PHP MySQL INSERTING - php

Question

I have a php with mysql that would insert some data to the database if there isn't the same information
<?php
$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];
$con = mysql_connect("localhost", "root", "");
if (!$con)
{die('Could not connect to mysql: ' . mysql_error());}
$mydb = mysql_select_db("gunbound");
if (!$mydb)
{die('Could not connect to database: ' . mysql_error());}
$dup = mysql_query("SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'");
if(mysql_num_rows($dup) >= 1){
echo '<b>You have already ask for guild request.</b>';
}
else
{
$dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");
}
Print "<center>You have requested to join the guild.</center>";
mysql_close($con);
?>
but its not adding the record to the database if there isn't a record equal
Nor executing this:
$dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");
even if the code:
if(mysql_num_rows($dup) >= 1){
says that he can do the action of inserting
please help me

Answer 1

Try this:
<?php
$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];
$con = mysql_connect("localhost", "root", "");
if (!$con)
{die('Could not connect to mysql: ' . mysql_error());}
$mydb = mysql_select_db("gunbound");
if (!$mydb)
{die('Could not connect to database: ' . mysql_error());}
$dup = mysql_query("SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'");
if(mysql_num_rows($dup) >= 1){
echo '<b>You have already ask for guild request.</b>';
}
else
{
$dup2 = mysql_query("INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')");
return $dup2;
}
Print "<center>You have requested to join the guild.</center>";
mysql_close($con);
?>
I have add return to your else statement. I will execute your $dub2 variable. You could if you want to leave the variable $dub2 out then you will intermediately execute your query. Another way would be to use mysql_execute() function.
This would be a MYSQLI equivalent:
<?php
$id=$_POST['id'];
$guildname=$_POST['guildname'];
$level=$_POST['level'];
$score=$_POST['score'];
$guildmaster=$_POST['guildmaster'];
$host = "hostname";
$user = "username";
$password = "password";
$database = "database";
$con = mysqli_connect($host, $user, $password, $database);
if (!$con)
{die('Could not connect to mysql: ' . mysql_error());}
$dup = "SELECT Id FROM guildrequest WHERE Id='".$_POST['id']."'";
mysqli_query($con, $dup);
if (!$dup)
{die('Could not connect to database: ' . mysql_error());}
if(mysqli_num_rows($dup) >= 1){
echo '<b>You have already ask for guild request.</b>';
}
else
{
$dup2 = "INSERT INTO guildrequest VALUES ('$id', '$guildname', '$level', '$score', '$guildmaster')";
mysqli_query($con, $dup2);
}
Print "<center>You have requested to join the guild.</center>";
mysqli_close($con);
?>