Parse error

vegas · April 21, 2013, 3:34am

Can someone please tell me what my problem is here? I keep getting this error “Parse error: syntax error, unexpected ‘=’ in C:\xampp\htdocs\images\insert2.php on line 7” and just can’t figure out why.

Here is my code:[php]

<?php $path="localhost"; $username="root"; $password="xxxxxx"; $database="test"; $field1-name=$_POST['Value1']; $field2-name=$_POST['Value2']; $field3-name=$_POST['Value3']; $field4-name=$_POST['Value4']; $field5-name=$_POST['Value5']; mysql_connect($path,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO person VALUES ('','$field1-name','$field2-name','$field3-name','$field4-name','$field5-name')"; mysql_query($query); mysql_close(); ?>

[/php]

m11 · April 21, 2013, 3:50am

You cannot use dashes in variable names. You can use underscores though

vegas · April 21, 2013, 3:59am

Thanks Matt that removed the parse error now I have the following errors.

Notice: Undefined index: Value1 in C:\xampp\htdocs\images\insert2.php on line 7

Notice: Undefined index: Value2 in C:\xampp\htdocs\images\insert2.php on line 8

Notice: Undefined index: Value3 in C:\xampp\htdocs\images\insert2.php on line 9

Notice: Undefined index: Value4 in C:\xampp\htdocs\images\insert2.php on line 10

Notice: Undefined index: Value5 in C:\xampp\htdocs\images\insert2.php on line 11

Notice: Undefined variable: field1 in C:\xampp\htdocs\images\insert2.php on line 17

Notice: Undefined variable: field2 in C:\xampp\htdocs\images\insert2.php on line 17

Notice: Undefined variable: field3 in C:\xampp\htdocs\images\insert2.php on line 17

Notice: Undefined variable: field4 in C:\xampp\htdocs\images\insert2.php on line 17

Notice: Undefined variable: field5 in C:\xampp\htdocs\images\insert2.php on line 17

m11 · April 21, 2013, 4:01am

This would indicate your $_POST variables are missing. Can you post your form HTML?

Likely because you forgot to update the dashes in the query. So instead of $field1_name it is looking for $field1

vegas · April 21, 2013, 4:01am

actually I figured out what most of those previous errors were from now I just have the following,

Notice: Undefined index: Value1 in C:\xampp\htdocs\images\insert2.php on line 7

Notice: Undefined index: Value2 in C:\xampp\htdocs\images\insert2.php on line 8

Notice: Undefined index: Value3 in C:\xampp\htdocs\images\insert2.php on line 9

Notice: Undefined index: Value4 in C:\xampp\htdocs\images\insert2.php on line 10

Notice: Undefined index: Value5 in C:\xampp\htdocs\images\insert2.php on line 11

vegas · April 21, 2013, 4:05am

here is my html code:

<form action="insert2.php" method="post">
Value1: <input type="text" name="field1_name"><br>
Value2: <input type="text" name="field2_name"><br>
Value3: <input type="text" name="field3_name"><br>
Value4: <input type="text" name="field4_name"><br>
Value5: <input type="text" name="field5_name"><br>
<input type="Submit">
</form>

vegas · April 21, 2013, 4:13am

and here is the corrected php code:

[php]

<?php $path="localhost"; $username="root"; $password="xxxxx"; $database="test"; $field1_name=$_POST['Value1']; $field2_name=$_POST['Value2']; $field3_name=$_POST['Value3']; $field4_name=$_POST['Value4']; $field5_name=$_POST['Value5']; mysql_connect($path,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO person VALUES ('','$field1_name','$field2_name','$field3_name','$field4_name','$field5_name')"; mysql_query($query); mysql_close(); ?>

[/php]

vegas · April 21, 2013, 4:32am

Thanks for your help Matt, I figured out what the remaining problem was, made code changes to this:

[php]

<?php $path="localhost"; $username="root"; $password="xxxxx"; $database="test"; $field1=isset( $_POST['Value1']) ? $_POST['Value1'] : ""; $field2=isset( $_POST['Value2']) ? $_POST['Value2'] : ""; $field3=isset( $_POST['Value3']) ? $_POST['Value3'] : ""; $field4=isset( $_POST['Value4']) ? $_POST['Value4'] : ""; $field5=isset( $_POST['Value5']) ? $_POST['Value5'] : ""; mysql_connect($path,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO person VALUES ('','$field1','$field2','$field3','$field4','$field5')"; mysql_query($query); mysql_close(); ?>

[/php]

Now it works fine!

m11 · April 21, 2013, 4:47am

That is good practice but I don’t believe that fixes your problem.

Value1: <input type="text" name="field1_name"><br>
Value2: <input type="text" name="field2_name"><br>
Value3: <input type="text" name="field3_name"><br>
Value4: <input type="text" name="field4_name"><br>
Value5: <input type="text" name="field5_name"><br>

These should be named to what you are looking for in the $_POST array. e.g.

