I have dropdown menus wich contains name of players and opposition team. When user picks team for opposition to play against he clicks the submit button and isset function is triggered where the data from the dropdown menus will be captured and uploaded to the database.
It is simple enough but I've got what I guess is a syntax error. I get the message in the title when page is opened. I've been trying to fix this with no luck. If someone can point me in the right direction it would be greatly appreciated.
if ( isset($_POST['submit']) ) {
$player_ids = array_map('intval', $_REQUEST['players']);
$opponents_id = $_REQUEST['players'];
var_dump($opponents_id);
$query = 'SELECT `name`, `position`
FROM `player_info`
WHERE `player_id` IN (' . implode(',', $player_ids) . ')';
$return_names = mysql_query($query) or die(mysql_error());
while ( $row = mysql_fetch_assoc($return_names) )
{
$selected[] = $row['name'];
$position[] = $row['position'];
}
$query = ("SELECT `fixture_id`
FROM `fixtures`
WHERE `fixture_id` = $opponents_id");
$result = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($query))
{
$fixture_id[] = $row['fixture_id'];
}
$query = ("SELECT `fixture_id`
^ See Here
FROM `fixtures`
WHERE `fixture_id` = $opponents_id");
$result = mysql_query($qury)
^ See here
You have mistakenly typed wrong variable in mysql_query.
I believe this is the problem: $qury
$result = mysql_query($qury) or die(mysql_error());
Change to:
$result = mysql_query($query) or die(mysql_error());
Related
This question already has answers here:
Passing an array to a query using a WHERE clause
(17 answers)
Closed 11 months ago.
I'm trying to query a MySQL database using an array but I'm having trouble!
I have a table called clients, I want to be able to select 'name' from all rows whose 'sector' column is equal to $sectorlink.
I then want to put all the names into an array so I can perform my next query: select all rows from another table whose 'client' column is equal to one of the names returned from the first query. I'm doing something wrong because it returns a fatal SQL error. I'm getting confused with all the variables!
$sectorlink and $connection are the only variables that are defined outside of this code
Any suggestions?
$query1 = "SELECT name FROM clients WHERE sector = '$sectorlink'";
$clientresult = mysql_query($query1, $connection) or trigger_error("SQL", E_USER_ERROR);
while($row = mysql_fetch_array($clientresult)){
foreach($row AS $key => $value){$temp[] = '"'.$value.'"';}
$thelist = implode(",",$temp);
$query = "SELECT count(*) FROM studies WHERE client IN ($row) ORDER BY (date) desc";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
}
The second query should use $thelist not $row, and it should be outside of the while loop. The foreach loop is unnecessary when processing a single row. You can access the name in $row with a simple $row[0]. Something like this (untested):
$query1 = "SELECT name FROM clients WHERE sector = '$sectorlink'";
$clientresult = mysql_query($query1, $connection) or trigger_error("SQL", E_USER_ERROR);
while($row = mysql_fetch_array($clientresult)){
$temp[] = '"'.$row[0].'"';
}
$thelist = implode(",",$temp);
$query = "SELECT count(*) FROM studies WHERE client IN ($thelist) ORDER BY (date) desc";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
Caution: Please be aware that your code is highly vulnerable to SQL injection attacks. It's fine for testing or internal development but if this code is going to be running the Fort Knox web site you're going to want to fix it up quite a bit. Just an FYI. :-)
Try This:
$query1 = "SELECT name FROM clients WHERE sector = '$sectorlink'";
$clientresult = mysql_query($query1, $connection) or trigger_error("SQL", E_USER_ERROR);
while($row = mysql_fetch_array($clientresult)){
$client = $row['name'];
$query = "SELECT * FROM studies WHERE client='$client' ORDER BY date DESC";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
/* echo results here */
}
Couple of things. First you have an unnecessary loop there. Try:
while (list($name) = mysql_fetch_row($clientresult)) {
$temp[] = $name;
}
To build your temporary array.
