help with some code

amarafzal

ive created a form which when submitted puts the data in a database.

The problem im having is because ive not declared a unique field within the database, when im recalling the information those people who have had the same reference number the results are displayed one after another on the same page.

Is there any type of validation that can can check the database, when the form is submitted and tells the user on the web page that "this reference exists, can you declare another one."?

does that make sense?

gnome354

Have you tried querying the database for that reference number before it is entered into the database? That why if the number is already taken, you can inform the user "That reference number is already taken, please try again".

amarafzal

thats what i want it to do, what code do i need? can you give me some i can use

gnome354

I would like to help you, but first I need more information.

How are you processing the form data?
What database are you using?
Are you using sessions?

amarafzal

How are you processing the form data?
The data is sent to an mysql database using a INSERT TAGS. The user fills out a form and when they click submit the information is inputted in to the database

What database are you using? MYSQL

Are you using sessions? No

gnome354

mytable = db table where this is taking place
ref_num = reference number field in mytable
mydb = mysql database
$refnum = user entered reference number

After the completed form is submitted...
1. Check if all required fields have been completed
2. Query the database for duplicate reference numbers

//connect to mysql
$conn = mysql_connect($dbhost,$dbuname,$dbpass) or die('Could not connect to db');
//select the db
mysql_select_db('mydb');
//query the db for duplicate ref numbers...
$result = mysql_query("select name from mytable where ref_num ='$refnum'",$conn);
//is this a duplicate reference number?
if(mysql_num_rows($result)>0){
   //yes it is, inform the user to choose a different number
  echo "This reference number is already in use. Please try again";
  <center><a href=\"javascript:history.go(-1)\">Go Back</a></center>  

}else{
  //no it is not, enter information into the database
  $result = mysql_query()...
} 
mysql_close($conn);

amarafzal

do i put this code in the page where the form is or on the page that displays the confirmation?

gnome354

the code should go wherever you process the information from the submitted form.

amarafzal

some code im already using for example:

$db_name = "web";
$table_name = "jobs";
$connection = @mysql_connect("dfsdf", "webadmin", "r2d2c6po") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());

$sql = "INSERT INTO $table_name (job_id, contact_name, contact_phone, date_posted, job_title, job_summary, location, job_status, salary, career_level) VALUES ('$POST[job_id]', '$POST[contact_name]', '$POST[contact_phone]', '$POST[date_posted]', '$POST[job_title]', '$POST[job_summary]', '$POST[location]', '$POST[job_status]', '$POST[salary]', '$POST[career_level]')";
$result = @($sql,$connection) or die(mysql_error());
?>

how do i put your code in to that as some of your code is same as mine?

will it make a difference if i put your code in? wont it mess it up?

i do appreciate your help

gnome354

$db_name = "web";
$table_name = "jobs";
$connection = @mysql_connect("dfsdf", "webadmin", "r2d2c6po") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());
//check for duplicate ref numbers
$sql = "SELECT contact_name from $table_name where job_id='$_POST[job_id]'";
$result = @mysql_query($sql,$connection) or die(mysql_error());
if(@mysql_num_rows($result)>0){
//inform user the job id is already taken
echo "This job id is already taken. Please try again";
}else{

$sql = "INSERT INTO $table_name (job_id, contact_name, contact_phone, date_posted, job_title, job_summary, location, job_status, salary, career_level) VALUES ('$_POST[job_id]', '$POST[contact_name]', '$POST[contact_phone]', '$POST[date_posted]', '$POST[job_title]', '$POST[job_summary]', '$POST[location]', '$POST[job_status]', '$POST[salary]', '$_POST[career_level]')";

$result = @mysql_query($sql,$connection) or die(mysql_error());
}
?>

amarafzal

that works!

one thing though, to display the following statement which tells the user that information has been inputted

echo "Thank you. Your job <?php echo $_REQUEST['job_title']; ?> has been posted on the talent website";

where would i put that, i put it above and below the insert statemnet and i get an error

any suggestions

amarafzal

JUST SORTED IT, MADE A MISTAKE

THANKS FOR THE HELP IT IS VERY MUCH APPRECIATED, YOU JUST DONT KNOW HOW MUCH!

IF I HAVE ANY OTHER PROBLEMS CAN I EMAIL YOU?

gnome354

sure.

amarafzal

one thing, if i want the user to go back to the page they were on, what code do i use so that the informaton they added is still there?

u gave me some code but it didnt work:

gnome354

The code I showed you is same has clicking the back button.

amarafzal

that works,

but is there any way of going back to the information they have added as at the moment it wipes it all off

stolzyboy

try this one instead

amarafzal

nope that does not work either, there is nothing in the form when you go back

thanks for trying, any other ideas?

stolzyboy

ok, well instead of using the back button idea, simply spit out the form again and autofill the fields with the variables back into the form so they can then keep the information and try to submit again

gnome354

There are two ways I know of which can solve your problem.

1a. Sessions - first modify your form such that in the "value" field in your <input> tags insert variable names.

For example:

<form action="executeform.php" method="post">
Name: <input type="text" name="myname" size="11" maxlength="10" value="$sessname">  

Job Title: <input type="text" name="mytitle" size="26" maxlength="25" value="$sesstitle">
</form>

1.b After the form is submitted, register the form variables as session variables.
$sessname = $POST['myname'];
$sesstitle = $POST['mytitle'];
session_register('sessname'); //without the $
session_register('sesstitle'); //without the $

If you submit the form and now go back the fields will be filled in. The draws backs to using session in this manner is you have to remember to unregister your session variables (i.e. session_unregister(<session variable name>))

The other option is messier...
2a. Follow step 1a above.
2b. When the form has been submitted, pass the variables back to the form. I would suggest enclosing the form in function tags if it is not currently. If there is an error call your form function passing along the form variables.
For example:

//original form with the form variables ($myname, $mytitle)initialized to nothing
//if there is an error the form variables ($myname and $mytitle) 
//will have values and replace the information
function theform($myname="", $mytitle=""){
echo "<form action=\"executeform.php\" method=\"post\">
Name: <input type=\"text\" name=\"myname\" size=\"11\" maxlength=\"10\" value=\"$myname\">  

Job Title: <input type=\"text\" name=\"mytitle\" size=\"26\" maxlength=\"25\" value=\"$mytitle\">
</form>";
}

Senerio: the form has been submitted. Form processing begins, there is an error, tell user and call the form function again...

echo "Error message<br>";
theform($myname,$mytitle);

I hope this makes sense.