[SOLVED] HELP! Getting "Query was empty" error!

DarkShadowWing · September 6, 2009

Hi all. I'm getting a "Query is empty" message on my index page.

Why, I do not know..

But here's the code:

index.php:

<html>
<head><title>Metal Detecting</title></head>
<?
<body onload="document.forms.test.s1.disabled = true;" oncontextmenu="return false;">
<style type="text/css">
a:hover{Color: #FFFFFF;text-decoration: none;}
a:link{Color: #FFFFFF;text-decoration: underline;}
a:visited{Color: #FF0000;text-decoration: underline;}
BODY{
Color: #FFFFFF;
}
.input {
padding-right: 1px; padding-left: 1px; padding-bottom: 1px; margin: 4px 0px 5px 8px; font: 11px/14px Arial, Helvetica, sans-serif; color: #5a698b; padding-top: 1px
}
#title {
font: bold 11px/18px Arial, Helvetica, sans-serif; text-transform: uppercase; width: 330px; color: #5a698b; padding-top: 10px; padding-bottom:10px; letter-spacing: 2px; height: 26px; text-align: center
}
#firstname{
margin 0.5em;
}
#lastname{
margin 0.5em;
}
#email{
margin 0.5em;
}
</style>
<table style="width: 1024; height: 50;" border="1" cellpadding="0" cellspacing="0"><td>
<a href="./itemsubmit.php">Submit Items</a><br/>
</td></table>
<script src="script01.js">
</script>
<?
$conn = "localhost";
$user = "myuser";
$pass = "mypass";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");

$site_con = mysql_query($sql_site) or die(mysql_error());
$numrows = mysql_num_rows($site_con);

if($numrows > 0){
while($row = mysql_fetch_array($site_con))
{
if(isset($warning)){
echo "<Center><H1>Signups</H1></Center><br>\n";
echo "Please contact us!<br /><br />\n";
echo "*Please note you can submit the form ONLY once. Any double form submissions will be deleted.<br />\n";
echo "*You will be emailed depending on your statement whether or not you've been approved to the group!<br />\n";
echo "*You MUST speak FLUENT english<br />\n";
echo "<center><font Color=\"#00FF00\">You have been warned!</font></center>\n";
echo "Please contact us!<br /><br />\n";
echo "*Please note you can submit the form ONLY once. Any double form submissions will be deleted.<br />\n";
echo "*You MUST speak FLUENT english<br />\n";
echo "<form name=\"test\" id=\"test\" method=\"POST\" onsubmit=\"return checkdata(this), emailCheck(this.email.value), checkCheckBox(this)\" action=\"send.php\">\n";
echo "<div id = \"div01\" style=\"width: 100; height: 25;\">";
echo "  Phone/Cell number: <input name=\"requirednumber\" id=\"number\" type=\"text\" />\n";
echo "  Firstname: <input name=\"requiredfirstname\" id=\"firstname\" type=\"text\" />\n";
echo "  Lastname: <input name=\"requiredlastname\" id=\"lastname\" type=\"text\" />\n";
echo "  Email: <input name=\"requiredemail\" id=\"email\" type=\"text\" /><br /><br />\n";
echo "</div>\n";
echo "<textarea readonly=\"readonly\" name=\"license\" cols=\"40\" rows=\"15\" id=\"license\">Blah</textarea><br/>\n";
echo "<input onclick=\"this.form.s1.disabled=! this.checked;\" name=\"agree\" id=\"agree\" type=\"checkbox\"> I have read & agree to the above<br/>\n";
echo "  <input name=\"s1\" id=\"s1\" value=\"Submit\" type=\"submit\" /> <input type=\"reset\" name=\"rset\" value=\"Reset\" /><br/>\n";
echo "</form>\n";
}else if( isset($banned)){
echo "<font Color=\"#FF0000\">Sorry, but you have been banned!</font>";
}else if(! isset($warning) || isset($banned)){
echo "Please contact us!<br /><br />\n";
echo "*Please note you can submit the form ONLY once. Any double form submissions will be deleted.<br />\n";
echo "*You MUST speak FLUENT english<br />\n";
echo "<form name=\"test\" id=\"test\" method=\"POST\" onsubmit=\"return checkdata(this), emailCheck(this.email.value), checkCheckBox(this)\" action=\"send.php\">\n";
echo "<div id = \"div01\" style=\"width: 100; height: 25;\">";
echo "  Phone/Cell number: <input name=\"requirednumber\" id=\"number\" type=\"text\" />\n";
echo "  Firstname: <input name=\"requiredfirstname\" id=\"firstname\" type=\"text\" />\n";
echo "  Lastname: <input name=\"requiredlastname\" id=\"lastname\" type=\"text\" />\n";
echo "  Email: <input name=\"requiredemail\" id=\"email\" type=\"text\" /><br /><br />\n";
echo "</div>\n";
echo "<textarea readonly=\"readonly\" name=\"license\" cols=\"40\" rows=\"15\" id=\"license\">Blah</textarea><br/>\n";
echo "<input onclick=\"this.form.s1.disabled=! this.checked;\" name=\"agree\" id=\"agree\" type=\"checkbox\"> I have read & agree to the above<br/>\n";
echo "  <input name=\"s1\" id=\"s1\" value=\"Submit\" type=\"submit\" /> <input type=\"reset\" name=\"rset\" value=\"Reset\" /><br/>\n";
echo "</form>\n";
}
}
}
?>
</body>
</html>

