[SOLVED] PHP script error

[Exc.BluePhoenix]

Hello

 

I am trying a simple search script by following a tutorial. However I get the following error:

 

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

C:\Program_Files\wamp\www\searchresults.php on line 18.

 

After that I read around how to try and fix it, and to my dismay nothing really worked, or more so nothing really fitted my example. Afterwards I added the line "or die(mysql_error())" to give me the problem. The result of this says "Query was empty", of course by knowing English I understand that the Query is empty but I don't know how to make it not empty or how to not give me the error. Can anyone point out to me the error in the script please, thank you.

 

<?php

        $db = mysql_connect("localhost","","password");
mysql_select_db("exampledb", $db);
$guery = "SELECT shops.name, shops.phone, shops.email, shops.website FROM shops WHERE name LIKE '%".$name."%'";
$result = mysql_query($query) or die( mysql_error() );
        while ($record = mysql_fetch_assoc($result)){
	while (list($fieldname, $fieldvalue) = each($record)){
		echo $fieldname.":<B>".$fieldvalue."</B><BR>";
		}
	echo "<BR>";
}
?>

[mrdamien]

$guery = "SELECT shops.name, shops.phone, shops.email, shops.website FROM shops WHERE name LIKE '%".$name."%'";
$result = mysql_query($query) or die( mysql_error() );

You have a typo:

$guery != $query  G/Q

 

$query = "SELECT shops.name, shops.phone, shops.email, shops.website FROM shops WHERE name LIKE '%".$name."%'";

Use that instead

[Exc.BluePhoenix]

Well thank you very much for the reply and the answer. It works now, I cannot believe that I spend 3 hours over a typo.

 

Thank you because I was never going to catch that.

 

 

However I now find myself with one more question. This is supposed to be a search script. I have a very simple form that goes with this. Now I wanted to see if its working properly before I move on. So I type in the search box the word "records" what this script is supposed to do is give back information on all the shops that match that name. However, when I do it, it sends back all the shops. Can anyone point out what might be the reason for this?

 

Here is the code again.

 

<?php

$db = mysql_connect("localhost","","inuyasha");
mysql_select_db("exampledb", $db);
$query = "SELECT shops.name, shops.phone, shops.email, shops.website 
		  FROM shops WHERE name LIKE '%".$name."%'";
$result = mysql_query($query) or die( mysql_error() );
while ($record = mysql_fetch_assoc($result)){
	while(list($fieldname, $fieldvalue) = each($record)){
		echo $fieldname.":<B>".$fieldvalue."</B><BR>";
	}
	echo "<BR>";
}
?> 

[revraz]

echo $query before you run the mysql_query command and make sure it looks like it should.  Maybe $name is empty so it returns everything (%%)

[Exc.BluePhoenix]

When you say echo $query before the mysql_query command and make sure it looks like it should, what do you mean? I dont understand what should it look like? Still i did it and it just adds the string. Furthermore when you say maybe $name is empty, do you mean in the db? if so its not; I just check with phpmyadmin.

 

I'm sorry for the trouble.

[revraz]

Copy and paste what it echos.

 

<?php

$db = mysql_connect("localhost","","inuyasha");
mysql_select_db("exampledb", $db);
$query = "SELECT shops.name, shops.phone, shops.email, shops.website 
		  FROM shops WHERE name LIKE '%".$name."%'";
echo $query;  //add this
exit;  //add this
$result = mysql_query($query) or die( mysql_error() );
while ($record = mysql_fetch_assoc($result)){
	while(list($fieldname, $fieldvalue) = each($record)){
		echo $fieldname.":<B>".$fieldvalue."</B><BR>";
	}
	echo "<BR>";
}
?> 

[Exc.BluePhoenix]

I think that I get what you meant by the name been empty now. Here is the return.

 

SELECT shops.name, shops.phone, shops.email, shops.website FROM shops WHERE name LIKE '%%'

 

'%%' there is nothing there is that what you meant before? Also anyway of fixing this that you may offer, or pointing the problem?

 

Thank you again, and also for your previous replies.

[revraz]

So that means $name is empty.  With the code you posted, it's not shown where you set it.

[Exc.BluePhoenix]

That is all my PHP code however, maybe the form html will help?

 

<head>
<title>Record Shops Search</title>
</head>
<body>
<H2> Search for a shop</H2>
<BR>
<FORM ACTION="searchresults.php" METHOD="POST">
Please enter the name, or part of the name, of the shop you are seeking:
<BR>
<INPUT NAME="name" TYPE=TEXT />
<BR>
<INPUT TYPE=SUBMIT VALUE="search">
</FORM>
</body>
</html>

 

If that does not help, then its fine and thanks for your help, really nice of you.

[Ken2k7]

<?php

$db = mysql_connect("localhost","","inuyasha");
mysql_select_db("exampledb", $db);
$query = "SELECT shops.name, shops.phone, shops.email, shops.website 
		  FROM shops WHERE name LIKE '%".$_POST['name']."%'";
$result = mysql_query($query) or die( mysql_error() );
while ($record = mysql_fetch_assoc($result)){
	while(list($fieldname, $fieldvalue) = each($record)){
		echo $fieldname.":<B>".$fieldvalue."</B><BR>";
	}
	echo "<BR>";
}
?> 

[revraz]

Or preferred

 

<?php

$db = mysql_connect("localhost","","inuyasha");
mysql_select_db("exampledb", $db);
             $name = mysql_real_escape_string($_POST['name']);
$query = "SELECT shops.name, shops.phone, shops.email, shops.website 
		  FROM shops WHERE name LIKE '%".$name."%'";
echo $query;  //dont forget to remove this
exit;  //dont forget to remove this
$result = mysql_query($query) or die( mysql_error() );
while ($record = mysql_fetch_assoc($result)){
	while(list($fieldname, $fieldvalue) = each($record)){
		echo $fieldname.":<B>".$fieldvalue."</B><BR>";
	}
	echo "<BR>";
}
?> 

[Exc.BluePhoenix]

LOL revraz thanks for that last one, first thing I was not going to do.

 

Thank you Ken2k7 that worked LIKE A CHARM, I appreciated, same goes to revraz and mrdamien.

 

Thanks a lot!!!

[revraz]

I also added this line in there so you dont let bad info come in

 

$name = mysql_real_escape_string($_POST['name']);