Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Your example shows the tablename as lowercase, so you seem to prove my point while claiming to disagree. (?) You are supposed to put the command reserved words in capitals and the table name in lowercase, just as you did.

I am simply suggesting that Chris might try renaming his table from 'Pet' to 'pet'. If you read about MySQL rules here: http://dev.mysql.com/doc/refman/5.1/...nsitivity.html you'll see why they recommend using all lowercase for compatibility, portability, and to avoid having to mess with the mySQL .ini file.

You are right they recommend. But it works ))) Actually i don't use them. But have used before...

I just noticed that Chris used capitals in his database name too. I really think that is asking for trouble. I would rename the database from PetCatalog to petcatalog and make sure all table names are lowercase. MySQL automatically converts database and table names to lowercase during backup and storage - not sure what happens when it restores. It is OK to use mixed case in field names but not database or table names.

I once used a hyphen in a table name (eg. users-new) and nothing worked until I changed it to an underscore (users_new). Won't do that again.

kuau, yes you are right. That is another problem of Chris. But problem what he asked, i answered in page 1 ^^.

P.S. I'm not sure too. But logically in restore, it must go back... :D

Quote:

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Originally Posted by allahverdi

umm. I think you all missed one thing.

change this:

Code:

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$row = mysql_fetch_array($result)

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to this:

Code:

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$row = mysql_fetch_array($result, MYSQL_ASSOC)

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How will changing the array type help? :confused: The error is not in the type of array, it is in the resource being turned into an array-- the query is not working.

BTW: there is nothing wrong with capital letters in Database names or table names. Think of them like variables-- you wouldn't call a variable $VARIABLE, but you could. I have used capital letters many times (not as common practice, mind, but sometimes it's just easier for readability).
But you should never use special chars, like the hyphen, because that does cause problems.

Hi all,

And thanks ever so much for everybody's responses! I'll address each of them!

Nile: I created a new page based on your cleaned-up code, but got the following error when I ran it:

Parse error: syntax error, unexpected $end in C:\wamp\www\chris\petdisplay2.php on line 45

Jas: I'm fairly sure my query is right - when I run the query by querying the MySQL db directly it returns the relevant records. I tried inserting your code:

PHP Code:

---

echo $query;
$results = MySQL_Query($query) or die(MySQL_Error()); 


---

But wasn't 100% sure where I should insert it, I placed it beneath the line:

PHP Code:

---

$result = mysql_query($query) 


---

I still received the same error as before with no more helpful info. Re privileges + permissions, I've checked this and have used the root@localhost user which has full permissions but it's still giving me the error.

Allahverdi: I tried adding MYSQL_ASSOC: as follows:

PHP Code:

---

$row = mysql_fetch_array($result, MYSQL_ASSOC) 


---

But again-no joy!

Kuau: Re the uppercase lowercase suggestion, the book I'm working from displays the databases/tables with an initial uppercase letter, and same for tables - I've copied this in my db and as you'll see further down below I've since discovered that capital letters are not an issue.

SO!!! Thanks for your responses. What I tried tonight after your suggestions was moving onto the next tutorial that my book offers based on near enough the same code... a slightly more advanced page which uses a for loop rather than a while loop.

Strangely the slightly different page loaded perfectly!! Which puts to rest any debate about uppercase/lowercase, incorrect login, privileges etc.

So, whilst this is good, my inquisitive brain isn't satisfied! Why does this code continue to fail:

PHP Code:

---

<?php
/*    Program:    petdisplay.php
*    Desc:        Displays all pets in selected category.
* /
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
    $user="root";
    $host="localhost";
    $password="";
    $database = "PetCatalog";
    $connection = mysql_connect($host,$user,$password)
        or die ("Couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("Couldn't select database");
    $pettype = "horse";  //horse was typed in a form by user
    $query = "SELECT * FROM Pet WHERE petType='$pettype'";
    $result = mysql_query($query)
        or die ("Couldn't execute query.");

    /* Display results in a table */
    $pettype = ucfirst($pettype)."s";
    echo "<h1>$pettype</h1>";
    echo "<table cellspacing='15'>";
    echo "<tr><td colspan='3'><hr></td></tr>";
    while ($row = mysql_fetch_array($result))
    {
        extract($row);
        $f_price = number_format($price,2);
         echo "<tr>\n
                <td>$petName</td>\n
                <td>$petDescription</td>\n
                <td align='right'>\$$f_price</td>\n
                </tr>\n";
            echo "<tr><td colspan='3'><hr></td></tr>\n";
    }
    echo "</table>\n";
?>
</body></html>

---

Whilst this code runs perfectly?!:

PHP Code:

---

<?php
/*    Program:    petDescripFor.php
*    Desc:        Displays a numbered list of all pets in
*                selected category.
*/
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
    $user="root";
    $host="localhost";
    $password="";
    $database = "PetCatalog";
    $connection = mysql_connect($host,$user,$password)
        or die ("couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("couldn't select database");
    $pettype = "horse";  //horse was typed in a form by user
    $query = "SELECT * FROM Pet WHERE petType='$pettype'";
    $result = mysql_query($query)
        or die ("Couldn't execute query.");
    $nrows = mysql_num_rows($result);

