[SOLVED] How to retreive and echo just one field from db?

[sebastiaandraaisma]

Hello I have question that might sound stupid, but how do I echo just one field from the database?

I tried the following:
// make the connection to the database
@require "connect.php";

// Retreive data from database for translation
$result = mysql_query('SELECT * FROM admins WHERE id = 29');// Select id 29 from table admins
$res = mysqli_query($dbh, $result);// I don't understand this
$newArray = mysqli_fetch_array($res, MYSQLI_ASSOC);// Something with an aray?
$name = $newArray['name'];//Extract the name from the aray and give it the variable $name?
echo "$name";// Show the name

What it does... actually what it doesn't do is echo (print) the name on the screen.

The table looks like this:

CREATE TABLE `admins` (
  `id` int(11) NOT NULL auto_increment,
  `name` varchar(50) default NULL,
  `user` varchar(50) default NULL,
  `passwd` varchar(50) default NULL,
  `type` int(11) default NULL,
  `language` int(11) default '0',
  PRIMARY KEY  (`id`)
) TYPE=MyISAM AUTO_INCREMENT=34 ;


INSERT INTO `admins` VALUES (29, 'John Doe', 'hello', 'test', 1, 0);

I just don't know how to get the name as a variable and than echo"$name"
I'm very new to php :)

Any help is welcome!

Kind regards,
Sebas.

[JP128]

[code]
<?php
//DB Connect above this...
$result = mysql_query("SELECT * FROM admins WHERE id = '29'");
while(mysql_fetch_assoc($result)){
echo $result['name'];
}
[/code]

[sebastiaandraaisma]

Wow, susch a quick reply! :)

Thank you so much, I will try it as soon as I come home :)
Thank you.

[JP128]

NP, I hope it helps.

[sebastiaandraaisma]

Hello again :)

I got an error saying:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\www.globalhome.eu\indextest.php on line 13

On line 13 is the following text:
while(mysql_fetch_assoc($result)){

I tried a couple of things but without result.

I don't know if it has something to do with the fact that I'm running on XP and off-line?

Kind regards,
Sebastiaan

[QuietWhistler]

Yes, you can only fetch data from a database when you are connected to the internet, or if you have mysql installed on localhost. Furthermore, are you sure that you have put:

[code]$result = $result = mysql_query("SELECT * FROM admins WHERE id = '29'");[/code]

instead of just:

[code]mysql_query("SELECT * FROM admins WHERE id = '29'");[/code]

[sebastiaandraaisma]

Hello, thank you for your reply!

I have on line 12:
$result = mysql_query("SELECT * FROM admins WHERE id = '29'");

I did try your version, which gave the same error
$result = $result = mysql_query("SELECT * FROM admins WHERE id = '29'");

To come back on your question wether I have installed MySQL, Yes MySQL is installed
I have also Apache 2.x installed and running. I am able to output other php scripts and in the MySQL command line I can see the database, the table and the value I try to extract.

I'm just very new to this and do not realy know what I'm doing wrong.
In one of my books I read that an error that includes a Warning has something to do with the file path or something? As far as I know I configured things corectly. Is there anything specific I should look for that you might be aware of?

Kind regards,
Sebastiaan

[taith]

you dont need
[code]$result = $result = mysql_query("SELECT * FROM admins WHERE id = '29'");[/code]

what that error is telling us is that this query is giving us errors, but not showing them... use this on every query
[code]$result = mysql_query("SELECT * FROM admins WHERE `id`='29'") or die(mysql_error());[/code]
that will give you an error, and then we can work from there

[QuietWhistler]

[quote author=QuietWhistler link=topic=118857.msg486063#msg486063 date=1166271566]
Yes, you can only fetch data from a database when you are connected to the internet, or if you have mysql installed on localhost. Furthermore, are you sure that you have put:

[code]$result = $result = mysql_query("SELECT * FROM admins WHERE id = '29'");[/code]

instead of just:

[code]mysql_query("SELECT * FROM admins WHERE id = '29'");[/code]

[/quote]

Woops my bad, must have copied it wrong.

[sebastiaandraaisma]

Wow, I'm realy dealing with experts here!!!

Thanks a lot guys for taking the time to help me out!

The error message I get now is:
No database selected

I always though that this would be done automaticaly?

My connection script looks like this:
<?php
$dbcnx = @mysql_connect('localhost', 'root', 'abc123');
if (!$dbcnx) {
echo( '<p>Unable to connect to the ' .
'database server at this time.</p>' );
exit();
}
?>

The name of the database is:
alfdis_global

I realy apriciate the help I'm getting here
I wish I had the knowledge to help someone else on the forum (maybe in 6 months from now if I keep studying hard ;) )

[taith]

you've connected the database, but you havent seleceted it...

[code]
mysql_connect("$host","$username","$password");
mysql_select_db("$database");
[/code]

[sebastiaandraaisma]

Now all errors are gone,
But it does not echo out the name.

It does do the rest of the php scripts on the page correct.

I now have:
<?php
$database = "alfdis_global";
$dbcnx = @mysql_connect('localhost', 'root', 'abc123');
mysql_select_db("$database");
if (!$dbcnx) {
echo( '<p>Unable to connect to the ' .
'database server at this time.</p>' );
exit();
}
?>

:)

[sebastiaandraaisma]

Everything seems to work on-line so I guess its some configuration in MySQL, PHP or Apache :)

Anyways, a different question.

What if I don't want to echo it out but convert the result into a new variable like this:

// This section retreives the country id number for the listed object
$result1 = mysql_query("SELECT * FROM objects WHERE `id`='$objid'") or die(mysql_error());
while(mysql_fetch_assoc($result1)){
$country_id = $result1['countryid'];// This is the new variable that holds the country id for the object
}

I got an error that the variable was not defined. What should I write instead?

Thanks in advance!!! :)

Sebas.

[kenrbnsn]

In the while statement, you need to assign the results of the mysql_fetch_assoc() function to an array variable and then index into that array. Since it seems like you're expect to get only one value you don't need the while loop.

[code]<?php
$result1 = mysql_query("SELECT * FROM objects WHERE `id`='$objid'") or die(mysql_error());
$row = mysql_fetch_assoc($result1)
$country_id = $row['countryid'];// This is the new variable that holds the country id for the object
?>[/code]

Ken

[sebastiaandraaisma]

Wow, such a quick reply!!!

I will try it imideatley! :)

Thank you

[sebastiaandraaisma]

It works!!!

Thank you so much. It might be a small thing to you but it means a lot to me.

I wish I had the knowledge to help others but in time this will be I hope :)

Thanks again!