Can i use isset() to control execution of mysql query - php

I have created an editable database to help me automate weekly member updates. There are 9 values that each member updates each week, these are controlled by $_POST submit to secondary php.
From that php, the post values are set as php var, then used to UPDATE sql db.
mysql_select_db("web_footy1") or die(mysql_error());
// The SQL statement is built
$strSQL = "UPDATE Round_6 SET ";
$strSQL = $strSQL . "Game1= '$Game1', ";
$strSQL = $strSQL . "Game2= '$Game2', ";
$strSQL = $strSQL . "Game3= '$Game3', ";
$strSQL = $strSQL . "Game4= '$Game4', ";
$strSQL = $strSQL . "Game5= '$Game5', ";
$strSQL = $strSQL . "Game6= '$Game6', ";
$strSQL = $strSQL . "Game7= '$Game7', ";
$strSQL = $strSQL . "Game8= '$Game8', ";
$strSQL = $strSQL . "Game9= '$Game9' ";
$strSQL = $strSQL . "WHERE Member = '$Member' ";
// The SQL statement is executed
mysql_query($strSQL) or die(mysql_error()) ;
Yes, i am aware this is subject to SQL injection, it is a private site so security can wait atm
The problem is that all values are update at same time, and to update one, you need to re enter all otherwise they are replaced with empty value.
So my question in twofold.
A) What is the cleanest way to be able to control each variable separately,
B) Can i, and if so, how to use isset($GameX) to control which queries are executed.
Eg
IF (isset($Game1)) {UPDATE Round6 SET Game1='$Game1' WHERE Member='$Member'} ;
Please keep in mind, 3 weeks ago i knew nothing about coding, and have crash coursed in html, php and sql in that time... Cheers

I really really really can't recommend enough that you FIX your SQL injection.
Having said that, you can programatically add conditions to your UPDATE clause.
An example might be the following snippet:
<?php
$Game3 = "things";
$Game5 = "stuff";
$Game6 = "awesome";
$Member = 'ben';
$update_parts = array();
for ($game_counter = 1; $game_counter < 10; $game_counter++) {
$variable_name = "Game" . $game_counter;
if ( isset($$variable_name) ) { // This is like isset($Game1)
$update_parts[] = "Game" . $game_counter . " = '" . $$variable_name . "'";
}
}
if ( sizeof($update_parts) > 0 ) {
$strSQL = "UPDATE Round_6 SET ";
$strSQL .= implode(", ", $update_parts);
$strSQL .= " WHERE Member = '$Member'";
echo $strSQL;
}
I've put in a couple of variables up the top there. This yields the following SQL:
UPDATE Round_6 SET Game3 = 'things',
Game5 = 'stuff', Game6 = 'awesome' WHERE Member = 'ben'
EDIT: If you want to use PDO, you need the Query and the Parameters separated. In my example below, I'm putting the Query parameters in
$conn = new PDO("mysql:host=localhost;dbname=database;","username","password"); // Your Connection String
$update_parts = array();
$query_params = array();
for ($game_counter = 1; $game_counter < 10; $game_counter++) {
$variable_name = "Game" . $game_counter;
if ( isset($$variable_name) ) { // This is like isset($Game1)
$update_parts[] = "Game" . $game_counter . " = ?";
$query_params[] = $$variable_name;
}
}
if ( sizeof($update_parts) > 0 ) {
$strSQL = "UPDATE Round_6 SET ";
$strSQL .= implode(", ", $update_parts);
$strSQL .= " WHERE Member = ?";
$query_params[] = $Member;
// Here is where you'd run the update
$stmt = $conn->prepare($strSQL);
$stmt->execute($query_params); // Notice I'm passing in the parameters separately
}

Related

Setting up SQL queries with multiple parameters

I need to set up a SQL query with multiple parameters that are being pulled from the URL. So far I can only get it to work with the there is only one item in the URL.
My default query to pull in all the content
$sql = "SELECT ";
$sql .= "* ";
$sql .= "FROM ";
$sql .= "cms_site_content ";
$sql .= "WHERE ";
$sql .= "1";
I then check if anything was passed through the URL and retrieve it.
if (isset($_GET["d"])) {
$d=$_GET["d"];
Inside the if statement, I break the values passed as "d" into separate items
$newD = explode(',',$d);
$countD = count($newD);
foreach($newD as $discipline) {
if ($countD == 1) {
$sql .= " AND";
$sql .= " discipline='".$discipline."'";
}
My problem is getting the SQL to work if there is more than one discipline value. It should read something like this:
SELECT * FROM cms_site_content WHERE 1 AND discipline="value"
however if there's more than one discipline value, it should read:
SELECT * FROM cms_site_content WHERE 1 AND discipline="value OR discipline="value2" OR discipline="value3"
Is there a more efficient way to write this? I can't figure out how to insert the OR into the foreach statement.
Save all discipline values in an array;
$discipline_arr = array();
foreach($newD as $discipline) {
$discipline_arr[] = $discipline;
// by the way, don't forget to escape for sql injection
// mysql_escape_string is the depracated one, u can use that if u have no
// other choice
}
Then in your sql, add them as discipline in ('value1','value2', 'etc ...') condition (that is for strings, for numeric types it would be like discipline in (1,2,3,4, etc)
$sql = " SELECT * FROM cms_site_content WHERE 1 " .
(empty($discipline_arr) ? "" : "and
discipline in ('". implode("','" , $discipline_arr). "') ") ;
Link to escaping
http://tr1.php.net/manual/en/function.mysql-escape-string.php
Assuming the rest of your query is in tact. Simply store all of your discipline values in an array as follows, then feed the $discipline_string to your $sql query:
$discipline_ary = array('option1', 'option2', 'option3');
$discipline_string = "";
for($i=0; $i < count($discipline_ary); $i++){
$discipline_string .= " discipline = '" . $discipline[$i] . "' ";
if($i+1 == count($discipline_ary)){
break;
}else{
$discipline_string .= " OR "
}
}

