can't get the right result from mySQL query

swidjaja

hi all

I've been trying for over 5 hours to get this thing working. I quite new to this database thing so please bare with me.

		
	$username = "user";
	$password = "pass";
	$host 	= "localhost";
	$dbname = "dbname";

$selectImg = "SELECT img FROM webBody WHERE sectionID = '$pgID'";
$selectPgName = "SELECT pageName FROM webBody WHERE sectionID = '$pgID'";

$dbconnect	= mysql_connect ($host, $username, $password) or die ('I cannot connect to the database.');	
//mysql_select_db ($dbname,$dbconnect);

$im = mysql_fetch_array (mysql_query($selectImg));
$pg	= mysql_fetch_array (mysql_query($selectPgName));

$img = $im["img"];
$pgName=$pg["pageName"]
echo "page name is $pgName<br>";
echo "image path is $img<br>";

The Outout I;ve been getting is

page name is B[/B]

image path is B[/B]

The output I've been trying to get is

page name is page 1

image path is /images/img.gif

could someone point me out to the right direction?

thanks

tomhath

First of all, you don't need two queries, just select both fields in one:

$selectImg = "SELECT img, pageName FROM webBody WHERE sectionID = '$pgID'";

Second, assign the result of the mysql_query to a $result variable and check for an error with an "or die..."

Third, check if the query returned anything with something like:

if ($row = mysql_fetch_array($result) ) ...

My guess is that what you think is in $pgID is incorrect, that the query simply isn't returning any rows.

swidjaja

Originally posted by tomhath
First of all, you don't need two queries, just select both fields in one:

$selectImg = "SELECT img, pageName FROM webBody WHERE sectionID = '$pgID'";

Second, assign the result of the mysql_query to a $result variable and check for an error with an "or die..."

Third, check if the query returned anything with something like:

if ($row = mysql_fetch_array($result) ) ...

My guess is that what you think is in $pgID is incorrect, that the query simply isn't returning any rows.

tom...

I'm practically quite new to this db thing, so I was hoping someone could correct the code so I can learn from it. I've done all the steps you suggested except it's abit unclear on the last step. So if you or someone else don't mind please help me to fix the code.

	$username = "selectas_sam";
	$password = "curt";
	$host 	= "localhost";
	$dbname = "selectas_website";

$selection = "SELECT img,pageName FROM webBody WHERE sectionID = '$pgID'";
$dbconnect	= mysql_connect ($host, $username, $password) or die ('I cannot connect to the database.');	

$result = mysql_query($selection) or die ('i am just stupid');

if ( $row = mysql_fetch_array($result)){
	$img = $row["img"];
	$pgName=$row["pageName"];
}

echo "page name is $pgName<br>";
echo "image path is $img<br>";

regards

fuji

Here's a few changes i made to your script.... I wrote in a little error checking for that result set... to make sure that you're getting a result.... because if you aren't, that mysql_fetch_array() function will either not fire (if in the if-statement) or will cause an error.... Hope this helps 🙂

$username = 'user';
$password = 'pass';
$host     = 'localhost';
$dbname   = 'dbname';

// db connection and selection
$dbconnect    = mysql_connect ($host, $username, $password) or die ('I cannot connect to the database.');    

mysql_select_db ($dbname,$dbconnect);


// I assume that $pgID is from the URL? such as /file.php?pgID=1 or
// something similar?  If so... i'd do this:

$pid=intval($HTTP_GET_VARS['pgID']) // or use $_GET['pgID'] depending on phpversion

$sql = "SELECT img,pageName FROM webBody WHERE sectionID = '$pid'";
$result = mysql_query($sql);


// Begin checking to see if we got results, if we didnt, that
// mysql_fetch_array() will stop our app producing a nice little error...

if (mysql_num_rows($result) < 1)
{
	echo ' My query did not return any results...';
	echo ' Here is my query '.$sql;
	exit;
}

// Our result is okay.... so let's continue
$pagedata = mysql_fetch_array($result);    

echo 'Page Name is '.$pagedata['pageName'].'<br />';
echo 'Image Path is '.$pagedata['img'].'<br />';

swidjaja

hi fuji,

I've tried your code but the server is giving me this error message

[B]Parse error:[/B] parse error, unexpected T_VARIABLE in [B] /home/selectas/public_html/page.php[/B] on line [B] 24[/B]

on line 24 is actually the var that contains the sql statement

$sql = "SELECT img,pageName FROM webBody WHERE sectionID = '$pid'";

I've tried changing

$HTTP_GET_VARS['pgID']

with

$_GET['pgID']

and still get the same result.

any sugestions?

MrAlaska

When you get any kind of 'unexpected' error, look at the line preceding the one that generated the error. A common syntax error is failing to terminate a line. In this case there seems to be a semicolon missing. Change:

$pid=intval($HTTP_GET_VARS['pgID'])
to:
$pid=intval($HTTP_GET_VARS['pgID']);

and see if that works better for you.

swidjaja

😃 it works!!!

I can't believe it's only because a small ';' (semicolon)