Second, the parts of the IN clause are strings, so when you implode, you'll need to enclose each value in quotes:
$thelist = "'". implode("','", $temp) . "'";
Lastly, in your query you are passing $row to the IN clause, you should be passing $thelist:
$query = "SELECT count(*) FROM studies WHERE client IN ($thelist) ORDER BY date desc";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
So altogether:
$query1 = "SELECT name FROM clients WHERE sector = '$sectorlink'";
$clientresult = mysql_query($query1, $connection) or trigger_error("SQL", E_USER_ERROR);
while (list($name) = mysql_fetch_row($clientresult)) {
$temp[] = $name;
}
$thelist = "'". implode("','", $temp) . "'";
$query = "SELECT count(*) FROM studies WHERE client IN ($thelist) ORDER BY date desc";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
I expect you'd be better off doing this in one query with a join:
$query = "SELECT COUNT(*) FROM `studies` INNER JOIN `clients` on studies.client = clients.name WHERE clients.sector = '$sectorlink' ORDER BY studies.date DESC";
$result = mysql_query($query, $connection) or trigger_error("SQL", E_USER_ERROR);
I'm making a form that submits a story into a MySQL table called 'work'. I want to later take the id of the newly created record and put the information into a different table.
But when I submit the story, it says:
$workid is undefined.
I can't see the problem though because I believe I've defined it?
<?php
if (!empty($_POST) && !empty($_POST['title']) && !empty($_POST['story']) && !empty($_POST['genre']) && !empty($_POST['rating'])) {
$title = strip_tags($_POST['title']);
$story = strip_tags($_POST['story']);
$title = mysqli_real_escape_string($db, $title);
$story = mysqli_real_escape_string($db, $story);
$genre = $_POST['genre'];
$rating = $_POST['rating'];
$query = "SELECT COUNT(*) AS count FROM works WHERE Title = '".$title."'";
$result = $db->query($query);
$data = $result->fetch_assoc();
if ($data['count'] > 0) {
echo "<p>Story already exists!</p>";
} else {
$query = "INSERT INTO works (author_id, login_id, Title, Story, Genre, Rating) VALUES ('".$userid."','".$authorid."','".$title."','".$story."','".$genre."','".$rating."')";
$query = "SELECT `id` FROM `works` WHERE `Title` = '".$title."'";
if ($result = $db->query($query)) {
while ($row = $result->fetch_assoc())
$workid = $row["id"]; //workid is written here but still considered undefined
}
$query = "INSERT INTO `author_work` (`author_id`) VALUES ('".$authorid."')";
$result = $db->query($query);
$query = "INSERT INTO `author_work` (`work_id`) VALUES ('".$workid."')";
$result = $db->query($query);
$query = "INSERT INTO `login_work` (`work_id`) VALUES ('".$workid."')";
$result = $db->query($query);
$query = "INSERT INTO `login_work` (`login_id`) VALUES ('".$userid."')";
$result = $db->query($query);
if ($result) {
echo "<p>Story submitted!</p>";
} else {
echo "SQL Error: " . $db->error;
}
}
}
?>
You never did a $db->query() on your INSERT INTO... query string, so it was never inserted, and was overwritten by your SELECT id ... query.
$query = "INSERT INTO works (author_id, login_id, Title, Story, Genre, Rating) VALUES ('".$userid."','".$authorid."','".$title."','".$story."','".$genre."','".$rating."')";
$db->query($query); // Missing this $db->query()
$query="SELECT `id` FROM `works` WHERE `Title` = '".$title."'";
if ($result = $db->query($query)) {
while ($row= $result->fetch_assoc())
$workid = $row["id"];}
Your $workid might not be initialized, depending on your condition and the result of your SQL query: so try to avoid next operations that will causes warnings/errors by using continue or else
I'm using the code below to query the database.
mysql_query ("SELECT * FROM _$symbol ORDER BY date DESC;");
$_10day = mysql_query ("SELECT AVG(close) FROM _$symbol limit 10;");
$_21day = mysql_query ("SELECT AVG(close) FROM _$symbol limit 21;");
$_50day = mysql_query ("SELECT AVG(close) FROM _$symbol limit 50;");
echo "$_10day\n";
echo "$_21day\n";
echo "$_50day\n";
mysql_query ("INSERT INTO _$symbol(_10day) VALUE ('$_10day');");
echo mysql_errno($sql) . ": " . mysql_error($sql). "\n";
I need to get the average of close from the database, and I'm using the AVG()function, then insert the return value into the database. The queries work in mysql however they do not work through the script. It does enter a value into the database, but it always enters 0. What am I doing wrong?
