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Hi, I'm working on a site that allows users to connect with each other. User can search, find friend, and then request a connection. It works but if user request a second time it duplicates the row in table. I only want INSERT if does not exists.
Here is code.

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form2")) { 
	$insertSQL = sprintf("INSERT INTO mystuff.contact(user_id, user_url, rec_id, contact_id) VALUES(%s, %s, %s, %s)",
                       GetSQLValueString($_POST['hiddenField2'], "int"),
					   GetSQLValueString($_POST['hiddenField'], "text"),
					   GetSQLValueString($_POST['hiddenField3'], "text"),
					   GetSQLValueString($_POST['hiddenField4'], "text"),
					   GetSQLValueString($_POST['hiddenField5'], "text"),
                       GetSQLValueString($_POST['contact_id'], "int"));
mysql_select_db($database_connAdmin, $connAdmin);
  $Result1 = mysql_query($insertSQL, $connAdmin) or die(mysql_error());

Only thing that really matters here is user_id and rec_id, which is other users user_id.
Thanks

3
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6 Years
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Last Post by phpDave
1

You can use REPLACE in SQL syntax which will update if does exist or insert if not.

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Thanks,
Although I am thinking that using REPLACE would reset my request col. from 1 back to 0.
0 if request, and 1 if request is accepted.
Anyway, I trying to just validate by using:

do {
	$already = $row_Recordset4['rec_id'];
	$test = $row_Recordset1['user_id'];
} while ($row_Recordset4 = mysql_fetch_assoc($Recordset4));
//
if ($already == $test) {
	$error['No'] = 'Already connected with.'; }

Problem is $test is Recordset1 and $already is Recordset4
Can not seem to get it to work right.

1

their r 2 ways of doing this:-
1. Use this code...

$result = mysql_query("SELECT * FROM table1", $link);
$num_rows = mysql_num_rows($result);
if($num_rows>0)
{
echo "Already Connected";
}

Another solution can be setting a variable and checking that variable before insertion...if that variable is set then display already connected...
Hope i m clear...

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Thanks, I like the second choice and I have tried it before but could not get it to work, any advice on the code.
I tried something like this before and also a few other things.

#
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form2")[B] &&(!$already)[/B]) {
#
$insertSQL = sprintf("INSERT INTO mystuff.contact(user_id, user_url, rec_id, contact_id) VALUES(%s, %s, %s, %s)",
0

whats the problem with ur code...it's looks fine ...
check this way....

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form2") &&($already!=1))

for first way i told...
use the select query that u want...i don't know ur exact query so i wrote the simplest one...

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Thanks, I tried but still not working, I have the $already var. in a do while loop.
Could that be why not working?
Each user can have more than one contact. I got the validation to work but only match one connection, not multiple.
???

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Almost there!
Here is what I have:

$var1_check = "-1";
if (isset($_POST['hiddenField9'])) {
  $var1_check = $_POST['hiddenField9'];
}

do {
	 $already = $row_Recordset4['rec_id'];
} while ($row_Recordset4 = mysql_fetch_assoc($Recordset4));
//
if ($var1_check ==($already) {
	$error['No'] = 'Already connected with.';}

Only problem is $already in the if statement is only bringing up one rec_id and not all as it does inside the loop. Not sure how to handle this.

0

Thanks for replies, Problem solved. Some of it was in SELECT QUERY.

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