[SOLVED] Why does this return only the first row in the database?

[surochek]

The database is OK. The query works. But for the rest of my script to function as it should, this part must return all the rows corresponding to the query. I only get the first one (should be 2 from my test database). What am I doing wrong?  ???

 

<?php

$query = "SELECT * FROM user WHERE DATE_ADD(regdate, INTERVAL 5 day) < CURDATE()";

 

$result=mysql_query($query)or die (mysql_error());

 

$num=mysql_num_rows($result);

 

$user_array = mysql_fetch_array($result);

 

for ($i=0; $i<$num; $i++) {

echo $user_array["username"]."<br>"; 

}

 

?>

 

Thank you from a very frustrated learner.

[Orio]

You are using mysql_fetch_array() in the wrong way.

 

<?php
$query = "SELECT * FROM user WHERE DATE_ADD(regdate, INTERVAL 5 day) < CURDATE()";

$result=mysql_query($query)or die (mysql_error());

while ($user_array = mysql_fetch_array($result)) {
  echo $user_array["username"]."
"; 
}

?>

 

Orio.

[Barand]

 

 

mysql_fetch_array fetch a row at a time, putting the fields of that row into an array

 

 

[SoulAssassin]

You might also want to try adding MYSQL_ASSOC in there like this:

<?

while ($user_array = mysql_fetch_array($result, MYSQL_ASSOC))

{

?>

 

 

[surochek]

It works! Thank you!

 

(Feeling dumb  ) Why? Does mysql_fetch_array stop itself, so to speak, therefore a while loop doesn't turn into an infinite loop?

 

Surochek.

[Orio]

From the manual:

 

Returns an array that corresponds to the fetched row and moves the internal data pointer ahead.

 

That means that it grabs a row, moves the internal pointer ahead. If the internal pointer reached the end ( => The last row was grabbed in the last run), it returns false.

 

See more info and examples in the manual.

 

Orio.