PHP MySQL INSERTING

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

Make sure that in final version you are escaping those values or using prepared statements instead of just interpolating them into the string unmodified. Very dangerous. — Brandon Buck, Feb 03 '13 at 23:39
@BackinaFlash - True. In the future, [link to this](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php#answer-12860140). — Joseph Silber, Feb 03 '13 at 23:50
"Not executing" is not enough information - what's the error ? — Nir Alfasi, Feb 03 '13 at 23:52
@thiago Marquezini Look at your variable dub2. You do not execute your insert query. — Mr. Radical, Feb 04 '13 at 00:12

Mr. Radical · Accepted Answer · 2013-02-04T00:29:35.977

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);
?>

PHP MySQL INSERTING

1 Answers1