    /* Display results in a table */
    echo "<h1>Horses</h1>";
    echo "<table cellspacing='15'>";
    echo "<tr><td colspan='4'><hr></td></tr>";
    for ($i=0;$i<$nrows;$i++)
    {
            $n = $i + 1;    #add 1 so that numbers don't start with 0
            $row = mysql_fetch_array($result);
            extract($row);
            $f_price = number_format($price,2);
                echo "<tr>\n
                    <td>$n.</td>\n
                    <td>$petName</td>\n
                    <td>$petDescription</td>\n
                    <td align='right'>\$$f_price</td>\n
                    </tr>\n";
                echo "<tr><td colspan='4'><hr></td></tr>\n";
    }
    echo "</table>\n";
?>
</body></html>

---

Are there any particularly clever people who want to offer their suggestions?! :confused:

Quote:

---

Jas: I'm fairly sure my query is right - when I run the query by querying the MySQL db directly it returns the relevant records. I tried inserting your code:

PHP Code:

---

echo $query;
$results = MySQL_Query($query) or die(MySQL_Error()); 


---

But wasn't 100% sure where I should insert it, I placed it beneath the line:

PHP Code:

---

$result = mysql_query($query) 


---

I still received the same error as before with no more helpful info. Re privileges + permissions, I've checked this and have used the root@localhost user which has full permissions but it's still giving me the error.

---

Instead of putting it under the line, replace it with the line.

You mean this is your full code?

Code:

---

<?php
/*    Program:    petdisplay.php
*    Desc:        Displays all pets in selected category.
* /
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
    $user="root";
    $host="localhost";
    $password="";
    $database = "PetCatalog";
    $connection = mysql_connect($host,$user,$password)
        or die ("Couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("Couldn't select database");
    $pettype = "horse";  //horse was typed in a form by user
    $query = "SELECT * FROM Pet WHERE petType='$pettype'";
    $result = mysql_query($query)
        or die ("Couldn't execute query.");

    /* Display results in a table */
    $pettype = ucfirst($pettype)."s";
    echo "<h1>$pettype</h1>";
    echo "<table cellspacing='15'>";
    echo "<tr><td colspan='3'><hr></td></tr>";
    while ($row = mysql_fetch_array($result))
    {
        extract($row);
        $f_price = number_format($price,2);
        echo "<tr>\n
                <td>$petName</td>\n
                <td>$petDescription</td>\n
                <td align='right'>\$$f_price</td>\n
                </tr>\n";
            echo "<tr><td colspan='3'><hr></td></tr>\n";
    }
    echo "</table>\n";
?>
</body></html>

---

Full copy of this???

I copied this to dw. And guess what i see? Full comment till

Code:

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/* Display results in a table */

---

If your code full like this. (I wish it won't be). Change to it:

Code:

---

<?php
/*    Program:    petdisplay.php
*    Desc:        Displays all pets in selected category.
*/
?>
<html>
<head><title>Pet Catalog</title></head>
<body>
<?php
    $user="root";
    $host="localhost";
    $password="";
    $database = "PetCatalog";
    $connection = mysql_connect($host,$user,$password)
        or die ("Couldn't connect to server");
    $db = mysql_select_db($database,$connection)
        or die ("Couldn't select database");
    $pettype = "horse";  //horse was typed in a form by user
    $query = "SELECT * FROM Pet WHERE petType='$pettype'";
    $result = mysql_query($query)
        or die ("Couldn't execute query.");

    /* Display results in a table */
    $pettype = ucfirst($pettype)."s";
    echo "<h1>$pettype</h1>";
    echo "<table cellspacing='15'>";
    echo "<tr><td colspan='3'><hr></td></tr>";
    while ($row = mysql_fetch_array($result))
    {
        extract($row);
        $f_price = number_format($price,2);
        echo "<tr>\n
                <td>$petName</td>\n
                <td>$petDescription</td>\n
                <td align='right'>\$$f_price</td>\n
                </tr>\n";
            echo "<tr><td colspan='3'><hr></td></tr>\n";
    }
    echo "</table>\n";
?>

---

Your problem was, you didn't close your comment correctly :(. But i think just you copied here incorrectly...

ALLAHVERDI!!!!!!!!!!!! I don't believe it!!! That's it! :) :) :)

I'm so happy! Pathetic I know!!! Woooooaooooo!!!!

So just for nice conclusion purposes, the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\chris\petdisplay.php on line 28

Was being caused by nothing other than.... an incorrectly closed code comment:

<?php
/* Program: petdisplay.php
* Desc: Displays all pets in selected category.
* /
?>

Which when changed to:

<?php
/* Program: petdisplay.php
* Desc: Displays all pets in selected category.
*/
?>

Worked perfectly! Thanks to everybody for their kind and helpful input! Very very strange that it showed the error as being a problem so far down the page (line 28!)

Bravo Allahverdi!!!! Talk about not seeing the forest for the trees haha. If you look back, the whole thing is bright orange. Duh!