Query was Empty

tobimichigan · May 13, 2009

Hi there,

Code gurus please help me check this code and probably help me identify what is wrong with it. I keep getting "Query was empty" and nothing is inserted into the table.

Here's the form:

<tr>

<p><font size='2' face='verdana'>WELCOME TO STAFFCAMS Registration FORM</font></p>

</div></td>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='28'><font size='2' face='verdana'>LEDGER NO.</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>FIRST NAME</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>OTHER NAMES</font><font size='2' face='verdana'></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>STATE OF ORIGIN</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>LOCAL GOVERNMENT</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>NATIONALITY</font></td>

</font></td>

</tr>

<tr>

<td height='26'><font size='2' face='verdana'>Email address</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>RESIDENTIAL ADDRESS</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>TELEPHONE NUMBER</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>DEPARTMENT</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>MARITAL STATUS</font></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>AMOUNT DEDUCTIBLE</font></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

</tr>

<tr>

</tr>

</table>

</form><div align='center'><font size='1' face='verdana'><a href='index.php'>Back to login page</a></font></div>

and here's the Register_Action.php action script code:

<?php

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

$amountd=$_POST['amountd'];

$dpt=$_POST['department'];

$email=$_POST['email'];

$fname =$_POST['fname'];

$id=$POST['id'];

$ledgerno =$_POST['ledgerno'];

$lga =$_POST['lga'];

$lname=$_POST['lname'];

$Nationalty=$_POST['Nationalty'];

$oname=$_POST['oname'];

$pfno=$_POST['pfno'];

$residentialaddress=$_POST['pfno'];

$soorigin=$_POST['soorigin'];

$today= date("Ymd H:i:s");

$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");

//$result_p_ticket= mysql_query("Insert into pawn_ticket(pawn_ticket_customer_number,pawn_ticket_ticket_number,pawn_ticket_ticket_check_digit,pawn_ticket_ticket_issue_date) values('$cust_no','$pawn_ticket_number','$check_digit_only','$today')");

if (!mysql_query($result))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($con)

?>

<html>

<body>

</body>

</html>

Please can someone identify for me why this form returns "Query was empty" each time I click Register Button?

jackpf · May 13, 2009

Jesus...put that in code tags...

And also, put or

die(mysql_error());

after every query.

merck_delmoro · May 13, 2009

#Your Code:

$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");

#Check every variables in your values you forgot to place semi quote on some variables.

tobimichigan · May 13, 2009

Jesus...put that in code tags...

And also, put or
die(mysql_error());
after every query.

What codes do you want me to put in tags. I implemented the die(mysql_error() after every query like you said but it gave a new error in sql saying:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'epe)' at line 1

The field value its refering to actually has to do with the LOCAL GOVERNMENT $lga variable.

Here's my modified Register_Action.php script:

<?php

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

$amountd=$_POST['amountd'];

$dpt=$_POST['department'];

$email=$_POST['email'];

$fname =$_POST['fname'];

$id=$POST['id'];

$ledgerno =$_POST['ledgerno'];

$lga =$_POST['lga'];

$lname=$_POST['lname'];

$Nationalty=$_POST['Nationalty'];

$oname=$_POST['oname'];

$pfno=$_POST['pfno'];

$residentialaddress=$_POST['pfno'];

$soorigin=$_POST['soorigin'];

$today= date("Ymd H:i:s");

$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)") or die(mysql_error());

if (!mysql_query($result))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($con)

?>

<html>

<body>

</body>

</html>

What am I doing wrong this time?

tobimichigan · May 13, 2009

#Your Code:

$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");

#Check every variables in your values you forgot to place semi quote on some variables.

There's been a little improvement in the code. Now its giving a new type_error saying:

"Column count doesn't match value count at row 1" what does this mean?

jackpf · May 13, 2009

It means that you've got

INSERT INTO (columnames...) VALUES (values...)

but you've got more column names than you have values.

tobimichigan · May 13, 2009

It means that you've got
INSERT INTO (columnames...) VALUES (values...)
but you've got more column names than you have values.

Please could you be more specific? Here's my sql.table:

CREATE TABLE `User_table` (

`id` bigint(20) NOT NULL auto_increment,

`pfno` varchar(255) NOT NULL,

`ledgerno`varchar(255) NOT NULL,

`fname` text NOT NULL,

`oname` text NOT NULL,

`lname` text NOT NULL,

`soorigin` text NOT NULL,

`lga` text NOT NULL,

`Nationalty` text NOT NULL,

`email` text NOT NULL,

`residentialadd` text NOT NULL,

`department` text NOT NULL,

`amountd` text NOT NULL,

PRIMARY KEY (`id`)

) TYPE=MyISAM COMMENT='cUser - User Table' AUTO_INCREMENT=1 ;

and her's my php $variables:

$amountd=$_POST['amountd'];

$dpt=$_POST['department'];

$email=$_POST['email'];

$fname =$_POST['fname'];

$id=$POST['id'];

$ledgerno =$_POST['ledgerno'];

$lga =$_POST['lga'];

$lname=$_POST['lname'];

$Nationalty=$_POST['Nationalty'];

$oname=$_POST['oname'];

$pfno=$_POST['pfno'];

$residentialaddress=$_POST['pfno'];

$soorigin=$_POST['soorigin'];

What could be responsible for

"Column count doesn't match value count at row 1"?

jackpf · May 13, 2009

Count your column names on the left side of "VALUES", and then count the values on the right. You'll find they don't match.