Value1: <input type="text" name="Value1"><br>
Value2: <input type="text" name="Value2"><br>
Value3: <input type="text" name="Value3"><br>
Value4: <input type="text" name="Value4"><br>
Value5: <input type="text" name="Value5"><br>

vegas · April 21, 2013, 6:16am

I spoke too soon, although the changes have cleared up my errors, it appears as though nothing is being posted in MySQL, all I get is a white page after submit is pressed on the input.html page. I check the tables in MySQL and they are empty and I get no errors at all. I made the changes in the html form you suggested with no change in the results. Below is what my current html and php codes is since the changes:

This is input.html

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>Input</title>
</head>

<body>
<form action="insert2.php" method="post">
Value1: <input type="text" name="value1"></input><br/>
Value2: <input type="text" name="value2"></input><br/>
Value3: <input type="text" name="value3"></input><br/>
Value4: <input type="text" name="value4"></input><br/>
Value5: <input type="text" name="value5"></input><br/>
<input type="submit"></input>
</form></body>

</html>

This is insert2.php
[php]

<?php $path="localhost"; $username="root"; $password="xxxxx"; $database="test"; $field1=isset( $_POST['Value1']) ? $_POST['Value1'] : ""; $field2=isset( $_POST['Value2']) ? $_POST['Value2'] : ""; $field3=isset( $_POST['Value3']) ? $_POST['Value3'] : ""; $field4=isset( $_POST['Value4']) ? $_POST['Value4'] : ""; $field5=isset( $_POST['Value5']) ? $_POST['Value5'] : ""; mysql_connect($path,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO person VALUES ('','$field1','$field2','$field3','$field4','$field5')"; mysql_query($query); mysql_close(); ?>

[/php]

and this would be output.php
[php]

output.php <?php $path="localhost"; $username="root"; $password="xxxxx"; $database="test";

mysql_connect($path,$username,$password);
@mysql_select_db($database) or die( “Unable to select database”);
$query=“SELECT * FROM person”;
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>

<?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"field1"); $f2=mysql_result($result,$i,"field2"); $f3=mysql_result($result,$i,"field3"); $f4=mysql_result($result,$i,"field4"); $f5=mysql_result($result,$i,"field5"); ?> <?php $i++; } ?>

Value1	Value2	Value3	Value4	Value5
<?php echo $f1; ?>	<?php echo $f2; ?>	<?php echo $f3; ?>	<?php echo $f4; ?>	<?php echo $f5; ?>

[/php]

Just a note, I’m just a beginner with all of this and this isn’t on a production server, These are just things I’m trying to strengthen my knowledge in.

m11 · April 21, 2013, 7:37am

Check your case. ‘value1’ is not the same as ‘Value1’

vegas · April 21, 2013, 5:40pm

I verified proper spelling and changed anything that was wrong but still no change.

m11 · April 22, 2013, 7:22am

Please post your table structure so that I can verify it matches your query.

vegas · April 22, 2013, 7:46am

person
Column Type Null Default Comments MIME
Value1 varchar(30) No
Value2 varchar(30) No
Value3 varchar(30) No
Value4 varchar(30) No
Value5 varchar(30) No

m11 · April 22, 2013, 7:52am

Right so do you see the problem here?

[php]
$f1=mysql_result($result,$i,“field1”);
$f2=mysql_result($result,$i,“field2”);
$f3=mysql_result($result,$i,“field3”);
$f4=mysql_result($result,$i,“field4”);
$f5=mysql_result($result,$i,“field5”);
[/php]

Your column names are not “field1” etc.

vegas · April 22, 2013, 7:58am

I figured it out Matt, in my insert2.php code on line 17 I had

(’’,’$Value1’,’$Value2’,’$Value3’,’$Value4’,’$Value5’)";

I changed it to:

(’$Value1’,’$Value2’,’$Value3’,’$Value4’,’$Value5’)";

and now the values are inserted in the database.

vegas · April 22, 2013, 8:13am

SQL result
Host: 127.0.0.1
Database: test
Generation Time: Apr 22, 2013 at 10:11 AM
Generated by: phpMyAdmin 3.5.2.2 / MySQL 5.5.27
SQL query: SELECT * FROM person LIMIT 0, 30 ;
Rows: 3

Value1 Value2 Value3 Value4 Value5
test test test test test
test2 test2 test2 test2 test2
test3 test3 test3 test3 test3

m11 · April 23, 2013, 5:10am

Yes that would have prevented your INSERT queries from working. If you add proper error_reporting you can save yourself some headaches

For example, adding or die(mysql_error()) to mysql_query

[php]
$query = “INSERT INTO person VALUES
(’$field1’,’$field2’,’$field3’,’$field4’,’$field5’)”;

mysql_query($query)or die(mysql_error());
[/php]

I also recommend always specifying column names. If you modify the table in the future (add/remove a column) this query will no longer work.

For example:

[php]
$query = “INSERT INTO person (Value1,Value2,Value3,Value4,Value5) VALUES
(’$field1’,’$field2’,’$field3’,’$field4’,’$field5’)”;
[/php]

And finally, this part of your code is likely still not working if you are still using the wrong column names.

[php]
$f1=mysql_result($result,$i,“field1”);
$f2=mysql_result($result,$i,“field2”);
$f3=mysql_result($result,$i,“field3”);
$f4=mysql_result($result,$i,“field4”);
$f5=mysql_result($result,$i,“field5”);
[/php]

vegas · April 23, 2013, 6:19am

Thanks for you help Matt, I learned a lot from that one mistake. As for the $field portion of the code, I don’t know what to say as the script is working perfectly, I can post to the database and display from it as well. It’s ok though as this was more of a learning tool for me as I didn’t write the script I got it from a tutorial on another site and found out it was riddled with errors so I used it as a learning tool and it worked out fairly well as it’s helped me debug the scripts that I am writing. Once again, Thanks for your help, I greatly appreciate it.

annaharris · April 24, 2013, 1:14pm

I think that there is nothing wrong with your table structure.