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MySQL Insert Help


Guest PcGamerz13

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Guest PcGamerz13
Why ain't this inserting into the Database no errors are showing up and i have it setup to do that. [code]@mysql_query("INSERT INTO clan_members SET Gamer_Tag = '$Gamer_Tag', Password = '$Password', Rank = '$Rank2', Rank_Name = '$Rank_Name', Disabled = '$Disabled'");[/code]
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[quote author=PcGamerz13 link=topic=107409.msg430891#msg430891 date=1157749495]
Why ain't this inserting into the Database no errors are showing up and i have it setup to do that. [code]@mysql_query("INSERT INTO clan_members SET Gamer_Tag = '$Gamer_Tag', Password = '$Password', Rank = '$Rank2', Rank_Name = '$Rank_Name', Disabled = '$Disabled'");[/code]
[/quote]

'@' means 'be quiet!'

If you use '@' in front of a php function, you're telling php [font=Verdana][size=10pt]NOT[/size][/font] to display errors.

You probably are recieving errors from the function, but you can't tell because you told PHP to be quiet.

Jeff
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Okay, maybe there's something we're assuming.

We could be assuming PHP and MySQL are installed correctly. :)

We could be assuming you have your mysql connection.

We could be assuming that PHP is reporting errors.

We could be assuming that MySQL is reporting errors.

We could be assuming that this code is actually being executed.

Maybe you could check these assumptions.  Beyond that, I can't begin to guess.

Jeff
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Guest PcGamerz13
heres the whole page and yes everyone MYSQL code but that works on that page [code]<?
include("Config.php");
session_start();
$Member_Gamer_Tag = $_SESSION['Member_Gamer_Tag'];
$sql="SELECT * FROM clan_template";
$mysql_result=mysql_query($sql);
$num_rows=mysql_num_rows($mysql_result);
while ($row=mysql_fetch_array($mysql_result))
{
$Header=$row["Header"];
$Footer=$row["Footer"];
}
echo $Header;
$Blank = "";
$Date = date('l dS \of F Y h:i:s A');
$Disabled = "0";
$Log_Reason = "Recruited $Gamer_Tag .";
$Two = "2";
$Rank2 = "1";
$Rank_Name = "Peon";
$sql2="SELECT * FROM clan_members WHERE Gamer_Tag='$Member_Gamer_Tag'";
$mysql_result2=mysql_query($sql2);
$num_rows2=mysql_num_rows($mysql_result2);
while ($row2=mysql_fetch_array($mysql_result2))
{
$Rank=$row2["Rank"];
}
if($Member_Gamer_Tag == $Blank){
?>
<form method="POST" action="Console.php?Login=Yes">
  <center>Gamer Tag: <input type="text" name="Member_Gamer_Tag2" size="20"></center>
  <center>Password:&nbsp;&nbsp; <input type="password" name="Password" size="20"></center>
  <center><input type="submit" value="Submit" name="B1"></center>
</form>
<?
} else {
if($Rank >= $Two){
if($_GET['Recruit_Member'] == Yes){
function generate_code($chars){
for($i=0;$i<=($chars-1);$i++){
$r0 = rand(0,1); $r1 = rand(0,2);
if($r0==0){$r .= chr(rand(ord('A'),ord('Z')));}
elseif($r0==1){ $r .= rand(0,9); }
if($r1==0){ $r = strtolower($r); }
}
return $r;
}
$Password = generate_code(10);
@mysql_query("INSERT INTO clan_members SET Gamer_Tag = '$Gamer_Tag', Password = '$Password', Rank = '$Rank2', Rank_Name = '$Rank_Name', Disabled = '$Disabled'");
@mysql_query("INSERT INTO clan_logs SET Date = '$Date', Gamer_Tag = '$Member_Gamer_Tag', Log = '$Log_Reason'");
echo "<center>His/Her Gamer Tag is $Gamer_Tag and his/her password is $Password .</center>";
}else{
?>
<form method="POST" action="Recruit_Member.php?Recruit_Member=Yes">
  <center>Gamer Tag:&nbsp;<input type="text" name="Gamer_Tag" size="20"></center>
    <center><input type="submit" value="Submit" name="B1"></center>
</form>
<?
}
}else{
echo "<center>Your rank is not high enought to do this.</center>";
}
}
echo $Footer;
?>[/code]
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How about modifying it to this:

[code]
$result = mysql_query("INSERT INTO clan_members SET Gamer_Tag = '$Gamer_Tag', Password = '$Password', Rank = '$Rank2', Rank_Name = '$Rank_Name', Disabled = '$Disabled'");
if (!$result) {
  die('Invalid query: ' . mysql_error());
}
[/code]

At least then you're super sure it's not a mysql error. 
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