Somehow, I need to read the record for warning and banned, already in the database

and i need to display the original website which is this part:

}else if(! isset($warning) || isset($banned)){
echo "Please contact us!<br /><br />\n";
echo "*Please note you can submit the form ONLY once. Any double form submissions will be deleted.<br />\n";
echo "*You MUST speak FLUENT english<br />\n";
echo "<form name=\"test\" id=\"test\" method=\"POST\" onsubmit=\"return checkdata(this), emailCheck(this.email.value), checkCheckBox(this)\" action=\"send.php\">\n";
echo "<div id = \"div01\" style=\"width: 100; height: 25;\">";
echo "  Phone/Cell number: <input name=\"requirednumber\" id=\"number\" type=\"text\" />\n";
echo "  Firstname: <input name=\"requiredfirstname\" id=\"firstname\" type=\"text\" />\n";
echo "  Lastname: <input name=\"requiredlastname\" id=\"lastname\" type=\"text\" />\n";
echo "  Email: <input name=\"requiredemail\" id=\"email\" type=\"text\" /><br /><br />\n";
echo "</div>\n";
echo "<textarea readonly=\"readonly\" name=\"license\" cols=\"40\" rows=\"15\" id=\"license\">Blah</textarea><br/>\n";
echo "<input onclick=\"this.form.s1.disabled=! this.checked;\" name=\"agree\" id=\"agree\" type=\"checkbox\"> I have read & agree to the above<br/>\n";
echo "  <input name=\"s1\" id=\"s1\" value=\"Submit\" type=\"submit\" /> <input type=\"reset\" name=\"rset\" value=\"Reset\" /><br/>\n";
echo "</form>\n";
}

i need it to read the original website code whether or not warning AND / OR banned is 0 OR 1.

the part that needs to be fixed, is this:

<?
$conn = "localhost";
$user = "myuser";
$pass = "mypass";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");

$site_con = mysql_query($sql_site) or die(mysql_error());
$numrows = mysql_num_rows($site_con);

if($numrows > 0){
while($row = mysql_fetch_array($site_con))
{

as i said before though, this is the part that is causing "Query is empty" error.

ANY help is GREATLY appreciated!

Philip · September 6, 2009

Where is $sql_site defined?

DarkShadowWing · September 6, 2009

ok, now im getting:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource on line 55.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource on line 56.

code:

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($warning);
$numrows2 = mysql_num_rows($banned);

if($numrows1 > 0 || $numrows2 > 0){
while($row = mysql_fetch_array($site_con))
{

DarkShadowWing · September 6, 2009

I've tried everything i can think of. can someone please help me?

kratsg · September 6, 2009

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($warning);
$numrows2 = mysql_num_rows($banned);

$warning and $banned aren't queries to count the number of rows if you look at it closely.

Change it to this:

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($site1_con);
$numrows2 = mysql_num_rows($site2_con);

DarkShadowWing · September 7, 2009

Ok, now it sorta works. BUT i cant seem to return a "1" OR "0" for $warning and $banned.

code:

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$sql = ("SELECT $warning1 AND $banned1 FROM $tbl;");
$site_con = mysql_query($sql) or die(mysql_error());
$warning = $site_con
$banned = $site_con
$numrows = mysql_num_rows($site_con);

if($numrows > 0){
while($row = mysql_fetch_array($site_con))
{
if($warning > 0){

kratsg · September 7, 2009

$sql = ("SELECT $warning1 AND $banned1 FROM $tbl;");
$site_con = mysql_query($sql) or die(mysql_error());
$warning = $site_con
$banned = $site_con
$numrows = mysql_num_rows($site_con);

Is not the same as your original

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($site1_con);
$numrows2 = mysql_num_rows($site2_con);

What's going on here?

