Help! Parse Error

[Anthony1312002]

I'm getting this error in my form and I don't know why.

Parse error: parse error, unexpected '=' in c:\program files\apache group\apache\htdocs\communityservice\dataentry.php on line 81

What I'm trying to do is prevent a person from entering the same ANumber and the same DateOfService twice in the database. Here is how I'm trying to do this. Please not, I'm totally new to PHP. Line 81 is in bold.

$sql="SELECT * FROM activitytracking WHERE ANumber=$ANumber && DateOfService=$DateOfService";
$query=mysql_query($sql);
$data=mysql_fetch_array($query);
$num=$data['num'];
if (strlen ($ANumber) = ANumber && strlen ($DateOfService) = DateOfService){
print "Sorry, this employee already exists for the date of service you are entering";

[laserlight]

From what I see, you want to query the database with a given ANumber and DateOfService and check if the activity tracking table already has the record.
In that case, a simple solution would be to check the number of rows returned.
if it is not 0, then "this employee already exists for the date of service you are entering"

$sql = "SELECT * FROM activitytracking WHERE ANumber='$ANumber' AND DateOfService='$DateOfService'";
$query = mysql_query($sql);
if (mysql_num_rows($query) > 0) {
	echo "Sorry, this employee already exists for the date of service you are entering";
}
else {
	//do what is needed
}

In your given code, the main problem is:
"strlen ($ANumber) = ANumber"
Is ANumber defined?
Also, you should use "==" to compare, not "="
To PHP, it appears that you are trying to set the undefined constant 'ANumber' to the integer length returned from strlen(), which is absurd.

[Anthony1312002]

Sorry about the lack of information. The ANumber is not numerical. Here is an example: A081512. This is an ID that they key in to identify an employee. The DateOfService field is the date they key into the database. Does this help? I'm just trying to keep them from entering the same employee twice for the same DateOfService.

Here is a correction I made.

$sql="SELECT * FROM activitytracking WHERE ANumber=$ANumber && DateOfService=$DateOfService";
$query=mysql_query($sql);
//$data=mysql_fetch_array($query);
if (!$ANumber) == ANumber && (!$DateOfService) == DateOfService)
{
$error = "Sorry, this employee already exists for the date of service you are entering";

	} else {



$Query = "Insert activitytracking (ANumber, FirstName, LastName, TeamLeader, Department, DateOfService,
          NumberOfHours, ActivityType, Org, Description, TimeTakenWhen, siteID)
          Values ('$ANumber', '$FirstName', '$LastName', '$TeamLeader', '$Department',
                 '$DateOfService', '$NumberOfHours', '$ActivityType', '$Org',
                 '$Description', '$TimeTakenWhen', '$siteid')";

print "Thank you. Your record has been submitted to the database.";

[laserlight]

Look at my code snippet again.
I have enclosed $ANumber is quotes in the query, which is even more important since it is a string.
Likewise for $DateOfService

Let's look through the logic of your correct code:
1. Select the row where column ANumber has the value of $ANumber and the column DateOfService has the value of $DateOfService
2. If the row returned has NOT $ANumber equal to undefined constant ANumber AND NOT $DateOfService equal to undefined constant DateOfService, then set an error message

It doesnt make sense, does it?

My code does this:
1. Select the row where column ANumber has the value of $ANumber and the column DateOfService has the value of $DateOfService
2. If the number of rows returned is greater than 1, then output (or set) an error message

If a row is returned, it must have ANumber equal to $ANumber and the DateOfService equal to $DateOfService
This means that the new row that would be inserted would be a duplicate.

[Anthony1312002]

Thanks for getting back to me. I tried what you suggested but I'm getting the error below. What could be the problem?

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\communityservice\dataentry.php on line 79
Thank you. Your record has been submitted to the database.

$sql = "SELECT * FROM activitytracking WHERE ANumber='$ANumber' AND DateOfService='$DateOfService'";
$query = mysql_query($sql);
if (mysql_num_rows($query) > 0) {
echo "Sorry, this employee already exists for the date of service you are entering";
} else {

[PyroX]

It means that there is no mysql data in $query.

It means your query failed.

Where is your connection to mysql? I assume you just did not post it.

$query = mysql_query($sql, <SOMETHING IS MISSING HERE>);

[Anthony1312002]

Here is the connection I'm using. It's an include file.

