What on earth is wrong with this?

[Tezread]

Im trying to run a simply Select * from query but the result is a blank page- here is my code- when I connect to the db using the same code but inserting data I get no problems- so Im having no server problems.

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = @mysql_select_db($database, $connection)
or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

or die (mysql_error());

//execute query and store the result
$res = mysql_query($sql,$connection);

//check to make sure the query actually ran

if(!$res) {
echo mysql_error().":".mysql_error()."";
return 0;

//$result = $db->query($sql);
echo $sql;
}
}

[leatherback]

try:

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = @mysql_select_db($database, $connection)
or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql) or die(mysql_error());

while($row=mysql_fetch_arrtay($result))
  {
  echo $row[0];
  }

[Weedpacket]

In other words, if you don't output anything when you have a result, don't be surprised when you don't get any output when you have a result 🙂

[leatherback]

Weedpacket wrote:

In other words, if you don't output anything when you have a result, don't be surprised when you don't get any output when you have a result 🙂

😃
Q.E.D.

[Tezread]

thx for the quick response- i tried the code you suggested but it still results in a blank page

[leatherback]

Hi,

Does the first colum hold data? (It now only gives you resulkts from the first colum in the table. You need to echo one that does hold some information. To check whether it is retrieving, do:

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = mysql_select_db($database, $connection) or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql) or die(mysql_error());

while($row=mysql_fetch_array($result))
  {
  echo "result found!";
  echo $row[0];
  } 

[Tezread]

still a blank page im afraid. The table has 21 records in it. Ive checked using Select * from trainingtopics; in MySQL. The first column is an auto_increment column- the others are standard columns

[leatherback]

Hi,

can you post the updated code please.

[Tezread]

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = mysql_select_db($database, $connection) or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql) or die(mysql_error());

while($row=mysql_fetch_array($result))
{
echo "result found!";
echo $row[0];
}
?>

[leatherback]

seems to me that

$connection = @mysql_connect($host, $root, $password)

should be:

$connection = @mysql_connect($host, $user, $password)

I am surprised you do not get the error message. Maybe you should remove the @ from that statement?

Please verify that you use the same code in the insert?

[Tezread]

To verify this is my insert code that works a treat:

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = @mysql_select_db($database, $connection)
or die ("Couldn't select database.");

if (isset($_POST['Submit'])) {
//insert the values entered into the courseentry form into the database

$coursename = $POST['coursename'];
$accreditingbody = $POST['accreditingbody'];
$briefdesc = $POST['briefdesc'];
$aimedat = $POST['aimedat'];
$level1 = $POST['level1'];
$level2 = $POST['level2'];
$level3 = $POST['level3'];
$elearning = $POST['elearning'];
$facetoface = $POST['facetoface'];
$blended = $POST['blended'];
$elearningtime = $POST['elearningtime'];
$facetofacetime = $POST['facetofacetime'];
$blendedtime = $_POST['blendedtime'];

$sql = "INSERT INTO trainingtopics (coursename, accreditingbody, briefdesc, aimedat, level1, level2, level3, elearning, facetoface, blendedlearning, elearntime, facetime, blendedtime)

VALUES ('$coursename'
,'$accredtingbody'
,'$briefdesc'
,'$aimedat'
,'$level1'
,'$level2'
,'$level3'
,'$elearning'
,'$facetoface'
,'$blended'
,'$elearningtime'
,'$facetofacetime'
,'$blendedtime'
)"
or die (mysql_error());

//execute query and store the result
$res = mysql_query($sql,$connection);

//check to make sure the query actually ran

if(!$res) {
echo mysql_error().":".mysql_error()."";
return 0;

//$result = $db->query($sql);
echo $sql;
}
}

echo $coursename;
echo $briefdesc;

?>

[Tezread]

This is most frustrating- I have a project due in on Friday and this keeps happening. Can't think what is wrong as my INSERT (see last post) query works fine

[leatherback]

In your insert you specifically use the connection. I never do this (Since I only have one db.)

You could try and see of that helps?

$res = mysql_query($sql,$connection);

[n2m]

leatherback wrote:

seems to me that

$connection = @mysql_connect($host, $root, $password)

should be:

$connection = @mysql_connect($host, $user, $password)

I am surprised you do not get the error message. Maybe you should remove the @ from that statement?

Please verify that you use the same code in the insert?

i see that $root is not assigned a value just like what leatherback noted
what is the value of $root? nothing?

[Tezread]

Tried this and it still doesn't work- this is doing my nut in :-)

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = @mysql_select_db($database,$connection);
or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql,$connection) or die(mysql_error());

while($row=mysql_fetch_array($result))

echo "result found!";
echo $row[0];
{
}
?>

[Tezread]

Im running PHP 4.3.9 with Apache Server if this info helps-

[leatherback]

Why did you change the code I suggested:

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = @mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = @mysql_select_db($database,$connection);
or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql,$connection) or die(mysql_error());

while($row=mysql_fetch_array($result))
{  // NOTE this one ABOVE the echo
echo "result found!";
echo $row[0];

}
?>

[Tezread]

Don't know- not sure If I have changed it- I have tried all suggestions- if I have changed one of your suggestions I apologise- what is the latest suggestion?

[leatherback]

see my comment in previous post.

And I would remove all the @ symbols, so you get errors when something goes wrong.

[Tezread]

I removed the @ symbols and references to $connection but I get a blank page still. Do you think is may be worth downloading something to install Apache/MySQL/PHP all over again and fix any possible config errors? or am I being a little hasty?. If not do I have to uninstall my current versions of Apache/PHP/MySQL first? thx for all your help by the way!

<?php
$user = "root";
$host = "localhost";
$password = "";
$connection = mysql_connect($host, $root, $password)
or die ("Couldn't connect to server.");
$database = "courses";
$db = mysql_select_db($database,);
or die ("Couldn't select database.");

$sql = "SELECT * FROM trainingtopics";

$result = mysql_query($sql) or die(mysql_error());

while($row=mysql_fetch_array($result))
{ // NOTE this one ABOVE the echo
echo "result found!";
echo $row[0];

}