Idk what doesn't match, you'll just have to look. You know your db/what it's supposed to be doing better than me.

tobimichigan · May 14, 2009

Count your column names on the left side of "VALUES", and then count the values on the right. You'll find they don't match.

Idk what doesn't match, you'll just have to look. You know your db/what it's supposed to be doing better than me.

Yea, you're right, the columns did not match. I've corrected it.

But now its saying

"Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1"

Here's my updated data:

<tr>

<p><font size='2' face='verdana'>WELCOME TO LASUSTAFFCAMS Registration FORM</font></p>

</div></td>

</tr>

<tr>

</tr>

<tr>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='28'><font size='2' face='verdana'>LEDGER NO.</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>FIRST NAME</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>OTHER NAMES</font><font size='2' face='verdana'></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>STATE OF ORIGIN</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>LOCAL GOVERNMENT</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>NATIONALITY</font></td>

</font></td>

</tr>

<tr>

<td height='26'><font size='2' face='verdana'>Email address</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>RESIDENTIAL ADDRESS</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>TELEPHONE NUMBER</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>DEPARTMENT</font></td>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>MARITAL STATUS</font></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

<td height='25'><font size='2' face='verdana'>AMOUNT DEDUCTIBLE</font></td>

</font></td>

</tr>

<tr>

</font></td>

</tr>

<tr>

</tr>

<tr>

</tr>

</table>

</form>

Here's now Register_Action.php,

<?php

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

$amountd=$_POST['amountd'];

$dpt=$_POST['department'];

$email=$_POST['email'];

$fname =$_POST['fname'];

$ledgerno =$_POST['ledgerno'];

$lga=$_POST['lga'];

$lname=$_POST['lname'];

$nationalty=$_POST['Nationalty'];

$oname=$_POST['oname'];

$pfno=$_POST['pfno'];

$residentialadd=$_POST['residentialadd'];

$sex=$_POST['sex'];

$soorigin=$_POST['soorigin'];

//$today= date("Ymd H:i:s");

$result = mysql_query("Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin')") or die(mysql_error());

if (!mysql_query($result))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($link)

?>

The error its giving now seems to be in along the insert query I guess its embeded in the php sql script,

""Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1"

Where could this be coming from?

jackpf · May 14, 2009

Try running addslashes on your post data.

tobimichigan · May 14, 2009

Try running addslashes on your post data.

Here's what I did,

For Register_Action.php

<?php

include("input_cl.php");

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

$amountd=$_POST['amountd'];

$dpt=$_POST['department'];

$email=$_POST['email'];

$fname =$_POST['fname'];

$ledgerno =$_POST['ledgerno'];

$lga=$_POST['lga'];

$lname=$_POST['lname'];

$nationalty=$_POST['Nationalty'];

$oname=$_POST['oname'];

$pfno=$_POST['pfno'];

$residentialadd=$_POST['residentialadd'];

$sex=$_POST['sex'];

$soorigin=$_POST['soorigin'];

//$today= date("Ymd H:i:s");

$result = mysql_query("Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin')") or die(mysql_error());

if (!mysql_query($result))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($link)

For input_cl.php (where the addslashes array was set:

<?php

//create array to temporarily grab variables

$input_arr = array();

//grabs the $_POST variables and adds slashes

foreach ($_POST as $key => $input_arr) {

$_POST[$key] = addslashes($input_arr);

}

?>

But that ugly error

"Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1" still comes up.

What's wrong now?

tobimichigan · May 14, 2009

I commented the line referring to the input_cl.php

Here's now the code:

<?php

//include("input_cl.php");

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

addslashes($amountd)=$_POST['amountd'];

addslashes($dpt)=$_POST['department'];

addslashes($email)=$_POST['email'];

addslashes($fname) =$_POST['fname'];

addslashes($ledgerno) =$_POST['ledgerno'];

addslashes($lga)=$_POST['lga'];

addslashes($lname)=$_POST['lname'];

addslashes($nationalty)=$_POST['Nationalty'];

addslashes($oname)=$_POST['oname'];

addslashes($pfno)=$_POST['pfno'];

addslashes($residentialadd)=$_POST['residentialadd'];

addslashes($sex)=$_POST['sex'];

addslashes($soorigin)=$_POST['soorigin'];

//$today= date("Ymd H:i:s");

$result = mysql_query("Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin')") or die(mysql_error());

if (!mysql_query($result))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($link)

?>

However it returned this error:

Fatal error: Can't use function return value in write context in C:\wamp\www\staffCams\Register_Action.php on line 8

Please I need a sharp pointer to resolve this problem...

jackpf · May 14, 2009

You're running addslashes on the variable. You're supposed to run it on the data.

tobimichigan · May 14, 2009

You're running addslashes on the variable. You're supposed to run it on the data.

