trying to insert

I get the following message:

with this code. What does it mean?

<?php
$apt=$_POST['search_term'];
$stat = mysql_connect("localhost","root","");
$stat = mysql_select_db("prerentdb");
$query = "SELECT name FROM payments WHERE late = 'L'";
$stat = @mysql_fetch_assoc(mysql_query($query));
echo $stat["name"];
$name=$_POST['name'];  
$apt=$_POST['apt']; 
$amtpaid=$_POST['amtpaid'];  
$rentdue=$_POST['rentdue'];  
$prevbal=$_POST['prevbal'];
$hudpay=$_POST['hudpay'];  
$tentpay=$_POST['tentpay']; 
$datepaid=$_POST['datepaid'];
$late=$_POST['late']; 
$comments=$_POST['comments'];  
$paidsum=$_POST['paidsum'];
$query = "
INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal,
hudpay,tentpay,datepaid,late,comments,paidsum)
VALUES('$name','$apt','$amtpaid','$rentdue','$prevbal',
'$hudpay','$tentpay','$datepaid','$late','$comments','$paidsum')";
$stat = mysql_query($query) or die('Query failed: ' . mysql_error());  
mysql_close();
echo "data inserted<br /><br />"; 
?>

Code (markup):

 

same issue i too had.waiting for the experts resolution.

 

I hope we get the issues resolved. I don't have a lot of time but I'm trying to learn. I've been told a couple of times how terrible my coding is but that is why I ask questions. Good luck and don't be discouraged by the bad eggs.

 

That's because `late` field (column) does not exists in DB
look into DB table `payhist` and you will see there is no `late` field

 

I changed the code per suggestons from another forum but can't get it to work. I sure need to get this to work. Below is the new code:

<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");
$result=mysql_query($query);
$num=mysql_numrows($result);
$query = "
INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal,
hudpay,tentpay,datepaid,late,comments,paidsum)
Select name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,late,comments,paidsum
From payments
WHERE paidsum= rentdue OR late = 'L'";
$stat = mysql_query($query) or die('Query failed: ' . mysql_error() . " - query: $query");  
echo "data inserted</font><br /><br />"; 
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

PHP:

This is the result:

 

You're newbie right?
just after connecting to mysql, you are calling mysql_query($query), but $query does not exists.
You also can't call 2 queries in one mysql_query(); & you didn't enter VALUES that should be inserted

It should be like this

<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");

$query = "
INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal,
hudpay,tentpay,datepaid,late,comments,paidsum) VALUES ('1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1')";
$query2 ="
Select *
From payments
WHERE paidsum= 'rentdue' OR late = 'L'";
mysql_query($query);
mysql_query($query2) or die('Query failed: ' . mysql_error() . " - query: $query");  
echo "data inserted</font><br /><br />";
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

PHP:

that

VALUES ('1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1')

PHP:

is just to fill what has to be filled in correct syntax

 

I used your below code and got the message below. it entered the integer fields but not the name, apt, datepaid, comments

<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");
$query = "INSERT INTO payhist
(name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,late,comments,paidsum)
VALUES ('$name', '$apt', '$amtpaid', '$rentdue', '$prevbal', '$hudpay', '$tentpay', '$datepaid',
'$late', '$comments', '$paidsum')";
$query2 ="Select *From payments WHERE late = 'L'";
mysql_query($query);
mysql_query($query2) or die('Query failed: ' . mysql_error() . " - query: $query");
echo "data inserted</font><br /><br />";
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

data inserted

query: INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,comments,paidsum)
VALUES ('', '', '', '', '', '', '', '', '', '')

 

I used your below code and got the message below. it entered the integer fields but not the name, apt, datepaid, comments

<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");
$query = "INSERT INTO payhist 
(name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,late,comments,paidsum) 
VALUES ('$name', '$apt', '$amtpaid', '$rentdue', '$prevbal', '$hudpay', '$tentpay', '$datepaid',
'$late', '$comments', '$paidsum')";
$query2 ="Select *From payments WHERE late = 'L'";
mysql_query($query);
mysql_query($query2) or die('Query failed: ' . mysql_error() . " - query: $query");  
echo "data inserted</font><br /><br />";
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

PHP:

data inserted

 

because $name, $apt, $amtpaid, .............. you did not set them.

