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LiveFree
Forum Contributor
Posts: 258 Joined: Tue Dec 06, 2005 5:34 pm
Location: W-Town
Post
by LiveFree » Sat Dec 10, 2005 12:33 pm
Hello, say I have this code...
Code: Select all
<?php
include('includes/connect.php');
if (isset($_POST['submitted'])){
if (isset($_POST['id'])){
if(isset($_POST['name'])){
$id=$_POST['id'];
$name=$_POST['name'];
$sql="INSERT INTO test (id, name) VALUES ('$id','$name')";
$result=mysql_query($sql);
if (mysql_affected_rows($result) == 1){
echo "The script has completed sucessfully!";
mysql_close();
}else{
echo "There seems to be an issue.";
}
}else{
echo "Please select your name";
}
}else{
echo "Please select your id";
}
}else{
echo '<form name="test" method="POST" action="test.php">
ID:<select name="id"><option value="1">1</option><option value="2">2</option></select>
<br>Name: <select name="name"><option value="Tucker">Tucker</option><option name="Haener">Haener</option></select><br>
<input type="submit" name="submit" value="Submit"><input type="hidden" name="submitted" value="TRUE"></form>';
}
http://www.aa-25thID.com/ATFCS/test.php is the link if you want to try it
When you do it, it says:
Warning: mysql_affected_rows(): supplied argument is not a valid MySQL-Link resource in /home/recon/public_html/ATFCS/test.php on line 11
There seems to be an issue.
I was just wondering if I am not follwing the syntax for inserting <option> values into mysql....
THanks for any and all help
twigletmac
Her Royal Site Adminness
Posts: 5371 Joined: Tue Apr 23, 2002 2:21 am
Location: Essex, UK
Post
by twigletmac » Sat Dec 10, 2005 12:38 pm
The first thing to do is to add some error handling when you run the query. So instead of:
try
Code: Select all
$result = mysql_query($sql) or die(mysql_error().'<br />'.$sql);
This will give you a clearer idea as to what's going wrong.
Mac
LiveFree
Forum Contributor
Posts: 258 Joined: Tue Dec 06, 2005 5:34 pm
Location: W-Town
Post
by LiveFree » Sat Dec 10, 2005 1:04 pm
Okay I did it but it always gives me the..: Warning: mysql_affected_rows(): supplied argument is not a valid MySQL-Link resource in /home/recon/public_html/ATFCS/test.php on line 11
Could I contact you on MSN to furthur help resolve this issue?
shiznatix
DevNet Master
Posts: 2745 Joined: Tue Dec 28, 2004 5:57 pm
Location: Tallinn, Estonia
Contact:
Post
by shiznatix » Sat Dec 10, 2005 1:12 pm
it will be better if you just use these forums to help other people who read posts to learn things.
the reason you get that error is because there is somthing wrong with that query, you have to do error checking on it.
echo out your $sql and tell us what it says
twigletmac
Her Royal Site Adminness
Posts: 5371 Joined: Tue Apr 23, 2002 2:21 am
Location: Essex, UK
Post
by twigletmac » Sat Dec 10, 2005 1:13 pm
From your test script it looks like the problem is that the record already exists:
Code: Select all
Duplicate entry '2-Haener' for key 1
INSERT INTO test (id, name) VALUES ('2','Haener')
if ID is an auto-increment field then you won't need to set it manually each time.
Mac
LiveFree
Forum Contributor
Posts: 258 Joined: Tue Dec 06, 2005 5:34 pm
Location: W-Town
Post
by LiveFree » Sat Dec 10, 2005 1:14 pm
INSERT INTO test (id, name) VALUES ('2','Haener')Duplicate entry '2-Haener' for key 1
INSERT INTO test (id, name) VALUES ('2','Haener')
LiveFree
Forum Contributor
Posts: 258 Joined: Tue Dec 06, 2005 5:34 pm
Location: W-Town
Post
by LiveFree » Sat Dec 10, 2005 1:16 pm
Aha!
Problem solved
Was an issue with the mysql_affected_rows
Thanks!