Jump to content

MySQL Help


Guest PcGamerz13

Recommended Posts

Guest PcGamerz13
It only shows 1 Gamer_Tag from the database instead of all the Gamer Tags. Can someone fix it plz.
I want all the Gamer Tags into one Option plz help

[code]$sql3="SELECT * FROM clan_members";
$mysql_result3=mysql_query($sql3);
$num_rows3=mysql_num_rows($mysql_result3);
while ($row3=mysql_fetch_array($mysql_result3))
{
$Gamer_Tag=$row3["Gamer_Tag"];
}
echo "<center><font size='4'>Report Member</font></center>";
?>
<form method="POST" action="Report_Member.php?Report_Member=Yes">
<center>Gamer Tag: <select size="1" name="Reported_Gamer_Tag"><option><? echo $Gamer_Tag; ?></option></select></center>
<center>Reason: <textarea rows="2" name="Reason" cols="20"></textarea></center>
<center><input type="submit" value="Submit" name="B1">
</form>[/code]
Link to comment
Share on other sites

[!--quoteo(post=356812:date=Mar 21 2006, 01:04 AM:name=PcGamerz13)--][div class=\'quotetop\']QUOTE(PcGamerz13 @ Mar 21 2006, 01:04 AM) [snapback]356812[/snapback][/div][div class=\'quotemain\'][!--quotec--]
It only shows 1 Gamer_Tag from the database instead of all the Gamer Tags. Can someone fix it plz.
I want all the Gamer Tags into one Option plz help

[code]
...
while ($row3=mysql_fetch_array($mysql_result3))
{
$Gamer_Tag=$row3["Gamer_Tag"];
}
...
[/code]
[/quote]

here's a problem. you re-assign the value of $Gamer_Tag every time. if you want $Gamer_Tag to be an array, so you can just store all the values from the database, just use $Gamer_Tag[] instead:


[code]
...
$Gamer_Tag = array();

while ($row3=mysql_fetch_array($mysql_result3))
{
$Gamer_Tag[]=$row3["Gamer_Tag"];
}
...
[/code]

and then for your dropdown/menu, use:

[code]
foreach ($Gamer_Tag as $tag)
{
   echo '<option>'.$tag.'</option>';
}
[/code]
Link to comment
Share on other sites

Guest PcGamerz13
i got this code now and all it shows in the drop down is Array. How do you make this code show the whole list of the Gamer Tags in the table?

[code]$sql3="SELECT * FROM clan_members";
$mysql_result3=mysql_query($sql3);
$num_rows3=mysql_num_rows($mysql_result3);
$Gamer_Tag = array();
while ($row3=mysql_fetch_array($mysql_result3))
{
$Gamer_Tag[]=$row3["Gamer_Tag"];
}
?>
<form method="POST" action="Report_Member.php?Report_Member=Yes">
<center>Gamer Tag: <select size="1" name="Reported_Gamer_Tag"><option><? echo $Gamer_Tag; ?></option></select></center>
<center>Reason: <textarea rows="2" name="Reason" cols="20"></textarea></center>
<center><input type="submit" value="Submit" name="B1">
</form>[/code]
Link to comment
Share on other sites

whoops sorry i may have added to my post after you first read it. use the 'foreach' command (check my post above) which will just cycle through each item in the array and make a dropdown item from it

Cheers
Mark

[b]EDIT:[/b] or more specifically:

[code]
<select size="1" name="Reported_Gamer_Tag">
   <?php
   foreach ($Gamer_Tag as $tag)
   {
      echo '<option>'.$tag.'</option>';
   }
   ?>
</select>
[/code]

hope that helps
Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.