Notice: Undefined index...

Imitation

Okay I managed to solve the t_string errors (somehow 😕 )... now I am getting a different error. This is "Notice: Undefined index: postid in C:\Program Files\Apache Group\Apache2\htdocs\bm\news.php on line 11" which refers to the line $postid = $row["postid"];. What I can't figure out is why the text from the mysql dbase that goes into $content will display but all the others will give that error. I'm sure it's really simple but I've been reading up in the manual.. and i get nothing. Sorry.. I am trying to figure things out for myself but somethings just aren't possible 🙂

<?php
    mysql_connect("localhost", "xxx", "xxx") or die("could not connect");
    mysql_select_db("bm"); 

$select = "select content from news";
$result = mysql_query($select);

while ($row = mysql_fetch_array($result)) {
$postid = $row["postid"];
//$userid = $row["userid"];
//$userid = $row["date"];
//$userid = $row["time"];
$content = $row["content"];
echo $postid;
//echo $userid;
//echo $date;
//echo $time;
echo $content;
}
mysql_free_result($result);
?>

hand

Because you only selected "content" field from your table.
$select = "select content from news";

If you want also retrieve the field "postid" from your table you should code:
$select = "select content, postid from news";

to get all fields
$select = "select content, postid, userid, date, time from news";
or
$select = "select * from news";

<?php
    mysql_connect("localhost", "xxx", "xxx") or die("could not connect");
    mysql_select_db("bm"); 

$select = "select * from news";
$result = mysql_query($select);

while ($row = mysql_fetch_array($result)) {
	$postid  = $row["postid"];
	$userid  = $row["userid"];
	$date    = $row["date"];
	$time    = $row["time"];
	$content = $row["content"];

	echo "postid: $postid<br>";
	echo "userid: $userid<br>";
	echo "date: $date<br>";
	echo "time: $time<br>";
	echo "content: $content<br>";
}
mysql_free_result($result);
?>

Imitation

Okay that is embarassing. oh well. heh 🙂 How silly is that..

/me bashes head on desk for eternity.

Thanks for that hand.

MalindaB

You so easily answered Imitation's question, I hope you can help me with mine. I have a similar problem, but mine is in inserting and not selecting. I get an "Undifined Index:" error on two inputs - hireDate and birthDate. The code is below.

<?php
$host="localhost";
$user="root";
$pass="xxxxxxxxx";
$db="demo";
$con = mysqli_connect($host,$user,$pass,$db);

$firstName=$POST['firstName'];
$lastName=$POST['lastName'];
$address=$POST['address'];
$city=$POST['city'];
$state=$POST['state'];
$zip=$POST['zip'];
$hPhone=$POST['hPhone'];
$pEmail=$POST['pEmail'];
$ssn=$POST['ssn'];
$department=$POST['department'];
$reportsTo=$POST['reportsTo'];
$hireDate=date("Y-m-d",strToTime($POST['hireDate']));
$jobTitle=$POST['jobTitle'];
$birthDate=date("Y-m-d",strToTime($POST['birthDate']));
$comments=$_POST['comments'];

$query = "INSERT INTO contacts(firstName,lastName,address,city,state,zip,hPhone,pEmail,ssn,department,reportsTo,jobTitle,comments,hireDate,birthDate)
VALUES('$firstName','$lastName','$address','$city','$state','$zip','$hPhone','$pEmail','$ssn','$department','$reportsTo','$jobTitle','$comments','$hireDate','$birthDate')";
$result = mysqli_query($con,$query);
if(!$result)
{
echo "There was an input error. Please try again.";
exit;
}
else
{
echo "Insert Successful, 1 record added";
}
mysqli_close($con);
?>

MalindaB

Please disregard my question. I just found the answer. I'm sure I will have more in the future though. Thanks.