MySQL Problem

Technical Website Design/Showcase
1

I am trying to learn working with MySQL databases through PHP but I’m having some problems. I’ve written this simple scipt which takes a table named “contacts” and first enters some information into the table. Then, it attempts to read and display the data in the table.

Here’s the full code:

<?php
$username = "username";
$password = "password";
$database = "hhsrobotics_circuitrunners_com_-_ContactList";

$link = mysql_connect(localhost,$username,$password);

@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO contacts VALUES ('','John','Smith','555-555-5555','444-444-4444','333-333-3333','[email protected]','http://www.johnsmith.com')";

mysql_query($query);

$result = mysql_query("SELECT * FROM contacts",$link);

$num = mysql_numrows($result);

$i=0;
while ($i < $num) {

$first = mysql_result($result,$i,"first");
$last = mysql_result($result,$i,"last");
$phone = mysql_result($result,$i,"phone");
$mobile = mysql_result($result,$i,"mobile");
$fax = mysql_result($result,$i,"fax");
$email = mysql_result($result,$i,"email");
$web = mysql_result($result,$i,"web");

echo "<b>$first $last</b><br>Phone: $phone<br>Mobile: $mobile<br>Fax: $fax<br>E-mail: $email<br>Web: $web<br>";

$i++;
}

mysql_close();
?>

The problem is that I keep getting argument errors with the lines

$result = mysql_query("SELECT * FROM contacts",$link);

$num = mysql_numrows($result);

and all of the

$first = mysql_result($result,$i,"first");

lines.

Oh, and there is a primary key in the first part of the table named “id” which is not displayed to the user, and its set to Auto_incriment so the data doesn’t have to be entered for it.

2

I believe the function is “mysql_num_rows”, not “mysql_numrows” (note the underscore between num and rows).

3

Try using:


$result = mysql_query("SELECT * FROM contacts",$link) or die ("Mysql error: ".mysql_error());

Also, all that mysql_result stuff is probably overkill thanks to mysql_fetch_array.


while ($array = mysql_fetch_array($result,MYSQL_ASSOC)) {
print "$array'row'] $array'otherrow'] $array'etc']";
}

That makes it a whole lot cleaner and I’d assume quicker (but I’m not positive). (Also, take a look at www.php.net/mysql_fetch_array to see what that MYSQL_ASSOC is about, if you’re wondering. You don’t really need it, I guess, but I always use it)

4 

while ($array = mysql_fetch_assoc($result)) {
print "$array'row'] $array'otherrow'] $array'etc']";
}

You can actually use this for the same effect as the mysql_fetch_array with MYSQL_ASSOC as the parameter. That’s what I’m using in a script I’m writing for my church’s web site…

5

Allright guys, thanks, I’ll try these.

6

OK, I found the problem, but I didn’t get a good error report until evulish suggested that line with the “or die()”. The real problem was that my table was not named “contacts”, it was contact. lol

Thanks alot guys.