PHP query does not return result

This query is not returning any result as there seems to be an issue with the sql.
$sql = "select region_description from $DB_Table where region_id='".$region_id."' and region_status =(1)";
$res = mysql_query($sql,$con) or die(mysql_error());
$result = "( ";
$row = mysql_fetch_array($res);
$result .= "\"" . $row["region_description"] . "\"";
while($row = mysql_fetch_array($res))
{
echo "<br /> In!";
$result .= " , \"" . $row["region_description"] . "\"";
}
$result .= " )";
mysql_close($con);
if ($result)
{
return $result;
}
else
{
return 0;
}
region_id is passed as 1.
I do have a record in the DB that fits the query criteria but no rows are returned when executed. I beleive the issue is in this part ,
region_id='".$region_id."'
so on using the gettype function in my php it turns out that the datatype of region_id is string not int and thus the failure of the query to function as my datatype in my tableis int. what would be the way to get parameter passed to be considered as an int in php. url below
GetRegions.php?region_id=1
Thanks
Try it like this:
$sql = "SELECT region_description FROM $DB_Table WHERE region_id = $region_id AND region_status = 1"
The region_id column seems to be an integer type, don't compare it by using single quotes.
Try dropping the ; at the end of your query.
First of all - your code is very messy. You mix variables inside string with escaping string, integers should be passed without '. Try with:
$sql = 'SELECT region_description FROM ' . $DB_Table . ' WHERE region_id = ' . $region_id . ' AND region_status = 1';
Also ; should be removed.
try this
$sql = "select region_description from $DB_Table where region_id=$region_id AND region_status = 1";
When you are comparing the field of type integer, you should not use single quote
Good Luck
Update 1
Use this.. It will work
$sql = "select region_description from " .$DB_Table. " where region_id=" .$region_id. " AND region_status = 1";
You do not need the single quotes around the region id i.e.
$sql = "SELECT region_description FROM $DB_Table WHERE region_id = $region_id AND region_status = 1"

PHP SQL database query error message

Is there anything wrong with this SQL code? I got it from a tutorial, but it's returning the following error message
Database query failed: 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 'LIMIT 1' at line 1
function get_subject_by_id($subject_id) {
global $connection;
$query = "SELECT * ";
$query .= "FROM subjects ";
$query .= "WHERE id=" . $subject_id ." ";
$query .= "LIMIT 1";
$result_set = mysql_query($query, $connection);
confirm_query($result_set);
// if no rows are returned, fetch array will return false
if ($subject = mysql_fetch_array($result_set)){
return $subject;
} else {
return NULL;
}
}
Best to echo the query and see what it looks like.
Probably $subject_id contains no value or an invalid value. If $subject_id is a string, you should escape it (using mysql_real_escape_string) and put it inside quotes in the query.
[Edit]
You know you can put enters in strings too, right?
// More readable
$query = "
SELECT *
FROM subjects
WHERE id = $subject_id
LIMIT 1";
$query .= "where id=" . $page_id . " ";
Needs to be put within single quotes. Replace the above statement by
$query .= "where id='" . $page_id . "' ";
Frankly, it's impossible to say what is exactly wrong with this code, not knowing what values are substituted in the query in place of variables.
Apart from that, the code in question may be subject to SQL injection attacks.
If I may put together other suggestions that will make sure no error is ever generated with this code:
function get_subject_by_id($subject_id) {
global $connection;
$query = "SELECT * ";
$query .= "FROM subjects ";
$query .= "WHERE id='" . mysql_real_escape_string($subject_id) ."' ";
// note the quotes and escaping wrapper
$query .= "LIMIT 1";
$result_set = mysql_query($query, $connection);
confirm_query($result_set);
// if no rows are returned, fetch array will return false
if ($subject = mysql_fetch_array($result_set)) {
return $subject;
} else {
return NULL;
}
}
Additionally, using global variables is a bad practice nowadays, so I suppose the example you're using is quite outdated.
Try to use mysql_real_escape_string()
function get_subject_by_id($subject_id) {
global $connection;
$query = "SELECT * ";
$query .= "FROM subjects ";
$query .= "WHERE id='" . $subject_id ."' "; //You need single quotes
$query .= "LIMIT 1";
$result_set = mysql_query($query, $connection);
confirm_query($result_set);
// REMEMBER:
// if no rows are returned, fetch_array will return false
if ($subject = mysql_fetch_array($result_set)) {
return $subject;
} else {
return NULL;
}
}
$query .= "WHERE id='" . $subject_id ."' "; //work
$query .= "WHERE id=" . $subject_id ." "; //not work