..and Sir, how may I do that?

nadeemshafi9 · May 14, 2009

     $result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");
       //$result_p_ticket= mysql_query("Insert into pawn_ticket(pawn_ticket_customer_number,pawn_ticket_ticket_number,pawn_ticket_ticket_check_digit,pawn_ticket_ticket_issue_date) values('$cust_no','$pawn_ticket_number','$check_digit_only','$today')");
if (!mysql_query($result))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

to

     
$sql = "Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)";

$result = mysql_query();
       //$result_p_ticket= mysql_query("Insert into pawn_ticket(pawn_ticket_customer_number,pawn_ticket_ticket_number,pawn_ticket_ticket_check_digit,pawn_ticket_ticket_issue_date) values('$cust_no','$pawn_ticket_number','$check_digit_only','$today')");
if (!mysql_query($result))
  {
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo "1 record added";

see what your query looks like

Jibberish · May 14, 2009

With the add slashes function i think you want it to be formatted like this

$amountd = addslashes($_POST['amountd']);

addslashes is a function that takes a string as an argument and returns it back with backslashes before characters that need to be quoted in database queries etc. (shamelessly stolen form php.net)

So you can't make it = a variable, but you can make a variable = to what it returns. Which is what I imagen you want to do.

Oh and on another not they seem to recommend using DBMS specific escape function ie. mysqli_real_eascape_string().

you can read more on : http://www.php.net/addslashes

nadeemshafi9 · May 14, 2009

AAHAHAHHAHAH

you are trying to execture a result a result of mysqlquery into mysqlquery

WRONG

<?php
$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");
       //$result_p_ticket= mysql_query("Insert into pawn_ticket(pawn_ticket_customer_number,pawn_ticket_ticket_number,pawn_ticket_ticket_check_digit,pawn_ticket_ticket_issue_date) values('$cust_no','$pawn_ticket_number','$check_digit_only','$today')");
if (!mysql_query($result))
?>

when in teh above you say if (!mysql_query($result)) - $result is an array derived from teh previouse $result = mysql_query("Insert into uaer_table(amountd,departmen ..... basicaly you already executed it and put teh result into result then you try to execute it again with the variable result which is not an sql statement but is infact a result array from the first mysql query result.

CORRECT

<?php
$sql= "Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");

if (!mysql_query($sql))
{
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo "1 record added";
?>

tobimichigan · May 14, 2009

AAHAHAHHAHAH

you are trying to execture a result a result of mysqlquery into mysqlquery

WRONG
<?php
$result = mysql_query("Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");
       //$result_p_ticket= mysql_query("Insert into pawn_ticket(pawn_ticket_customer_number,pawn_ticket_ticket_number,pawn_ticket_ticket_check_digit,pawn_ticket_ticket_issue_date) values('$cust_no','$pawn_ticket_number','$check_digit_only','$today')");
if (!mysql_query($result))
?>
when in teh above you say if (!mysql_query($result)) - $result is an array derived from teh previouse $result = mysql_query("Insert into uaer_table(amountd,departmen ..... basicaly you already executed it and put teh result into result then you try to execute it again with the variable result which is not an sql statement but is infact a result array from the first mysql query result.

CORRECT
<?php
$sql= "Insert into uaer_table(amountd,department,email,fname,ledgerno,lga,lname,nationalty,oname,pfno,residentialadd,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$nationalty','$oname',$pfno,$residentialadd,$soorigin)");

if (!mysql_query($sql))
{
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo "1 record added";
?>

Well code guru, I did as you said:

<?php

//include("input_cl.php");

$link = mysql_connect("localhost", "root", "") or die("Could not connect");

$db = mysql_select_db("lasustaffcams", $link) or die("Could not select database");

$amountd = addslashes($_POST['amountd']);

$dpt=addslashes($_POST['department']);

$email=addslashes($_POST['email']);

$fname =addslashes($_POST['fname']);

$ledgerno =addslashes($_POST['ledgerno']);

$lga=addslashes($_POST['lga']);

$lname=addslashes($_POST['lname']);

$nationalty=addslashes($_POST['Nationalty']);

$oname=addslashes($_POST['oname']);

$pfno=addslashes($_POST['pfno']);

