number as second parameter in mysql_result

echo mysql_result($result, 1);

or

use print_r(mysql_fetch_array($result)); where $result is your result resource

answered Dec 25, 2013 at 10:53

Adarsh Nahar

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Comments

Ajitha Ms · Accepted Answer · 2013-12-25 12:07:06Z

0

if you have any doubt you may refer my following example program.

<?php
$connect = mysql_connect('localhost', 'user_name', 'password', 'db_name');
if (!$connect) 
{
    die('Could not connect: ' . mysql_error());
}
if (!mysql_select_db('db_name')) 
{
    die('Could not select database: ' . mysql_error());
}
$row = mysql_query('SELECT (Tamil+English+Maths) as total FROM table_name');
if (!$row) 
{
    die('Could not query:' . mysql_error());
}
while($fetched_row = mysql_fetch_array($row)){
echo $fetched_row['total'];
}

mysql_close($connect);

?>

here Tamil,English and Maths are the column values of my table

answered Dec 25, 2013 at 12:07

Ajitha Ms

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