tutorial problem

[ianpitter]

I am having trouble printing out results using your tutorial php format. I can't see any errors of copying but they must be there somewhere! The database 'mail' has only one table: 'info' with four entries, one of which has 'good' in the 'opinion' field.

My php file is:

<html>
<head><title> How to Grab your MySQL Data </title></head>
<body bgcolor="#FFFFF F">

<php?
/* declare some relevant variables */
$DBhost = "localhost" ;
$DBuser = "root";
$DBpass = "";
$DBName = "mail";
$table = "info";

mysqli_connect( $DBhost,$DBuser ,$DBpass) or die("Unable to connect to database");
@mysqli_select_ db("$DBName") or die("Unable to select database $DBName");
$sqlquery = "SELECT * FROM $table WHERE opinion = 'good'";
$result = mysqli_query($s qlquery);
$number = mysqli_num_rows ($result);

$i = 0;

if ($number < 1) {
print "<CENTER><P>The re Were No Results for Your Search</CENTER>";
}
else {
while ($number > $i) {
$thename = mysqli_result($ result,$i,"name ");
$theemail = mysqli_result($ result,$i,"emai l");

<p><b>Name:</b><?php $thename?><br /><b>E-Mail:</b><?php>$theema il?></p>";

$i++;
}
}
?>
</BODY></HTML>

The result:

$i) { $thename = mysqli_result($ result,$i,"name "); $theemail = mysqli_result($ result,$i,"emai l");
Name:
E-Mail:$theemail? >

"; $i++; } } ?>

There are no log errors in Apache22.

[bevort]

Instead of <?php $thename?> try to use
<?php echo($thename); ?>
Also set a close tag ?> befor the HTML part and a <?PHP after it

Now is seems to be a part of the PHP code thus generating an error and showing a blanc page

[ianpitter]

Instead of <?php $thename?> try to use
<?php echo($thename); ?>
Also set a close tag ?> befor the HTML part and a <?PHP after it

Now is seems to be a part of the PHP code thus generating an error and showing a blanc page

Many thanks for your prompt reply; you were right about the echo additions. The tags aroung the HTML do not seem to make a difference.
Also I was using mysqli_ functions which did not produce an error but there was no output. Changing them to mysql_ functions produced a result.
I am appending the working script in case it is helpful.

[ianpitter]

Many thanks for your prompt reply; you were right about the echo additions. The tags aroung the HTML do not seem to make a difference.
Also I was using mysqli_ functions which did not produce an error but there was no output. Changing them to mysql_ functions produced a result.
I am appending the working script in case it is helpful.

I didn't! Here it is:

<html>
<head><title> How to Grab your MySQL Data </title></head>
<body bgcolor="#FFFFF F">

<?php
/* declare some relevant variables */
$DBhost = "localhost" ;
$DBuser = "root";
$DBpass = "";
$DBName = "mail";
$table = "info";

mysql_connect($ DBhost,$DBuser, $DBpass) or die("Unable to connect to database");
@mysql_select_d b("$DBName") or die("Unable to select database $DBName");
$sqlquery = "SELECT * FROM $table WHERE opinion = 'good'";
$result = mysql_query($sq lquery);
$number = mysql_num_rows( $result);

$i = 0;

if ($number < 1) {
print "<CENTER><P>The re Were No Results for Your Search</CENTER>";
}
else {
while ($number > $i) {
$thename = mysql_result($r esult,$i,"name" );
$theemail = mysql_result($r esult,$i,"email ");

?>
<p><b>Name:</b><?php echo $thename;?><br /><b>E-Mail:</b><?php echo $theemail;?></p>";

<?php
$i++;
}
}
?>
</body></html>

[Georges1979]

hi everybody !
if anybody can help me to get a program using php-mysql that will tell me ok after login in so i will be able to display data from any document saved on my folders

[bevort]

George
you can go to www.scribax.com. There you can find a free complete password script to protect pages on your site. It is a PHP-MySQL based solution with a MD5 javascript encription

[bevort]

Iapiiter

I rewrote the while loop as it normaly works with me with a simple echo to show the number of rows found.

here is the code

<html>
<head><title> How to Grab your MySQL Data </title></head>
<body bgcolor="#FFFFF F">

<?php
/* declare some relevant variables */
$DBhost = "localhost" ;
$DBuser = "root";
$DBpass = "";
$DBName = "mail";
$table = "info";

mysql_connect($ DBhost,$DBuser, $DBpass) or die("Unable to connect to database");
@mysql_select_d b("$DBName") or die("Unable to select database $DBName");
$sqlquery = "SELECT * FROM $table WHERE opinion = 'good'";
$result = mysql_query($sq lquery);

echo(mysql_num_ rows($result) . "rows found");

if (mysql_num_rows ($result) < 1) {
print "<CENTER><P>The re Were No Results for Your Search</CENTER>";
}
else {
while ($result = mysql_fetch_arr ay($results)) {
$thename = mysql_result($r esult,$i,"name" );
$theemail = mysql_result($r esult,$i,"email ");
echo("<p><b>Nam e:</b>" . $thename . "<br /><b>E-Mail:</b>" . $theemail . "</p>";
}
}
?>
</body></html>

Vincent

[bevort]

If this dies not work please describe what you get, step by step.

[ianpitter]

If this dies not work please describe what you get, step by step.

I saved your program as display_query.p hp. When I ran it I got a blank screen and an error message in Apache22\log\er ror:parse error, unexpected ';' in C:\\www\\webroo t\\display_quer y.php on line 27. It looks alright to me!
Also I do not understand the '.' s which are inserted in the echo line; I have not found an explanation of their relevance anywhere.