php question

Keewee6 · #1 (**permalink**) 17-08-2007, 09:50 AM

i am learning php at the moment and when i go to show a recordset if i have five records in it - but it is only showing four and if add a sixth it only shows up five - why here is the code

<?php
mysql_connect("localhost", "emerald", "emerald247") or die(mysql_error());
mysql_select_db("emerald_db1") or die(mysql_error());

$query = ("SELECT * FROM names");

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

while($row = mysql_fetch_array($result))
{
echo $row['name']. " - ". $row['surname'];
echo "<br />";
}

any help thanks

louie · #2 (**permalink**) 17-08-2007, 10:48 AM

try this:

PHP Code:

  mysql_connect("localhost", "emerald", "emerald247") or die(mysql_error());
mysql_select_db("emerald_db1") or die(mysql_error());

$query = ("SELECT * FROM names"); 

$result = mysql_query($query) or die(mysql_error());
$num_row = mysql_num_rows($result);
  echo "we found $num_row records in the database<br />";
//start loop
while($row = mysql_fetch_array($result)){
  echo $row['name']. " - ". $row['surname']. "<br />";
}
//end while

Forbairt · #3 (**permalink**) 17-08-2007, 12:51 PM

$row = mysql_fetch_array($result);

while($row = mysql_fetch_array($result))

You've already pulled one row out ... then you start into your while loop and you're pulling the second ...

so get rid of your first $row = mysql_fetch_array($result);

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