[SOLVED] Major posting problem... ???

kaimason1 · January 22, 2009

This is the second of my two php problems. I am creating a post page in php and I am getting the following error message :

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

Here is the script that gives me said error message:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE> Populate Database </TITLE>
  <META NAME="Generator" CONTENT="EditPlus">
  <META NAME="Author" CONTENT="">
  <META NAME="Keywords" CONTENT="">
  <META NAME="Description" CONTENT="">
</HEAD>
<BODY>
  <H1>Add Contact</H1>
  <FORM METHOD="GET" action="post.php">
   <P>First Name</P><input type="text" name="first" size=20 maxlength=50>
   <P>Last Name</P><input type="text" name="last" size=20 maxlength=50>
   <P>Cell Phone</P><input type="text" name="cell" size=20 maxlength=50>
   <P>Email</P><input type="text" name="email" size=20 maxlength=50>
   <INPUT type="hidden"
   <P></P><INPUT type="submit" value="ADD CONTACT">
  </FORM>
</BODY>
</HTML>
<?
//Retrieve Variables//
$first=@$_GET['first'];
$last=@$_GET['last'];
$cell=@$_GET['cell'];
$email=@$_GET['email'];
//Make sure all variables submitted//
if($first==""){
die("Enter First Name... please?");
}else if($last==""){
die("Enter Last Name... please?");
}else if($cell==""){
die("Enter Cell Phone... please?");
}else if($email==""){
die("Enter Email... please?");
}else{
//Connect to database//
$username="mekam2_kai";
$password="753159";
$database="mekam2_kai1";
mysql_connect(localhost,$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?");
@mysql_select_db($database)or die("The database you are trying to access is currently unavailable. Curious...");
//Query statement//
$query="INSERT INTO TEST1_contacts VALUES(,$first,$last,$cell,$email)";
mysql_query($query);
$query="SELECT * FROM `TEST1_contacts` WHERE `First_Name`=$first";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0){
die("Could not insert data");
}else if($num>1){
$query="SELECT * FROM TEST1_contacts WHERE Last_Name=$last";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0){
	die("Could not insert data");
}else if($num>1){
	$query="SELECT * FROM TEST1_contacts WHERE Phone_Number=$cell";
	$result=mysql_query($query);
	$num=mysql_num_rows($result);
	if($num==0){
		die("Could not insert data");
	}else if($num>1){
	$query="SELECT * FROM TEST1_contacts WHERE Email=$email";
	$result=mysql_query($query);
	$num=mysql_num_rows($result);
	if($num==0){
		die("Could not insert data");
		}
	}
}
}
}
$i=0;
while($i<$num);
$first=mysql_result($result,$i,"First_Name");
$last=mysql_result($result,$i,"Last_Name");
$cell=mysql_result($result,$i,"Phone_Number");
$email=mysql_result($result,$i,"Email");
echo "Inserted $first $last Cell: $cell Email: $email";
?>

haku · January 22, 2009

First, as I mentioned in your other thread, get rid of the @ marks.

You are using mysql_num_rows in a few spots in that script. Which one is throwing up the error?

kaimason1 · January 22, 2009

O.K.... my posting page is error message central now... help? ??? ??? ???

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'censored'@'localhost' (using password: NO) in ... on line 24

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in ... on line 24

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'censored'@'localhost' (using password: NO) in ... on line 25

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in ... on line 25

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'censored'@'localhost' (using password: NO) in ... on line 26

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in ... on line 26

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'censored'@'localhost' (using password: NO) in ... on line 27

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in ... on line 27

Enter First Name... please?

