[RESOLVED] Reporting duplicate entry?

linda_thomas

I have several tables where the product code is set to unquie, how would I go about reporting duplicate entrys when new products are added?

At the moment this is the AddNew_product.php:

<?php 
//log in 


@mysql_connect($host,$user,$password) or die("Unable to connect with server!"); 
@mysql_select_db($database)or die("Unable to connect database, please try later!"); 

$id=mysql_real_escape_string($_POST['id']); //This is set to unique. 
$name=mysql_real_escape_string($_POST['name']); 
$info=mysql_real_escape_string($_POST['info']); 
$pic=mysql_real_escape_string($_POST['pic']); 
$lg_pic=mysql_real_escape_string($_POST['lg_pic']); 
$price=mysql_real_escape_string($_POST['price']); 


$query = "INSERT INTO floral VALUES ('$id','$name','$info','$pic','$lg_pic','$price')"; 
mysql_query($query); 
    header("Location: added.php"); 

mysql_close(); 
?>

I realise it probably involves the if statement but I'm not sure how to write it.
Many thanks in advance,
Linda

devinemke

before you do your INSERT query do a SELECT query to see if that $id already in the table. if it is then abort the INSERT and display an error.

linda_thomas

Thank you very much devinemke, I'll have a go at that.

Best wishes,
Linda

linda_thomas

Ok i can't seem to get it to work, which doesn't surprise me, lol.... The trouble is I'm so new to php that I'm just getting more and more confuse. I've probably written this wrong but would someone mind telling me what I've done wrong!

<?php
//db conection

@mysql_connect($host,$user,$password) or die("Unable to connect to server!");
@mysql_select_db($database)or die("Unable to connect to database, please try later!");

$id=mysql_real_escape_string($_POST['id']);
$name=mysql_real_escape_string($_POST['name']);
$info=mysql_real_escape_string($_POST['info']);
$pic=mysql_real_escape_string($_POST['pic']);
$lg_pic=mysql_real_escape_string($_POST['lg_pic']);
$price=mysql_real_escape_string($_POST['price']);


$query = "SELECT $id FROM headstone WHERE id = '$id'"; 
$result = mysql_query($query); 

$num_rows = mysql_numrows($result); 

if ($num_rows>0) { 
    echo ("Sorry this ID all ready exhists in the database!"); 
} 
else { 

$query = "INSERT INTO headstone VALUES ('$id','$name','$info','$pic','$lg_pic','$price')";
mysql_query($query);
    header("Location: added.php");

  } 


mysql_close();
?>

Many thanks in advance,
Linda

devinemke

change your SELECT query to:

$result = mysql_query("SELECT id FROM headstone WHERE id = '$id'") or exit(mysql_error());

linda_thomas

I finaly got it working with this:

<?php
//db connection here

@mysql_connect($host,$user,$password) or die("Unable to connect to server!");
@mysql_select_db($database)or die("Unable to connect to database, please try later!");

$id=mysql_real_escape_string($_POST['id']);
$name=mysql_real_escape_string($_POST['name']);
$info=mysql_real_escape_string($_POST['info']);
$pic=mysql_real_escape_string($_POST['pic']);
$lg_pic=mysql_real_escape_string($_POST['lg_pic']);
$price=mysql_real_escape_string($_POST['price']);


$query = "SELECT id FROM floral WHERE id='$id'";
$result = mysql_query($query);

if (mysql_num_rows($result)>0) {
echo ("already in database");
} else {
$query = "INSERT INTO floral VALUES ('$id','$name','$info','$pic','$lg_pic','$price')";
$result = mysql_query($query);
  header("Location: added.php");
}


mysql_close();
?>

Gosh I feel like i'm climbing mount everest learning php 🙂 it's not as quick to learn as html and css.... sigh!

linda_thomas

I forgot to say thank you for all your help! I'm very greatful.

Have a great week.
Linda