EDIT--solution found
$get10 = mysql_query ("SELECT AVG( close ) AS CloseAverage from ( SELECT close FROM _$symbol ORDER BY date DESC limit 10 ) sub1 ;");
$row = mysql_fetch_assoc($get10);
$_10day = $row['CloseAverage'];
if (!mysql_query ("UPDATE _$symbol SET _10day = $_10day, _21day = $_21day, _50day = $_50day, _100day = $_100day, _120day = $_120day, _150day = $_150day, _200day = $_200day, _240day = $_240day, _20dayVol = $_20dayVol, _50dayVol = $_50dayVol where date = '$date';"))
{
echo "Update query failed";
}
I was querying it incorrectly apparently it needed to be selected as a sub query, solution is above, it is working now.
Here's an example on how to manage multiple result-rows:
function connect()
{
$db = mysql_connect("localhost","username","password") or die("your error msg");
mysql_select_db("database_name");
return $db;
}
$db = connect();
$query = "SELECT value1 FROM my_table";
$result = mysql_query($query, $db);
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
// This one will loop through the data in your output//
$outputValue = $row['value1'];
}
For inserting data:
// Continued from the state above //
$query = "INSERT INTO table_name (column1) VALUES (value1)";
$result = mysql_query($query, $db) or die(mysql_error());
$_10day is a mysql result, not a value. You can't just put a result into a query string and expect it to work.
You need to get the result data first, eg;
while($row = mysql_fetch_assoc($_10day)) {
$10day = $row['AVG(close)'];
}
Then you can execute your insert query;
mysql_query ("INSERT INTO _$symbol (_10day) VALUES ('$10day')");
Hope this helps.
I am trying to store a mysql value into a php variable. I have the following query which I know works. However, I the value for $count is always 0. Can someone explain what I need to do to get the count value? The count should be the count of x's w here name_x=.$id.
$query = "SELECT COUNT(name_x) FROM Status where name_x=.$id.";
$result = mysql_query($query);
$count = $result;
Is first letter in table name is really capital. Please check it first.
or Try :
$query = "SELECT COUNT(*) as totalno FROM Status where name_x=".$id;
$result = mysql_query($query);
while($data=mysql_fetch_array($result)){
$count = $data['totalno'];
}
echo $count;
$query = "SELECT COUNT(*) FROM `Status` where `name_x`= $id";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
$count = $row[0];
please try it
$query = "SELECT COUNT(*) FROM Status where name_x=$id";
$result = mysql_query($query);
$count = mysql_result($result, 0);
You are missing single quotes around $id. Should be
name_x = '" . $id . "'";
I have the table name in the form as a hidden field, i have to do it this way for some reason but it's not working when i do the following:
// This is being processed from the previous page in a form
$table = mysql_real_escape_string($_POST['just_field']);
//Here i am using the $table value to use in place of a table name
$sql = "select `Field` from `$table` where `another_field` = '$id'
and `another_another_field` = '$something'";
$query = mysql_query($sql) or die ("Error: ".mysql_error());
while ($row = mysql_fetch_array($query)){
$somefield = $row['field'];
}
mysql_free_result($query);
How can i make it work so the $table is being read by the command...
No need for quotes around the table assignment.
$table = mysql_real_escape_string($_POST['just_field']);
try to include the connection string
$sql = "select `Field` from `$table` where `another_field` = '$id'
and `another_another_field` = '$something'";
$query = mysql_query($sql,$conn) or die ("Error: ".mysql_error());
while ($row = mysql_fetch_array($query)){
$somefield = $row['field'];
}
mysql_free_result($query);
if you are not using any framework
Does this work for you
$sql = "select `Field` from ".$table." where `another_field` = '$id'
and `another_another_field` = '$something'";
Your problem is case-sensitivity.
In your query:
select `Field` from `$table`....
^ Capital F
And the later
$somefield = $row['field'];
^ lower case f
An example
php > $cxn = mysql_connect( 'localhost', '****', '****' );
php > mysql_select_db( 'test' );
php > $res = mysql_query( 'select * from genres' );
php > $row = mysql_fetch_assoc( $res );
php > print_r( $row );
Array
(
[id] => 1
[name] => Pop //notice the field name is lower case
)
php > echo $row['name']; //this works
Pop // <- see
php > echo $row['Name']; //this doesn't
php > // <- look, nothing
Try printing the contents of $table before sending it through mysql_real_escape_string() to make sure it contains the value you think it does.