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE> Populate Database </TITLE>
  <META NAME="Generator" CONTENT="EditPlus">
  <META NAME="Author" CONTENT="">
  <META NAME="Keywords" CONTENT="">
  <META NAME="Description" CONTENT="">
</HEAD>
<BODY>
  <H1>Add Contact</H1>
  <FORM METHOD="GET" action="post.php">
   <P>First Name</P><input type="text" name="first" size=20 maxlength=50>
   <P>Last Name</P><input type="text" name="last" size=20 maxlength=50>
   <P>Cell Phone</P><input type="text" name="cell" size=20 maxlength=50>
   <P>Email</P><input type="text" name="email" size=20 maxlength=50>
   <INPUT type="hidden"
   <P></P><INPUT type="submit" value="ADD CONTACT">
  </FORM>
</BODY>
</HTML>
<?
//Retrieve Variables//
$first=trim(mysql_real_escape_string($_GET['first']));
$last=trim(mysql_real_escape_string($_GET['last']));
$cell=trim(mysql_real_escape_string($_GET['cell']));
$email=trim(mysql_real_escape_string($_GET['email']));
//Make sure all variables submitted//
if($first==""){
die("Enter First Name... please?");
}else if($last==""){
die("Enter Last Name... please?");
}else if($cell==""){
die("Enter Cell Phone... please?");
}else if($email==""){
die("Enter Email... please?");
}else{
//Connect to database//
$username="";
$password="";
$database="";
mysql_connect(localhost,$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?");
mysql_select_db($database)or die("The database you are trying to access is currently unavailable. Curious...");
//Query statement//
$query="INSERT INTO TEST1_contacts VALUES(,$first,$last,$cell,$email)";
mysql_query($query);
$query="SELECT * FROM `TEST1_contacts` WHERE `First_Name`=$first";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0){
die("Could not insert data");
}else if($num>1){
$query="SELECT * FROM TEST1_contacts WHERE Last_Name=$last";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0){
	die("Could not insert data");
}else if($num>1){
	$query="SELECT * FROM TEST1_contacts WHERE Phone_Number=$cell";
	$result=mysql_query($query);
	$num=mysql_num_rows($result);
	if($num==0){
		die("Could not insert data");
	}else if($num>1){
	$query="SELECT * FROM TEST1_contacts WHERE Email=$email";
	$result=mysql_query($query);
	$num=mysql_num_rows($result);
	if($num==0){
		die("Could not insert data");
		}
	}
}
}
}
$i=0;
while($i<$num);
$first=mysql_result($result,$i,"First_Name");
$last=mysql_result($result,$i,"Last_Name");
$cell=mysql_result($result,$i,"Phone_Number");
$email=mysql_result($result,$i,"Email");
echo "Inserted $first $last Cell: $cell Email: $email";
?>

mike12255 · January 22, 2009

i dont think your connected to a database?

try setting those variables after your connected

haku · January 22, 2009

That is correct. mysql_real_escape_string() takes a database connection as a variable (it takes the default connection if this is left blank), and checks the collocation of the database. If you haven't connected, it cannot work.

kaimason1 · January 22, 2009

Back to the basics: I'm getting this simpler error again.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ... on line 49

Could not insert data

haku · January 22, 2009

which one is line 49?

kaimason1 · January 22, 2009

lines 48, 49, and 50 are as follows:

(this is the first set of the repitition near the end)

$result=mysql_query($query);
$num=mysql_num_rows($result);
if($num==0){

haku · January 22, 2009

Probably a query error. Try this:

$result=mysql_query($query);
$num=mysql_num_rows($result) OR die(mysql_error());
if($num==0){

kaimason1 · January 22, 2009

the revised error is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ... on line 49

Unknown column '(what i enter for First Name)' in 'where clause'

haku · January 22, 2009

$query="SELECT * FROM `TEST1_contacts` WHERE `First_Name`=$first";

Remove the single quotes from around First_Name.

kaimason1 · January 22, 2009

did that... problems still there...

kaimason1 · January 23, 2009

RRRRGGHHHH.... I'm getting zero... nothing... here's my script...

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE> Populate Database </TITLE>
  <META NAME="Generator" CONTENT="EditPlus">
  <META NAME="Author" CONTENT="">
  <META NAME="Keywords" CONTENT="">
  <META NAME="Description" CONTENT="">
</HEAD>
<BODY>
  <H1>Add Contact</H1>
  <FORM METHOD="GET" action="post.php">
   <P>First Name</P><input type="text" name="first" size=20 maxlength=50>
   <P>Last Name</P><input type="text" name="last" size=20 maxlength=50>
   <P>Cell Phone</P><input type="text" name="cell" size=20 maxlength=50>
   <P>Email</P><input type="text" name="email" size=20 maxlength=50>
   <INPUT type="hidden"
   <P></P><INPUT type="submit" value="ADD CONTACT">
  </FORM>
</BODY>
</HTML>
<?
//Connect to database//
$username="mekam2_kai";
$password="753159";
$database="mekam2_kai1";
mysql_connect(localhost,$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?");
mysql_select_db($database)or die("The database you are trying to access is currently unavailable. Curious...");
//Retrieve Variables//
$first=trim(mysql_real_escape_string($_GET['first']));
$last=trim(mysql_real_escape_string($_GET['last']));
$cell=trim(mysql_real_escape_string($_GET['cell']));
$email=trim(mysql_real_escape_string($_GET['email']));
//Make sure all variables submitted//
if($first == "") {
die("Enter First Name... please?");
} else if($last == "") {
die("Enter Last Name... please?");
} else if($cell == "") {
die("Enter Cell Phone... please?");
} else if($email == "") {
die("Enter Email... please?");
}else{
//Query statement//
$query="INSERT INTO TEST1_contacts VALUES(,$first,$last,$cell,$email)";
mysql_query($query);
$query="SELECT * FROM TEST1_contacts WHERE First_Name LIKE '$first'";
$result=mysql_query($query);
$num=mysql_num_rows($result) OR die(mysql_error());
if($num==0) {
die("Could not insert data");
} else if($num>1) {
$query="SELECT * FROM TEST1_contacts WHERE Last_Name LIKE '$last'";
$result=mysql_query($query);
$num=mysql_num_rows($result) OR die(mysql_error());
if($num==0) {
	die("Could not insert data");
} else if($num>1) {
	$query="SELECT * FROM TEST1_contacts WHERE Phone_Number LIKE '$cell'";
	$result=mysql_query($query);
	$num=mysql_num_rows($result) OR die(mysql_error());
	if($num==0) {
		die("Could not insert data");
	} else if($num>1) {
		$query="SELECT * FROM TEST1_contacts WHERE Email LIKE '$email'";
		$result=mysql_query($query);
		$num=mysql_num_rows($result) OR die(mysql_error());
		if($num==0) {
			die("Could not insert data");
		}
	}
}
}
}
$i=0;
while($i<$num){
$firstname=mysql_result($result,$i,"First_Name");
$lastname=mysql_result($result,$i,"Last_Name");
$cellphone=mysql_result($result,$i,"Phone_Number");
$emailaddress=mysql_result($result,$i,"Email");
echo "Inserted $firstname $lastname Cell: $cellphone Email: $emailaddress";
++$i;
}
?>

