Php/Mysql search returning only 1 result when there is more

Question

I have a double search that happens within the script. When it searches for a players gear, it will take the player UID and bring back the name of that player. The only issue is that there may be more than 1 result that will return and it will only display the last UID with the name.

$ip = "localhost";
$user = "******";
$pass = "*******";
$db = "hivemind";
$ill1 = $_POST['search'];

//Database Connection
$con = @mysql_connect("$ip:3316", "$user", "$pass")
            or die(mysql_error());

//Select Database
$dbcon = @mysql_select_db($db, $con)
            or die(mysql_error());

$sql = mysql_query("select PlayerUID, Inventory, Backpack from character_data where Inventory like '%$ill1%'");

while ($row = mysql_fetch_array($sql)) {
    $puid = $row['PlayerUID'];
    $inv = $row['Inventory'];
    $back = $row['Backpack'];
    ?>
    <html>
    <body>
    <table>
    <tr>

                    <td><?php echo "$puid"; ?></td>
                    <td><?php echo "$inv"; ?></td>
                    <td><?php echo "$back"; ?></td>

    </tr>
    </table>
    </body>
    </html>
    <?php }?>

    <?php

//Database Connection
$con = @mysql_connect("$ip:3316", "$user", "$pass")
            or die(mysql_error());

//Select Database
$dbcon = @mysql_select_db($db, $con)
            or die(mysql_error());

$sql = mysql_query("select PlayerUID, PlayerName from player_data where PlayerUID like '%$puid%'");

while ($row = mysql_fetch_array($sql)) {

    $puid2 = $row['PlayerUID'];
    $plnm = $row['PlayerName'];
    ?>
    <html>
    <body>
    <table>
    <tr>    
                    <td><?php echo "$puid"; ?></td>
                    <td><?php echo "$plnm"; ?></td>


    </tr>
    </table>
    </body>
    </html>
     }

Osiris · Accepted Answer · 2012-11-09 06:43:23Z

1

It is returning all the results. You're not printing them properly.

Move the html,body,table tags outside the while loop.

Change the loop to:

while ($row = mysql_fetch_array($sql)) {

    $puid2 = $row['PlayerUID'];
    $plnm = $row['PlayerName'];
    echo "<tr><td>".$puid."</td>";
    echo "<td>".$plnm."</td></tr>";
}

edited Nov 9, 2012 at 6:43

answered Nov 9, 2012 at 6:36

Osiris

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6 Comments

Ryahn Over a year ago

Just tried it. It still only pulls up the last PUID from above.

<html> 	<body> 	<table> 	<tr> <? while ($row = mysql_fetch_array($sql)) {  	$puid2 = $row['PlayerUID']; 	$plnm = $row['PlayerName']; 	> 		 					<td><?php echo "$puid"; ?></td> 					<td><?php echo "$plnm"; ?></td> 					 								 	 	<? }> </tr> 	</table> 	</body> 	</html>

Osiris Over a year ago

Sorry, just made an edit. The <tr></tr> needs to be inside the loop too. The loop will be executed for every row returned, so you'll need to create a <tr> element in each instance of the loop.

Ryahn Over a year ago

I just tried. Still only showing 1

<? while ($row = mysql_fetch_array($sql)) {  	$puid2 = $row['PlayerUID']; 	$plnm = $row['PlayerName']; 		 					echo "<tr><td>" .$puid. "</td>"; 					echo "<td>" .$plnm. "</td></tr>"; 					 								 	  }>  	</table> 	</body> 	</html>

Osiris Over a year ago

Paste the complete php page on pastebin.

Osiris Over a year ago

You still have html,body,table inside the body of the first while loop. And you don't need to use html,head,body twice. I'd recommend you put html,head and body outside all php code.

|

Nirav Ranpara · Accepted Answer · 2012-11-09 06:34:13Z

0

print below code in loop

<td><?php echo "$puid"; ?></td>
  <td><?php echo "$plnm"; ?></td>

you print only one time , so overwrite value.

answered Nov 9, 2012 at 6:34

Nirav Ranpara

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