Also, you can't retrieve just "1" or "0" from queries.. unless it's an UPDATE, or DELETE, or INSERT (where it returns true on success or false on failure). I suggest having a look here:

http://us2.php.net/manual/en/function.mysql-query.php

DarkShadowWing · September 7, 2009

i need to be able to pull a 1 or 0 from the database. i dont care what kind of query it takes. i just need to be able to ban people or give people a warning if i find it necessary.

DarkShadowWing · September 7, 2009

i decided to change back to what i had before.

but NONE of these debug variables return 1 or 0.

code:

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($warning);
$numrows2 = mysql_num_rows($banned);

echo "warning:".$warning."<br>\n";
echo "banned:".$banned."<br>\n";
echo "site1_con:".$site1_con."<br>\n";
echo "site2_con:".$site2_con."<br>\n";
echo "numrows1:".$numrows1."<br>\n";
echo "numrows2:".$numrows2."<br>\n";
if($numrows1 > 0 || $numrows2 > 0){
while($numrows1 = mysql_fetch_array($site1_con) && $numrows2 = mysql_fetch_array($site2_con))
{
if(isset($warning)){

it returns:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\myxampp\htdocs\detect\index.php on line 55

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\myxampp\htdocs\detect\index.php on line 56
warning:SELECT `warning` FROM users0001;
banned:SELECT `banned` FROM users0001;
site1_con:Resource id #3
site2_con:Resource id #4
numrows1:
numrows2:

DarkShadowWing · September 7, 2009

and by "1" or "0" i mean a value pulled from the database. not an ACTUAL 1 or 0.

kratsg · September 7, 2009

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($warning);
$numrows2 = mysql_num_rows($banned);

$warning and $banned aren't queries to count the number of rows if you look at it closely.

Change it to this:

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($site1_con);
$numrows2 = mysql_num_rows($site2_con);

Again, fix by using the second code instead of the first before we talk about debugging >.<

mikesta707 · September 7, 2009

a bunch of things.

$numrows1 = mysql_num_rows($warning);
$numrows2 = mysql_num_rows($banned);

neither of those variables are mysql resources. If you notice, when you echo $site_con1 and 2, they say mysql resource. You have to use those in the numrows functions, like

$numrows1 = mysql_num_rows($site1_con);
$numrows2 = mysql_num_rows($site2_con);

thats whats causing these errors


Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:myxampphtdocsdetectindex.php on line 55

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:myxampphtdocsdetectindex.php on line 56

second, again, mysql_query only returns a mysql resource. You have to use other mysql functions to get the data out of those resources. If you want the information, you have to use mysql_fetch_array or mysql_fetch_assoc. you can also trim your queries down to 1. Try something like


$query = "SELECT $warning1, $banned1 FROM $tbl";

$result = mysql_query($query) or die(mysql_error());
$rows = mysql_num_rows($result);

echo "$rows rows returned!<br />";
//now lets go through every row returned.
while($row = mysql_fetch_assoc($result)){
$warning = $row[$warning1];
$banned = $row[$banned1];

echo "Warning: $warning<br />";
echo "Banned: $banned<br />";
}

try that and see what happens

DarkShadowWing · September 7, 2009

ok, ok.. i fixed it.

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";
$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$warning = ("SELECT $warning1 FROM $tbl;");
$banned = ("SELECT $banned1 FROM $tbl;");
$site1_con = mysql_query($warning) or die(mysql_error());
$site2_con = mysql_query($banned) or die(mysql_error());
$numrows1 = mysql_num_rows($site1_con);
$numrows2 = mysql_num_rows($site2_con);

echo "warning:".$warning."<br>\n";
echo "banned:".$banned."<br>\n";
echo "site1_con:".$site1_con."<br>\n";
echo "site2_con:".$site2_con."<br>\n";
echo "numrows1:".$numrows1."<br>\n";
echo "numrows2:".$numrows2."<br>\n";
if($numrows1 > 0 || $numrows2 > 0){
while($numrows1 = mysql_fetch_array($site1_con) && $numrows2 = mysql_fetch_array($site2_con))
{
if(isset($warning)){

kratsg · September 7, 2009

second, again, mysql_query only returns a mysql resource. You have to use other mysql functions to get the data out of those resources. If you want the information, you have to use mysql_fetch_array or mysql_fetch_assoc. you can also trim your queries down to 1. Try something like

This is completely untrue. http://us.php.net/manual/en/function.mysql-query.php

From php.net

Return Values

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.