MadTechie · January 23, 2009

try this

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE> Populate Database </TITLE>
  <META NAME="Generator" CONTENT="EditPlus">
  <META NAME="Author" CONTENT="">
  <META NAME="Keywords" CONTENT="">
  <META NAME="Description" CONTENT="">
</HEAD>
<BODY>
  <H1>Add Contact</H1>
  <FORM METHOD="GET" action="post.php">
   <P>First Name</P><input type="text" name="first" size=20 maxlength=50>
   <P>Last Name</P><input type="text" name="last" size=20 maxlength=50>
   <P>Cell Phone</P><input type="text" name="cell" size=20 maxlength=50>
   <P>Email</P><input type="text" name="email" size=20 maxlength=50>
   <INPUT type="hidden"
   <P></P><INPUT type="submit" value="ADD CONTACT">
  </FORM>
</BODY>
</HTML>
<?
//Connect to database//
$username="mekam2_kai";
$password="753159";
$database="mekam2_kai1";
mysql_connect(localhost,$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?");
mysql_select_db($database)or die("The database you are trying to access is currently unavailable. Curious...");
//Retrieve Variables//
$first=trim(mysql_real_escape_string($_GET['first']));
$last=trim(mysql_real_escape_string($_GET['last']));
$cell=trim(mysql_real_escape_string($_GET['cell']));
$email=trim(mysql_real_escape_string($_GET['email']));
//Make sure all variables submitted//
if($first == "") {
   die("Enter First Name... please?");
} else if($last == "") {
   die("Enter Last Name... please?");
} else if($cell == "") {
   die("Enter Cell Phone... please?");
} else if($email == "") {
   die("Enter Email... please?");
}else{
//Query statement//
$query="INSERT INTO TEST1_contacts VALUES('$first','$last','$cell','$email')";
mysql_query($query);
$query="SELECT * FROM TEST1_contacts WHERE First_Name = '$first'";
$result=mysql_query($query);
$num=mysql_num_rows($result) OR die(mysql_error());
if($num==0) {
   die("Could not insert data");
} else if($num>1) {
   $query="SELECT * FROM TEST1_contacts WHERE Last_Name = '$last'";
   $result=mysql_query($query);
   $num=mysql_num_rows($result) OR die(mysql_error());
   if($num==0) {
      die("Could not insert data");
   } else if($num>1) {
      $query="SELECT * FROM TEST1_contacts WHERE Phone_Number = '$cell'";
      $result=mysql_query($query);
      $num=mysql_num_rows($result) OR die(mysql_error());
      if($num==0) {
         die("Could not insert data");
      } else if($num>1) {
         $query="SELECT * FROM TEST1_contacts WHERE Email = '$email'";
         $result=mysql_query($query);
         $num=mysql_num_rows($result) OR die(mysql_error());
         if($num==0) {
            die("Could not insert data");
         }
      }
   }
}
}
$i=0;
while($i<$num){
$firstname=mysql_result($result,$i,"First_Name");
$lastname=mysql_result($result,$i,"Last_Name");
$cellphone=mysql_result($result,$i,"Phone_Number");
$emailaddress=mysql_result($result,$i,"Email");
echo "Inserted $firstname $lastname Cell: $cellphone Email: $emailaddress";
++$i;
}
?>