"Unknown column 'geir' in 'field list'"

I wrote this query:
$query = "UPDATE encodage_answer
SET Answer = geir
WHERE encodage_question_ID = 128
AND encodage_ID = 305
AND Extra = NULL";
$insert = mysql_query($query, $connection) or die(mysql_error());
But if I run this code I always get the same error:
Unknown column 'geir' in 'field list'
It's probably me but I think I am not saying geir is a column/field; what's the issue?
When I run this query directly in my PHPMyAdmin it works great.
Update: Full code:
The answer exists, $Extra variable is Null
$AnswerExists = answer_exists($Question_ID, $encodage_ID, $Extra);
if($AnswerExists <> ""){
if($Answer != NULL){
$correctAnswer = mysql_prep($Answer);
if($Extra != NULL){
$query = "UPDATE `encodage_answer` SET `Answer` = '" . mysql_prep($Answer) . "' WHERE `ID` = '" . $AnswerExists . "'";
$insert = mysql_query($query, $connection) or die(mysql_error());
$query2 = "UPDATE `encodage_answer` SET `Extra` = '" . $Extra . "' WHERE `ID` = '" . $AnswerExists . "'";
$insert = mysql_query($query2, $connection) or die(mysql_error());
}else{
$querytest = "UPDATE `encodage_answer` SET Answer = " . $Answer . " WHERE ID = " . $AnswerExists;
$insert = mysql_query($querytest, $connection) or die(mysql_error());
}
}
}
function answer_exists($Question_ID, $encodage_ID, $Extra){
global $connection;
$trfa = false;
echo $Question_ID . " - " . $encodage_ID . "<br />";
if($Extra <> ""){
$query = "SELECT *
FROM encodage_answer
WHERE encodage_ID = {$encodage_ID} AND encodage_question_ID = {$Question_ID} AND Extra = {$Extra}";
}else{
$query = "SELECT *
FROM encodage_answer
WHERE encodage_ID = {$encodage_ID} AND encodage_question_ID = {$Question_ID}";
}
Try putting single quotes around geir. By not quoting the string you want to set the column to, the SQL backend thinks you want to set the value of the Answer column to the value of the geir column. Since the geir column doesn't exist in your table, it throws an error.
Edit: I suspect that PHPMyAdmin has some kind of SQL statement filtering to catch cases like this, and automatically puts quotes around the string for you.
Thanks for the help to everyone! I'm changing all queries to a safer format! SQL-Injection treats are no longer an issue! Thanks for the tip!
Concerning my question:
I'am a complete idiot! After searching for a solution for 20 hours I found my error! The error was for another query. I'm very sorry for wasting your time but I'm a newbie (ergo, the sql-injection issue), so I hope I am allowed to make a few mistakes.
Thanks
Jens

Why is this SQL query not working?

this script have to update things on every refresh but not working. lend me a hand
$yp = mysql_query("select id from yyy where twitterid = '$tid'");
$qq = "update yyy set twitterid = '$tid',
twitterkullanici = '$twk',
tweetsayisi = '$tws',
takipettigi = '$tkpettigi',
takipeden = '$tkpeden',
nerden = '$nerden',
bio = '" . mysql_real_escape_string($bio) . "',
profilresmi ='$img',
ismi = '$isim'
where id = '$yp'";
$xx = mysql_query($qq);
Looks like you are not getting the value out of the variable $yp.
You need to do
$row = mysql_fetch_row($yp);
then
id = '.$row[0] .'
in your update query
$yp - is a result of mysql_query (resource). You have to read id from database (mysql_fetch_array or mysql_fetch_row).
$yp = mysql_query("select id from yyy where twitterid = '$tid'");
if ($yp)
{
if ($row = mysql_fetch_array($yp,MYSQL_ASSOC))
$id = $row["id"];
}
Now use $id in WHERE clause.
To make debugging SQL easier in PHP add the following after to your mysql_query(0 call.
mysql_query($qq) or die("A MySQL error has occurred.<br />Your Query: " . $qq. "<br /> Error: (" . mysql_errno() . ") " . mysql_error())
Just make sure you remove it before you go into prod, as it can give useful info away to any hackers attempting Sql Injection.

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