[Resolved] mysql_fetch_array problems...

curtisdw

Here is my deal - I'm very new to PHP and MySql stuff - so please bear with me - the following code is returning the error::

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/amcampp/public_html/app_test/ReturnRecords.php on line 60

So here is my code:

$result = mysql_query("SELECT FirstName,LastName,StreetAddress,City,State,Zip,Phone,Fax,Email,Website,PractitionerTyp,Specialty FROM Practitioners");

echo "<table width='95%' border='0' cellspacing='2' cellpadding='2' class='body'>
<tr>
<td>";

while ($row = mysql_fetch_array($result)) <<--This is line 60
{

/* $row = mysql_fetch_assoc($result);*/
	echo $row[FirstName];
	echo $row[LastName]."<br>";
	echo $row[StreetAddress]."<br>";
	echo $row[City].", ";
	echo $row[State].".  ";
	echo $row[Zip]."<br>";
	echo $row[Phone]. " - ";
	echo "Fax:  ". $row[Fax] . "<br>";
	echo "<a href=\"mailto:".$row[Email]."\">" . $row[Email] . "</a><br>";
	echo "<a href=\"http://". $row[Website] . "\">" . $row[Website] . "</a><br>";
	echo "Type of Practitioner: " . $row[PractitionerType] . "<br>";
	echo "Specialty:  " . $row[Specialty] . "<br>";
	echo "<hr>";
	echo "</td>";
	echo "</tr>";

}
echo "</table>";

Any suggestions is appreciated.

bastien

have you linked it to the db?

i use


$username = "root";
        $pwd ="";
        $host = "localhost";
        $dbname = "test";



    //connect to db
    //$conn=mysql_connect($host, $username, $pwd) or die ("Unable to connect to database");

    if (!($conn=mysql_connect($host, $username, $pwd)))  {
        printf("error connecting to DB by user = $username and pwd=$pwd");
        exit;
    }
   $db=mysql_select_db($dbname,$conn) or die("Unable to connect to database1");

curtisdw

Yeah I have my connect statements in and they are connecting as far as I know - username and password are correct. The same statements connect fine in another page that runs a program.

The previous page gets a value from a select list of states then calls the program I listed in my orginal post-

This is the 1st page:

<form name="form1" method="post" action="ReturnRecords.php">
<select name="State" class="main-body">
<option value="" selected>Choose a State</option>
<option value="AL">Alabama </option>
<option value="AK">Alaska </option>
<option value="AZ">Arizona </option>
<option value="AR">Arkansas </option>
<option value="CA">California </option>
<option value="CO">Colorado </option>
...</select> you get the idea ... all 50 states here

Is there a better way to get all corresponding records?

Thanks,
Curtis

bastien

double check the connection strings..it is not hitting the db

Tracer

Errr... DOuble Check your sql as well...

It may be connecting, but if you have errors in your sql sattement the returned resource will be invalid.

Try typing you SQL directly into MySQL at a command prompt and see what happens.

tekky

this this....

$sql = 'SELECT FirstName,LastName,StreetAddress,City,State,Zip,Phone,Fax,Email,Website,PractitionerTyp,Specialty FROM Practitioners';

//this will give you an error if the query fails for some reason, rather than continuing on and erroring at the next call to the $result variable

$result = mysql_query($sql) or die('error in SQL statement: '. $sql . ' -- ' . mysql_error());

echo "<table width='95%' border='0' cellspacing='2' cellpadding='2' class='body'>
		<tr>
			<td>";


   while ($row = mysql_fetch_array($result))  [B][COLOR=blue]<<--This is line 60[/COLOR] [/B]
{

/* $row = mysql_fetch_assoc($result);*/
	echo $row[FirstName];
	echo $row[LastName]."<br>";
	echo $row[StreetAddress]."<br>";
	echo $row[City].", ";
	echo $row[State].".  ";
	echo $row[Zip]."<br>";
	echo $row[Phone]. " - ";
	echo "Fax:  ". $row[Fax] . "<br>";
	echo "<a href=\"mailto:".$row[Email]."\">" . $row[Email] . "</a><br>";
	echo "<a href=\"http://". $row[Website] . "\">" . $row[Website] . "</a><br>";
	echo "Type of Practitioner: " . $row[PractitionerType] . "<br>";
	echo "Specialty:  " . $row[Specialty] . "<br>";
	echo "<hr>";
	echo "</td>";
	echo "</tr>";
}
	echo "</table>";

curtisdw

I changed my query from:

mysql_query("SELECT FirstName,LastName,StreetAddress,City,State,Zip,Phone,Fax,Email,Website,PractitionerTyp,Specialty FROM Practitioners");

to:

mysql_query("SELECT FirstName,LastName,StreetAddress,City,State,Zip,Phone,Fax,Email,Website,PractitionerTyp,Specialty FROM Practitioners WHERE State='$State'");

Still get the same error - Is there a chance that I'm not passing the variable from the previous page correctly?

I use that State select list (name="State") which corresponds to the field I'm searching in the database.

The error leads me to believe that my query statement is wrong - any more suggestions - by the way thanks for all your responses thus far

laserlight

Are you sure there are no typos in say, a column name?

What error does mysql_error() return?

Tracer

if you are using this

$result = mysql_query($sql) or die('error in SQL statement: '. $sql . ' -- ' . mysql_error());

Your sql var will be echoed out, and you will see all of your other variables in your sql sattement. You can even try copying (just cut and paste the echoed sql) the sql into something like PhpMyAdmin which will helkp you correct your sql error. I think windows may also have a built in sql style query builder

curtisdw

The original error was:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/amcampp/public_html/app_test/ReturnRecords.php on line 60

HOWEVER:

I'm a dumb dumb -- no a dumb dumb dumb -
Thank you - I should have checked my spelling. One of my column names was misspelled <--in fact misspelled might be misspelled. Thanks for your help. Abundant blessings on you all