kaimason1 · January 23, 2009

tried it, no result STILL

:-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\ :-\

MadTechie · January 23, 2009

okay

update this line

$query="INSERT INTO TEST1_contacts VALUES('$first','$last','$cell','$email')";

to

$query="INSERT INTO TEST1_contacts (`First_Name`, `Last_Name`, `Phone_Number`, `Email`)VALUES('$first','$last','$cell','$email')";

kaimason1 · January 23, 2009

tried it, didn't work. reverted to earlier version

gevans · January 23, 2009

Just a personal note, I think that's the most annoying footer I've ever seen. It's kind of put me off reading the thread. (I might have a browse though)

kaimason1 · January 23, 2009

sorry... couldn't think of anything else ::) : :-[

redarrow · January 23, 2009

can you post your database please for this project.....

im not going to guess your database format, i can tell that others are trying to guess the database

names ?

gevans · January 23, 2009

Well, I've got your code re-written a bit, if it works happy days, if not it might give you a decent error, try this;

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE> Populate Database </TITLE>
  <META NAME="Generator" CONTENT="EditPlus">
  <META NAME="Author" CONTENT="">
  <META NAME="Keywords" CONTENT="">
  <META NAME="Description" CONTENT="">
</HEAD>
<BODY>
  <H1>Add Contact</H1>
  <FORM METHOD="GET" action="post.php">
   <P>First Name</P><input type="text" name="first" size=20 maxlength=50>
   <P>Last Name</P><input type="text" name="last" size=20 maxlength=50>
   <P>Cell Phone</P><input type="text" name="cell" size=20 maxlength=50>
   <P>Email</P><input type="text" name="email" size=20 maxlength=50>
   <INPUT type="hidden"
   <P></P><INPUT type="submit" value="ADD CONTACT">
  </FORM>
</BODY>
</HTML>
<?php
//Retrieve Variables//
$first = $_GET['first'];
$last = $_GET['last'];
$cell = $_GET['cell'];
$email = $_GET['email'];

//Make sure all variables submitted//
if(empty($first)) {
die("Enter First Name... please?");
} else if(empty($last)) {
die("Enter Last Name... please?");
} else if(empty($cell)) {
die("Enter Cell Phone... please?");
} else if(empty($email)) {
die("Enter Email... please?");
} else {

//Connect to database//
$username = "mekam2_kai";
$password = "753159";
$database = "mekam2_kai1";
mysql_connect('localhost',$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?");
mysql_select_db($database) or die("The database you are trying to access is currently unavailable. Curious...");

//Query statement//
$query = "INSERT INTO TEST1_contacts VALUES(NULL,'$first','$last','$cell','$email')";
mysql_query($query) or die(mysql_error());
$query = "SELECT * FROM `TEST1_contacts` WHERE `First_Name`='$first'";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_num_rows($result);
if(!$num) {
	die("Could not insert data");
} else if($num>=1) {
	$query = "SELECT * FROM TEST1_contacts WHERE Last_Name='$last'";
	$result = mysql_query($query) or die(mysql_error());
	$num = mysql_num_rows($result);
	if(!$num) {
		die("Could not insert data");
	} else if($num>=1) {
		$query = "SELECT * FROM TEST1_contacts WHERE Phone_Number='$cell'";
		$result = mysql_query($query) or die(mysql_error());
		$num = mysql_num_rows($result);
		if(!$num) {
			die("Could not insert data");
		} else if($num>=1) {
			$query = "SELECT * FROM TEST1_contacts WHERE Email='$email'";
			$result = mysql_query($query) or die(mysql_error());
			$num = mysql_num_rows($result);
			if(!$num){
				die("Could not insert data");
			} else {
				for($i=0;$i<$num;$i++) {
					$first = mysql_result($result,$i,"First_Name");
					$last = mysql_result($result,$i,"Last_Name");
					$cell = mysql_result($result,$i,"Phone_Number");
					$email = mysql_result($result,$i,"Email");
					echo "Inserted $first $last Cell: $cell Email: $email";
				}
			}
		}
	}
}
}
?>

redarrow · January 23, 2009

if you post the database info then we can do a proper test otherwise where all guessing should off worked buy now.

kaimason1 · January 23, 2009

AND THAT SOLVES IT, THANK YOU GEVANS! AND THANK YOU EVERYONE WHO SPENT VALUABLE TIME HELPING ME!

redarrow · January 23, 2009

so it was the database connection then 'localhost'

and variable names all wrong.

gevans · January 23, 2009

There were a few problems, localhost not wrapped in quotes, suppresing the select database error (so could run the script without a database selected), variables not wrapped with quotes in queries, setting id to NULL (I'm guessing it's id), checking the count is greater than one rather that greater than or equal and a very strange syntax for a while loop at the end!

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[SOLVED] Major posting problem... ???

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