Use mysql_num_rows() to find out how many rows were returned for a SELECT statement or mysql_affected_rows() to find out how many rows were affected by a DELETE, INSERT, REPLACE, or UPDATE statement.

mysql_query() will also fail and return FALSE if the user does not have permission to access the table(s) referenced by the query.

1 and 0 are true/false.

DarkShadowWing · September 7, 2009

0 rows returned!

code:

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";

$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$query = "SELECT $warning1, $banned1 FROM $tbl";

$result = mysql_query($query) or die(mysql_error());
$rows = mysql_num_rows($result);

$query = "SELECT $warning1, $banned1 FROM $tbl";

echo "$rows rows returned!<br />";
//now lets go through every row returned.
while($row = mysql_fetch_assoc($result)){
$warning = $row[$warning1];
$banned = $row[$banned1];

echo "Warning: $warning<br />";
echo "Banned: $banned<br />";

if(isset($warning)){

mikesta707 · September 7, 2009

...

Return Values

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data.

mysql query will return 0 when it fails any time (because of an error, an invalid connection, etc.) . it returns 1 when it preforms inserts, deletes, etc. (as specified on the quote) successfully.

However, in this case, he wants a 0 or 1 from his database, not from the query itself. the 0 or 1 from the query are only useful during debugging, as it tells whether or not the query you did was successful or not. However, as far as the data from the actual table the query is accessing goes, the returned value is rather useless unless you use one of the functions for use with the result tables.

But yes, I suppose I should have explained myself better.

When the query is SUCCESSFUL, mysql_query only returns a mysql_resource

DarkShadowWing · September 7, 2009

I know i accidently put query twice. sue me. im tired. lol

DarkShadowWing · September 7, 2009

so what should i change the code to?

it returned 0 rows..

mikesta707 · September 7, 2009

0 rows returned!

code:

<?
$conn = "localhost";
$user = "root";
$pass = "Gen0sis";
$dbname = "metaldetect01";
$tbl = "users0001";
$warning1 = "`warning`";
$banned1 = "`banned`";

$con = mysql_connect($conn,$user,$pass);
if (!$con)
  {
  die('Could not connect to database: "' . $dbname . '" because ' . mysql_error());
  }

mysql_select_db($dbname, $con);

$query = "SELECT $warning1, $banned1 FROM $tbl";

$result = mysql_query($query) or die(mysql_error());
$rows = mysql_num_rows($result);

$query = "SELECT $warning1, $banned1 FROM $tbl";

echo "$rows rows returned!<br />";
//now lets go through every row returned.
while($row = mysql_fetch_assoc($result)){
$warning = $row[$warning1];
$banned = $row[$banned1];

echo "Warning: $warning<br />";
echo "Banned: $banned<br />";

if(isset($warning)){

try

$query = "SELECT * FROM $tbl";

if it still returns 0 rows, then your table is empty, or you are using the wrong table name

DarkShadowWing · September 7, 2009

it still returns 0..

but i have the database filled..

Field Type Collation Attributes Null Default Extra Action 
  id int(11)   No None auto_increment               
  number varchar(16) latin1_swedish_ci  No                 
  Firstname varchar(32) latin1_swedish_ci  No                 
  Lastname varchar(32) latin1_swedish_ci  No                 
  email varchar(50) latin1_swedish_ci  No                 
  statement varchar(255) latin1_swedish_ci  No                 
  warning varchar(1) latin1_swedish_ci  No 0                
  banned varchar(1) latin1_swedish_ci  No 0

mikesta707 · September 7, 2009

are you sure you are using the right table name and connected to the right database then?

DarkShadowWing · September 7, 2009

yes and yes. i use a variable for both..

$tbl and $dbname

DarkShadowWing · September 7, 2009

so u have no other ideas? what do u wanna see? my database file? my php file? what?

kratsg · September 7, 2009

do the following:

var_dump($query) after each $query that you define and copy/paste those results into phpMyAdmin to see what happens.

DarkShadowWing · September 7, 2009

var_dump($query) returned:

string(37) "SELECT `warning`$ FROM users0001"

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[SOLVED] HELP! Getting "Query was empty" error!

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