<?
$Host = "localhost";
$DBName = "CommunityServiceTracking";
$User = "Anthony";
$dbpass = "";
$Link = mysql_connect ("$Host", "$User", "$dbpass", "$DBName");
//$Password = "";
$TableName = "activitytracking";
?>

[PyroX]

Try this: $query = mysql_query($sql, $Link);

[Anthony1312002]

I tried it but I'm still getting the same error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\communityservice\dataentry.php on line 79
Thank you. Your record has been submitted to the database.

It's odd that the record is still being posted to the database even though it's a duplicate. What am I missing?

$sql = "SELECT * FROM activitytracking WHERE ANumber=$ANumber AND DateOfService=$DateOfService";
$query = mysql_query($sql, $Link);
if (mysql_num_rows($query) > 0) {
$error = "Sorry, this employee already exists for the date of service you are entering";

[laserlight]

you can leave the link out, but you need to select your database

[Anthony1312002]

Where and how would I do this? I'm new to php. Do I do this in the include file or in the form?

[laserlight]

Here's an example, based on your own:

<?php
$Host = "localhost"; 
$DBName = "CommunityServiceTracking"; 
$User = "Anthony"; 
$dbpass = ""; 
mysql_connect($Host, $User, $dbpass);
mysql_select_db($DBName);
//$Password = ""; 
$TableName = "activitytracking"; 
?>

conversely you can use a $link = mysql_connect(/ etc /); but it isnt really needed if you arent opening more than 1 link

[Anthony1312002]

Okay, for the sake of keeping my sanity, here is what I've done in the include file.

<?
$Host = "localhost";
$DBName = "CommunityServiceTracking";
$User = "Anthony";
$dbpass = "";
$Link = mysql_connect ("$Host", "$User", "$dbpass", "$DBName");
$SQLStatement = mysql_select_db($DBName);
//$Password = "";
$TableName = "activitytracking";
?>

How do I now get the code block below to recognize the connection?

$sql = "SELECT * FROM activitytracking WHERE ANumber=$ANumber AND DateOfService=$DateOfService";
$query = mysql_query($sql, $Link);
if (mysql_num_rows($query) > 0) {
$error = "Sorry, this employee already exists for the date of service you are entering";

[laserlight]

if the file is included, the connection should be made.

I think there is a problem with the code snippet you just wrote.

Instead of telling it to you (I think I mentioned it earlier), change to this:
$query = mysql_query($sql, $Link) or die(mysql_error());

[Anthony1312002]

We must be getting close. I'm now getting this message which doesn't appear to be an error.

Unknown column 'A111111' in 'where clause'

Here is the adjusted code.

$sql = "SELECT * FROM activitytracking WHERE ANumber=$ANumber AND DateOfService=$DateOfService";
$query = mysql_query($sql, $Link) or die(mysql_error());
if (mysql_num_rows($query) > 0) {
$error = "Sorry, this employee already exists for the date of service you are entering";

These are the fields in my table:

ANumber
DateOfService

What am I missing?

[laserlight]

yep, that's it.
when parsed, the query looks like:
SELECT * FROM activitytracking WHERE ANumber=A111111 AND ...

So mysql runs around looking for the row where column ANumber has the same value as column A111111
then it notices that column A111111 doesnt even exist.
The reason is simple: A111111 is the string that is used to check, not the column.
Hence we must enclose this string in (single) quotes.
so,

$sql = "SELECT * FROM activitytracking WHERE ANumber='$ANumber' AND DateOfService='$DateOfService'"; 
$query = mysql_query($sql, $Link) or die(mysql_error()); 
if (mysql_num_rows($query) > 0) { 
	$error = "Sorry, this employee already exists for the date of service you are entering";

[Anthony1312002]

Thank you, thank you, thank you!!! Words cannot convey the relief I feel at this moment. I was about to ger really frustrated with this. You really came through. You deserve 2000 stars!!! Thanks again!!!

[PyroX]

conversely you can use a $link = mysql_connect(/ etc /); but it isnt really needed if you arent opening more than 1 link

I would advise you to always put the link ref in the query. That way if you ever need to make changes or add new connections, you do not have to go back and edit all your existing code.

The way you are doing it now is good:

$query = mysql_query($sql, $Link) or die(mysql_error());