mysql_fetch_array

fnbcprog

Has anyone ever experienced a problem with select information from a table and it only returns one record. Then the script wont print it out using mysql_fetch_array?

Does this make any sense or is there some trick?

Thanks

peppeto

Go to
http://www.php.net/manual/en/function.mysql-fetch-array.php
You wil finde your answers

Mordecai

Yeah, yeah, all the time. You need to specify either MYSQL_NUM or MYSQL_ASSOC to mysql_fetch_array():

while ($array = mysql_fetch_array($result,MYSQL_NUM)) {
  print_r($array);
}

fnbcprog

That didnt work lemme show you my script

fnbcprog

// Gets values

<?
	// Check for authorization
	require_once("../../include/authentication.php");
	// Connection File
	require_once('../../Connections/nbmm.php');

$result = mysql_query("SELECT id_nbcmdf,name,pol_num,ssn FROM nbpmdf WHERE $_GET[searchPar] LIKE '$_GET[searchstring]%' ORDER BY $_GET[searchPar]");
$row = mysql_fetch_array($result);

// Number of records returned
$numRecs = mysql_num_rows($result);
?>
// Prints values in html table
					<?
						while ( $row = mysql_fetch_array( $result, MYSQL_BOTH ) ) {

						// Alternate row colors
						if ( $x % 2 )
							echo "<tr bgcolor=\"#CCCCCC\">";
						else
							echo "<tr bgcolor=\"#FFFFFF\">";

						echo "<td width=\"125\" class=\"printID\">$row[ssn]</td>";
						echo "<td width=\"125\" class=\"fontsize12\">$row[pol_num]</td>";
						echo "<td width=\"475\" class=\"fontsize12\"><a href=\"../index.php?clientID=$row[id_nbcmdf]\">$row[name]</a></td>";
						echo "<tr>";
					}
				?>

fnbcprog

found my error

thanks for your help though

ucbones

I know you found your error, and it may well have been this, if not...

your LIKE statement is missing the first % sign:

<?
$result = mysql_query("SELECT id_nbcmdf,name,pol_num,ssn FROM nbpmdf WHERE $_GET[searchPar] LIKE '$_GET[searchstring]%' ORDER BY $_GET[searchPar]");
?>

But:

<?
$result = mysql_query("SELECT id_nbcmdf,name,pol_num,ssn FROM nbpmdf WHERE $_GET[searchPar] LIKE '%$_GET[searchstring]%' ORDER BY $_GET[searchPar]");
?>

Sorry if that was the same error that you had found

kevinsequeira

there are two calls to mysql_fetch_array($result);
so if u have one row, it is overwritten the second time ...

reg
kevin