PHP and MySQL for Dummies

flintar · August 31, 2007, 8:25am

Have some problems with the code that they gave me. Can anyone please let me know what is going wrong?

Here is the code:

[php]

<?php /*Program: mysql_send.php *Desc: PHP program that sends an SQL query to the * MySQL server and displays the results. */ echo " SQL Query Sender "; if(ini_get("magic_quotes_gpc") == "1") { $_POST['query'] = stripslashes($_POST['query']); } $host=""; $user=""; $password=""; /* Section that executes query and displays the results */ if(!empty($_POST['form'])) { mysql_connect($host,$user,$password); mysql_select_db($_POST['database']); $result = mysql_query($_POST['query']); echo "Database Selected: {$_POST['database']}
Query: {$_POST['query']}

Results

"; if($result == false) { echo "

Error: ".mysql_error($cxn)."

"; } elseif(@mysql_num_rows($result) == 0) { echo "

Query completed. No results returned.

"; } else { /* Display results */ echo ""; for($i = 0;$i < mysql_num_fields($result);$i++) { echo ""; } echo ""; for ($i=0;$i < mysql_num_rows($result);$i++) { echo ""; $row = mysql_fetch_row($result); foreach($row as $value) { echo ""; } echo ""; } echo "

".mysql_field_name($result,$i). "
".$value."

"; } /* Display form with only buttons after results */ $query = str_replace("'","%&%",$_POST['query']); echo "

"; exit(); } /* Displays form for query input */ if (@$_POST['queryButton'] != "Edit Query") { $query = " "; } else { $query = str_replace("%&%","'",$_POST['query']); } ?>

Type in database name	>
Type in SQL query	<?php echo $query ?>

[/php]

Here is the error msg that I am getting:

Database Selected: Query: hi Results

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource (( took out something for Secirty)) on line 28

Error:

**Mod Edit - Added PHP tags for better readability.

Q1712 · August 31, 2007, 1:37pm

where does the $cxn come form?
it’s neither defined nor needed.

use mysql_error() insted of mysql_error($cxn)

flintar · August 31, 2007, 3:01pm

Now I am having the problem were the “Query was empty” I do not understand why it is not passing the Query back to the DB.

If someone can help please do.

Here is the error. Code is on first post. Min the $cxn.

Database Selected: test1_db
Query: Hello
Results

Error: Query was empty

Q1712 · August 31, 2007, 3:45pm

have u tryed a real query?

i don’t know in witch cases mysql returns “Query was empty”, but “Hello” isn’t a vaild MySQL-query at all.

flintar · September 1, 2007, 7:29am

Yes I did use a real Query and I get Error: Query was empty it is from the code…


 if($result == false)
  {
     echo "Error: ".mysql_error()."";
  }
  elseif(@mysql_num_rows($result) == 0)
  {
     echo "Query completed.
            No results returned.";
  }
  else
  {
</end code section>
From what I see the query is not being passed from the form to the db so that it can give a  answer.
Please help make sure that the code is passing eveything correctly.
Thank you!

Flintar

flintar · September 1, 2007, 6:13pm

Having some passing problems…hope someone can help!

The problem is when I enter in a Query in the textarea it does not pass it to the DB. It give a error of “Query was empty” even if I enter in a true Query for it. If anyone can help please do! I don’t understand the
$_POST[‘query’] and what it is userd for? or how it is passed…

Please help if you can!
NOOB Flintar

here is the code if you need it.

 
<?php 
/*Program: mysql_send.php 
*Desc: PHP program that sends an SQL query to the 
* MySQL server and displays the results. 
*/ 
echo " 
SQL Query Sender 
"; 
if(ini_get("magic_quotes_gpc") == "1") 
{ 
$_POST['query'] = stripslashes($_POST['query']); 
} 
$host=""; 
$user=""; 
$password=""; 
/* Section that executes query and displays the results */

if(!empty($_POST[‘form’]))

{

mysql_connect($host,$user,$password);

mysql_select_db($_POST[‘database’]);

$result = mysql_query($_POST[‘query’]);

echo "Database Selected: {$_POST[‘database’]}


Query: {$_POST[‘query’]}
Results
"; 
if($result == false) 
{ 
echo "Error: ".mysql_error().""; 
} 
elseif(@mysql_num_rows($result) == 0) 
{ 
echo "Query completed. 
No results returned."; 
} 
else 
{ 
/* Display results */ 
echo ""; 
for($i = 0;$i < mysql_num_fields($result);$i++) 
{ 
echo ""; 
} 
echo ""; 
for ($i=0;$i < mysql_num_rows($result);$i++) 
{ 
echo ""; 
$row = mysql_fetch_row($result); 
foreach($row as $value) 
{ 
echo ""; 
} 
echo ""; 
} 
echo "
".mysql_field_name($result,$i). 
" 
".$value."
"; 
} 
/* Display form with only buttons after results */ 
$query = str_replace("'","%&%",$_POST['query']); 
echo "

 
 
 
 
 
 
"; 
exit(); 
} 
/* Displays form for query input */

if (@$_POST[‘queryButton’] != “Edit Query”)

{

$query = " “;

}

else

{

$query = str_replace(”%&%","’",$_POST[‘query’]);

}

?>
 
 

 
Type in database name 
 > 
 

Type in SQL query 
<?php echo $query ?>

Q1712 · September 2, 2007, 1:24am

i still cant find any errors.

can u put error_reporting(E_ALL); at the beginning of ur script and tell us if there is a Notice diplayed?

Zyppora · September 7, 2007, 1:05pm

Or try echoing the value of the variable that is passed as a parameter to mysql_query()?