Display mySQL info

[gazmania]

I am attempting to write a script that searches a database for a given string (in this case subsno) and then all info kept in that record it then places in a form, I get and error saying "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource" but I don't understand to why.

Could someone please offer some help, thanks.

if ($search) // perform search only if a string was entered. 
{ 
mysql_connect("localhost","****","****") or die ("Problem connecting to Database"); 

$query = "select * from contacts WHERE subsno LIKE '$subsno'";
$result = mysql_db_query("moneyfacts", $query); 

}
while ($r = mysql_fetch_array($result)) { // Begin while 
$contactname = $r["contactname"]; 
$contactphone = $r["contactphone"]; 
$contactemail = $r["contactemail"]; 
$subsno = $r["subsno"]; 
// $resolved = $r["resolved"]; 
echo "<tr> 
<td>$contactname</td> 
<td>$contactphone</td> 
<td>$contactemail</td> 
<td>$subsno</td> 
<tr> 
</tr>"; 
} // end while  echo "</table>"; 
?>

[the_Igel]

There's something not clear:

You're looking for subsno EXACTLY matching $subsno? then why not use '=' instead of LIKE.
And if you're looking for subsno CONTAINING $subsno, then you're missing wildcards :

WHERE subsno LIKE '%$subsno%'

[gazmania]

Duely noted, cheers, however still doesn't solve wierd problem 🙁

[Flameeyes]

move the } of the if AFTER the while, because your script try the while also if $search is false, in this case it return error because $result is not defined...

also check if there is errors in the query

the new code sounds like this:

if ($search) // perform search only if a string was entered. 

{ 

mysql_connect("localhost","****","****") or die ("Problem connecting to Database"); 



$query = "select * from contacts WHERE subsno LIKE '$subsno'";

$result = mysql_db_query("moneyfacts", $query); 

// check if there are errors in the query
	if ( mysql_errno() )
		die("Error executing query $query\n".mysql_errno().": ".mysql_error());


while ($r = mysql_fetch_array($result)) { // Begin while 

$contactname = $r["contactname"]; 

$contactphone = $r["contactphone"]; 

$contactemail = $r["contactemail"]; 

$subsno = $r["subsno"]; 

// $resolved = $r["resolved"]; 

echo "<tr> 
<td>$contactname</td> 
<td>$contactphone</td> 
<td>$contactemail</td> 
<td>$subsno</td> 
<tr> 
</tr>"; 

} // end while

echo "</table>"; 
}

[Flameeyes]

also this

$contactname = $r["contactname"]; 
$contactphone = $r["contactphone"]; 
$contactemail = $r["contactemail"]; 
$subsno = $r["subsno"]; 
// $resolved = $r["resolved"]; 

echo "<tr> 
<td>$contactname</td> 
<td>$contactphone</td> 
<td>$contactemail</td> 
<td>$subsno</td> 
<tr> 
</tr>";

can be simplified like this:

echo "<tr> 
<td>{$r["contactname"]}</td> 
<td>{$r["contactphone"]}</td> 
<td>{$r["contactemail"]}</td> 
<td>{$r["subsno"]}}</td> 
<tr> 
</tr>";

if you put the variable name into { and }, you can use associative arrays into strings and echo.

[gazmania]

Thanks a lot ppl, I am slowley learning thanks to your help 🙂

I shall try it at work tomorrow and let you know if its ok.

Cheers 😃