 

you mean like this?
<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");
$name = $_POST['name'];
$apt = $_POST['apt'];
$amtpaid = $_POST['amtpaid'];
$rentdue = $_POST['rentdue'];
$prevbal = $_POST['prevbal'];
$hudpay = $_POST['hudpay'];
$tentpay = $_POST['tentpay'];
$datepaid = $_POST['datepaid'];
$comments = $_POST['comments'];
$paidsum = $_POST['paidsum'];
$query = "INSERT INTO payhist
(name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,comments,paidsum)
VALUES ('$name', '$apt', '$amtpaid', '$rentdue', '$prevbal', '$hudpay', '$tentpay', '$datepaid',
'$comments', '$paidsum')";
$query2 ="Select * From payments WHERE late = 'L'";
mysql_query($query);
mysql_query($query2) or die('Query failed: ' . mysql_error() . " - query: $query");
echo "data inserted</font><br /><br />";
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

 

yes, exactly, works?

 

Do you mean like this?

<?php
mysql_connect(localhost,root,"");
mysql_select_db(prerentdb) or die( "Unable to select database");
$name = $_POST['name'];
$apt = $_POST['apt'];
$amtpaid = $_POST['amtpaid'];
$rentdue = $_POST['rentdue'];
$prevbal = $_POST['prevbal'];
$hudpay = $_POST['hudpay'];
$tentpay = $_POST['tentpay'];
$datepaid = $_POST['datepaid'];
$comments = $_POST['comments'];
$paidsum = $_POST['paidsum'];
$query = "INSERT INTO payhist 
(name,apt,amtpaid,rentdue,prevbal,hudpay,tentpay,datepaid,comments,paidsum) 
VALUES ('$name', '$apt', '$amtpaid', '$rentdue', '$prevbal', '$hudpay', '$tentpay', '$datepaid',
 '$comments', '$paidsum')";
$query2 ="Select * From payments WHERE late = 'L'";
mysql_query($query);
mysql_query($query2) or die('Query failed: ' . mysql_error() . " - query: $query");  
echo "data inserted</font><br /><br />";
mysql_close();
// echo $query for debugging purposes
echo "query: $query<br />";
?>

PHP:

 

this is more safe

$name = mysql_real_escape_string($_POST['name']);
$apt = mysql_real_escape_string($_POST['apt']);
$amtpaid = mysql_real_escape_string($_POST['amtpaid']);
$rentdue = mysql_real_escape_string($_POST['rentdue']);
$prevbal = mysql_real_escape_string($_POST['prevbal']);
$hudpay = mysql_real_escape_string($_POST['hudpay']);
$tentpay = mysql_real_escape_string($_POST['tentpay']);
$datepaid = mysql_real_escape_string($_POST['datepaid']);
$comments = mysql_real_escape_string($_POST['comments']);
$paidsum = mysql_real_escape_string($_POST['paidsum']);

PHP:

 

Parse error: syntax error, unexpected '[' in C:\xampp\htdocs\hofiles\lateinsert.php on line 12

 

I had mistake in my code,
replace these lines with

$name = mysql_real_escape_string($_POST['name']);
$apt = mysql_real_escape_string($_POST['apt']);
$amtpaid = mysql_real_escape_string($_POST['amtpaid']);
$rentdue = mysql_real_escape_string($_POST['rentdue']);
$prevbal = mysql_real_escape_string($_POST['prevbal']);
$hudpay = mysql_real_escape_string($_POST['hudpay']);
$tentpay = mysql_real_escape_string($_POST['tentpay']);
$datepaid = mysql_real_escape_string($_POST['datepaid']);
$comments = mysql_real_escape_string($_POST['comments']);
$paidsum = mysql_real_escape_string($_POST['paidsum']);

PHP:

you have something like

$comments = $mysql_real_escape_string(_POST['comments']);
hehe

 

no problem, it caught that. the result's still the same

 

no problem, I caught it. still the same results

 

some results? Which results?

 

correction, no error messages, nothing

 

after

add

and try it. show input here