PHP - Mysql problem - PHPBuilder Forums

PHP - Mysql problem

Anon

Hello guys & gals,

Please help me with this small mysql problem.
I had written the following code to display all the records in my table.

<?php
$link = mysql_connect("localhost");
mysql_select_db("test",$link);
$result = mysql_query("select * from mysqltest",$link);
$num = mysql_num_rows($result);
print $num;
for($i = 0;$i<$num;$i++)
{
$rrow = mysql_fetch_object($result);
print " ".$rrow->name." ==> ".$rrow->age;
}
mysql_close();
?>

The code gave a perfect result when I ran it in my PWS(ie.,locally).But when I uploaded this into my remote server, it gave the following error.Iused phpmyadmin to administer my database on the server.There were 3 records in the table.My remote server is Linux.

"Not a valid result source--Line 5"

Please suggest me a remedy..
regards,
vipin.

Anon

Hi,
I'm Kaloyan Petkov from Bulgaria.
Sorry for my poor English.
I'm not familiar with PHP but I'll try to help you.

First: Be sure that the MySQL database is on the same Linux server where your PHP scripts are executed.
I prefer to use the actual IP address or host name of the database server.
Example: $link = mysql_connect("MySQLserver_host_name");

Second: Check if the SQL query is executed on the database server.
Example: <?php $result = mysql_query("select * from mysqltest",$link) or die("Your error message"); ?> -- This will show the error message if any error occur.
You can check if the query is executing by using the MySQL console tool.
You can print the '&result'. I thinks that it must be a positive number.

Good luck.

Anon

Try this an correct for your MySQL DB :

// Invertion of backgroud color

function usecolorbkg()
{
static $colorbkg;

if($colorbkg == "#0080C0")
{
	$colorbkg = "#FEFAF1";
}
else
{
	$colorbkg = "#0080C0";
}

return($colorbkg);

}

//
// Invertion of font color
//
function usecolorfont()
{
static $colorfnt;

if($colorfnt == "#FEFAF1")
{
	$colorfnt = "#004080";
}
else
{
	$colorfnt = "#FEFAF1";
}

return($colorfnt);

}

//
// Create the cell in table for First row
//
function printtitle($printarg)
{
print("<TH align=center bgcolor=\"#004080\">");
print("$printarg");
print("</TH>\n");
}

//
// Create the cell in table for each line
//
function printline($printarg, $alignement, $bkg, $fnt, $wrap)
{
print("<TD align=$alignement $wrap bgcolor=\"$bkg\">");
print("$printarg");
print("</TD>\n");
}

//BEGIN OF PHP Programm

//
// Imprimer le Debut de la page HTML !! (pour faire croire a du HTML Reel)
//

//Data for TEST

$rechref = "%CPU%";
$wheresearch = "REF";
$searchvalue = "";

if($rechercheavancee <> "")
{
$wheresearch = "DESIGNATION";
$rechref = "%$rechercheavancee%";
}
elseif($rechref == "")
{
$rechref = "%";
}

$errcont = 1; //Boolean to check errors

// Ouvrir la database avec un utilisateur "lecture seule"

$mysqllink = mysql_connect("YOURSERVER OR IP", "USERNAME", "PASSWORD");

// Traitement d'erreur d'ouverture...

$errno = mysql_errno($mysqllink);
$error = mysql_error($mysqllink);

if($errno <> 0)
{
$errcont = 0;
print ("Error $errno : $error \n");
print ("Veuillez contacter le WEBMASTER par e-mail... \n");
}
else
{
$mysqlselect = mysql_select_db("ecs", $mysqllink);

$errno = mysql_errno($mysqllink);
$error = mysql_error($mysqllink);

if ($errno <> 0)
{
	$errcont = 0;
	print ("Error $errno : $errot<BR>\n");
	print ("Veuillez contacter le WEBMATER par e-mail...<BR>\n");
}

}

//If NO ERROR EXECUTE ELSE GOTO END !

if($errcont == 1)
{
// CREATE THE QUERY
//
$query = "SELECT * FROM LISTE_PRIX where $wheresearch like \"$rechref\"";

$mysqlresult = mysql_query($query, $mysqllink);

// MEMORIZE NUBER OF AFFECTED ROWS

$nbrarg = mysql_affected_rows($mysqllink);