$residentialadd=addslashes($_POST['residentialadd']);

$sex=addslashes($_POST['sex']);

$soorigin=addslashes($_POST['soorigin']);

//$today= date("Ymd H:i:s");

$sql = mysql_query("Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin')") or die(mysql_error());

if (!mysql_query($sql))

{

die('Error: ' . mysql_error()." SQL: ".$sql);

}

echo "1 record added";

?>

But its giving

"Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 SQL: 1" any further suggestions?

nadeemshafi9 · May 14, 2009

still wrong you are inserting an array into your mysqlquer in if

the if actualy executes teh function

i said do this

$sql = "Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin')";

if (!mysql_query($sql))

{

die('Error: ' . mysql_error()." SQL: ".$sql);

}

stop being a homo and listen

jackpf · May 14, 2009

Oh yeah I didn't notice that

tobimichigan · May 14, 2009

Oh yeah I didn't notice that

Well thank God I'm not alone....guess I didn't see that too.

Nad, I'm so sorry I didn't mean to hurt your feelings. One more thing Nad, since I've now got a working form, how can I introduce a file upload (picture precisely) into these variables and also insert them into the same table? I mean these variables:

<?php
$amountd = addslashes($_POST['amountd']);
$department=addslashes($_POST['department']);
$email=addslashes($_POST['email']);
$fname =addslashes($_POST['fname']);
$ledgerno =addslashes($_POST['ledgerno']);
$lga=addslashes($_POST['lga']);
$lname=addslashes($_POST['lname']);
$nationalty=addslashes($_POST['Nationalty']);
$oname=addslashes($_POST['oname']);
$pfno=addslashes($_POST['pfno']);
$residentialadd=addslashes($_POST['residentialadd']);
$sex=addslashes($_POST['sex']);
$soorigin=addslashes($_POST['soorigin']);
$regdate= date("Ymd H:i:s");			


$sql= "Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin,regdate) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin','$regdate')";  
	                            
if (!mysql_query($sql))
{
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo ("1 record added");
?>

Thanks for your help guys...you folks are great

nadeemshafi9 · May 14, 2009

use a file feild post it lets you pick a file and then it gets posted to teh server , that feild then is not liek normal feild but more complex

jackpf · May 14, 2009

You will need to make use of <input type="file" /> and the $_FILES array. There are undobtedly some good tutorials on google.

tobimichigan · May 15, 2009

use a file feild post it lets you pick a file and then it gets posted to teh server , that feild then is not liek normal feild but more complex

<?php
$amountd = addslashes($_POST['amountd']);
$department=addslashes($_POST['department']);
$email=addslashes($_POST['email']);
$fname =addslashes($_POST['fname']);
$ledgerno =addslashes($_POST['ledgerno']);
$lga=addslashes($_POST['lga']);
$lname=addslashes($_POST['lname']);
$nationalty=addslashes($_POST['Nationalty']);
$oname=addslashes($_POST['oname']);
$pfno=addslashes($_POST['pfno']);
$residentialadd=addslashes($_POST['residentialadd']);
$sex=addslashes($_POST['sex']);
$soorigin=addslashes($_POST['soorigin']);
$regdate= date("Ymd H:i:s");         
         

$sql= "Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin,regdate) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin','$regdate')";  
                                  
if (!mysql_query($sql))
{
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo ("1 record added");
?>

Nad, please kindly demonstrate what exactly you mean in code tags. So grateful...

nadeemshafi9 · May 15, 2009

<?php
$amountd = addslashes($_POST['amountd']);
$department=addslashes($_POST['department']);
$email=addslashes($_POST['email']);
$fname =addslashes($_POST['fname']);
$ledgerno =addslashes($_POST['ledgerno']);
$lga=addslashes($_POST['lga']);
$lname=addslashes($_POST['lname']);
$nationalty=addslashes($_POST['Nationalty']);
$oname=addslashes($_POST['oname']);
$pfno=addslashes($_POST['pfno']);
$residentialadd=addslashes($_POST['residentialadd']);
$sex=addslashes($_POST['sex']);
$soorigin=addslashes($_POST['soorigin']);
$regdate= date("Ymd H:i:s");         
         

$sql= "Insert into user_table(amountd,department,email,fname,ledgerno,lga,lname,Nationalty,oname,pfno,residentialadd,sex,soorigin,regdate) values('$amountd','$department','$email','$fname','$ledgerno','$lga','$lname','$nationalty','$oname','$pfno','$residentialadd','$sex','$soorigin','$regdate')";  
                                  
if (!mysql_query($sql))
{
  die('Error: ' . mysql_error()."  SQL: ".$sql);
  }
echo ("1 record added");
?>

inside yout form put this
<input type="file" name="thefile">

then upload your file like this and referenc eit in the sql to identify teh file with teh record

http://us3.php.net/features.file-upload

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