//CRONTROL DATA TO BE SUPPRESSED IN FINAL VERSION

print("QUERY : $query  <BR>");
print("Result : $mysqlresult <BR>");
print("NBRARG : $nbrarg  <BR>");
print("MYSQLLINK : $mysqllink <BR>");
print("MYSQLSELECT : $mysqlselect <BR>");

//BEGIN OF TABLE


print("<TABLE width=\"500\" cellpadding=\"5\" cellpadding=\"5\" align=\"center\">\n");
print("<TR>\n");
	printtitle("REF");
	printtitle("DESIGNATION");
	printtitle("Luf TTC");
	printtitle("Euro TTC");
	printtitle("FrF TTC");
	printtitle("Info");
print("</TR>\n");

//READ EACH RECORD AND CREATE EACH ROW IN TABLE

for($compteur = 1; $compteur <= $nbrarg; $compteur++)
{
	$row = mysql_fetch_array($mysqlresult);

	$bkgcolor = usecolorbkg();
	$fntcolor = usecolorfont();

	print("<TR>\n");
		printline($row["REF"], "left", $bkgcolor, $fntcolor, "nowrap");
		printline($row["DESIGNATION"], "left", $bkgcolor, $fntcolor, "");
		printline($row["Luf_ttc"], "right", $bkgcolor, $fntcolor, "");
		printline($row["Euro_ttc"], "right", $bkgcolor, $fntcolor, "");
		printline($row["FrF_ttc"], "right", $bkgcolor, $fntcolor, "");
		if($row["Comment"] <> "")
		{
			printline($row["Comment"], "center", $bkgcolor, "#FF0000", "");
		}
		else
		{
			printline("-", "center", $bkgcolor, $fntcolor, "");
		}
	print("</TR>\n");
}

// END OF TABLE

print("</TABLE>\n");

}
mysql_close($mysqllink);

If You want, send me an e-mail to get the code !

To see it working, you can go on my test server :

http://test.sit-net.lu

(sorry this site is in french)

Andre,
Luxembourg
Europe

Anon

In My last Explain, my Apache server is on One server, running PHP too.

The MySQL server is on another one !

Anon

In my code you can see How I crated the FOR statement...

The condition is assmued by useing $nbrarg

$nbrarg = mysql_affected_rows($mysqllink);

I Thing that the error in your code is located there.

Anon

Hi,

You gave a good answer...Ichecked out my mysql_connect() and mysql_select_db() functions by adding the die() function..The mysql_select_db() statement returned the error message which I had given..what to do next..How do I check out whether my php program and mysql are in the same servers..please respond positively..

regards,
vipin.

n.b...I am from India.

Anon

REMARK : Sorry for my poor english, Better in french, but...

Where are you uploading ? OS, Server ?

Remark, In your $link, have you add the User and Password !

<?php
$link = mysql_connect("localhost", "USER_NAME", "PASSWORD");
mysql_select_db("test",$link);

you have to declare in your My_SQL db this user...

You also have to look if you had applied restriction machine (host) who are quering the DB.

In my case, I have suppressed annonymous users.

In the User table, I've created one new user how can only log into My SQL from my Apache server (HTTP) in the host IP adress. This user has NO GENERAL authority.

In the TABLE DB, I've created the same user who can only access the client DB, whith only the authority to SELECT.

(Granting protection).

I also see that you use the following argument ...

$num = mysql_num_rows($result);
print $num;
for($i = 0;$i<$num;$i++)
{
$rrow = mysql_fetch_object($result);
print " ".$rrow->name." ==> ".$rrow->age;
}

Why don't you change for some thing like this (better to understand !

$nbrarg = mysql_affected_rows($mysqllink);

for($i = 1; $i <= $nbrarg; $i++)
{
$row = mysql_fetch_array($mysqlresult);
print($row["name"]);
print($row["age"]);
}

I Hope that could Help you !

André
Luxembourg !

Anon

Hi!

Sorry for my late reply but I did not have access to Internet.

You say that there is a error in mysql_select_db() function.
The function mysql_connect() is successful - this is good.

In this context I can suppose that the database name is not correct.
You must be sure that the database "test" or the database you wish to connect to, exist on the MySQL server.

About last question: When you make a program which access some database you must know where your program will be execute and where the database is hosted.

If my above answer did not help you:
There are other reasons which can prevent you to connect to MySQL database. I think that it would be better to ask the man who administer the MySQL host.Only this man have the ability to make the all things which prevent you to connect to database now.

Good luck!