Using a good SQL query instead of PHP code - php

I have a page that is taking a kind of long time to load, and I'm almost sure that this is caused by too many sql requests (AKA caused by my bad SQL skills). Is there anyway to join these 3 queries into one?
What I want to do with this query is to try to select a specific id from cardapios and, if there is anything there (if $num_rows > 0) the only thing I want to do is select that id. If there is nothing there, then I want to insert something and then select the id of that.
$query = "SELECT id FROM cardapios WHERE nome='$nome'";
$sql = mysqli_query($con,$query);
$num_rows = mysqli_num_rows($sql);
if ($num_rows > 0){
while ($row = mysqli_fetch_array($sql)){
$_SESSION['id_cardapio'] = $row['id'];
$num_rows = 0;
}}else{
$query = "INSERT INTO cardapios (nome, kcal, semana)
VALUES('$nome', '$kcal', '$semana')" or die(mysqli_error($con));
$sql = mysqli_query($con,$query);
$query = "SELECT id FROM cardapios WHERE nome='$nome' ";
$sql = mysqli_query($con, $query);
while ($row = mysqli_fetch_array($sql)){
$_SESSION['id_cardapio'] = $row['id'];
}
}
I am trying to put all of this into one query but getting nowhere. Is there anyway to use just one query for doing all of this?
Thanks in advance!

You can replace the last query by getting the mysqli_insert_id($con); as you already have the insert id available after the insert
$query = "SELECT id FROM cardapios WHERE nome='$nome'";
$sql = mysqli_query($con,$query);
$num_rows = mysqli_num_rows($sql);
if ($num_rows > 0){
while ($row = mysqli_fetch_array($sql)){
$_SESSION['id_cardapio'] = $row['id'];
$num_rows = 0;
}
}else{
$query = "INSERT INTO cardapios (nome, kcal, semana)
VALUES('$nome', '$kcal', '$semana')" or die(mysqli_error($con));
$sql = mysqli_query($con,$query);
if ( $sql !== false) { // did insert work
$_SESSION['id_cardapio'] = mysqli_insert_id($con);
} else {
// insert did nto work??
}
}

Related

Error when adding FIRST record to table

Ok. I ran into an issue. My code is able to insert records into my merchandise table. I truncated the table and a record is still inserted into the table but with an error "Undefined variable last_id". I assume that this is because when the table was truncate, there isn't a previous id since the record being inserted is the FIRST. Can someone help me resolve this issue. Thanks!
$sql = "SELECT m_id FROM merchandise";
$result = mysqli_query($connection, $sql);
while($row = mysqli_fetch_assoc($result)) {
$last_id = $row["m_id"];
}
$next_id = $last_id+1;
$conc = number_format($next_id/100,2,'-','');
$query = "INSERT INTO merchandise (mfr,type,description,mer_sku,price,qty) ";
$query .="VALUES ('$mfr','$type','$desc','MR{$mfr}{$conc}','$price','$qty')";
$add_sku_query = mysqli_query($connection, $query);
Declare last id as zero in case there are no rows.
$last_id = 0;
$sql = "SELECT m_id FROM merchandise";
$result = mysqli_query($connection, $sql);
while($row = mysqli_fetch_assoc($result)) {
$last_id = $row["m_id"];
}
$next_id = $last_id+1;

why does my while loop stop after executed another query inside?closed

my while loop stops after executed another query inside... can you correct my codes? I want to update the column status in table ordered_items_supplier to "Pending" when the pi_number is found in the table purchased_items_supplier and if not found the column status is "Active".
$sql2 = "select * from ordered_items_supplier";
$result = $connect->query($sql2);
if($result->num_rows > 0){
while ($row = $result->fetch_assoc()) {
$pi_number = $row['pi_number'];
$sql = "select * from purchased_items_supplier where pi_number = '$pi_number'";
$result = $connect->query($sql);
if($result->num_rows > 0){
while ($row2 = $result->fetch_assoc()) {
$pi_number = $row2['pi_number'];
$sql = "update ordered_items_supplier set status = 'Pending' where pi_number = '$pi_number'";
$query = $connect->query($sql);
}
}else{
$sql = "update ordered_items_supplier set status = 'Delivered' where pi_number = '$pi_number'";
$query = $connect->query($sql);
}
}
}
here's my mysql.. it should update the status "Delivered" in ID 11
The problem is overwriting the same variable each time.
Check that you use $result for the outer and inner query both.That's why the problem occur. So don't overwriting the $result variable.

Sending information from a newly created record to a different MySQL table

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

restrict database row insert by a number?

Based on my codes, i need to restrict the insertion of the data by 3, i mean is like after the insertion of 3 data row, it will be restricted from inserting in data. Is that possible? For more information, is like the borrow inserting 3 times, then it cannot be inserted anymore. Is there anyway to do so? I am still learning php by the way, thank you.
if(isset($_POST['selector']))
$id=$_POST['selector'];
else
$id = '';
$member_id = $_POST['member_id'];
$due_date = $_POST['due_date'];
$isbn = $_POST['due_date'];
if ($id == '' ){
//header("location: borrow.php");
if(isset($_POST['isbn'])){
$isbn = $_POST['isbn'];
$query = mysql_query("select book_id from book WHERE isbn = '$isbn'")or die(mysql_error());
$count = mysql_num_rows($query);
if($count > 0){
$row = mysql_fetch_array($query);
$bookid = $row['book_id'];
$date = date('Y-m-d');
}
mysql_query("insert into borrow (member_id,book_id,date_borrow,due_date) values ('$member_id','$bookid','$date','$due_date')")or die(mysql_error());
}
else{
header("location: borrow.php");
}
}else{
mysql_query("insert into borrow (member_id,date_borrow,due_date) values ('$member_id',NOW(),'$due_date')")or die(mysql_error());
$query = mysql_query("select * from borrow order by borrow_id DESC")or die(mysql_error());
$row = mysql_fetch_array($query);
$borrow_id = $row['borrow_id'];
}else{
mysql_query("insert into borrow (member_id,date_borrow,due_date) values ('$member_id',NOW(),'$due_date')")or die(mysql_error());
$query = mysql_query("select * from borrow order by borrow_id DESC")or die(mysql_error());
$row = mysql_fetch_array($query);
$borrow_id = $row['borrow_id'];
$N = count($id);
for($i=0; $i < $N; $i++)
{
mysql_query("insert borrowdetails (book_id,borrow_id,borrow_status)
values('$id[$i]','$borrow_id','pending')")or die(mysql_error());
}
header("location: borrow.php");
}
You just have to count number of user row before to make a new insert :
$query = mysql_query("SELECT COUNT(*) AS count FROM borrow WHERE member_id = '".$member_id."'");
$row = mysql_fetch_assoc($query);
if ( $row['count'] >= 3 )
echo('Max insert');
Also, check this : Why shouldn't I use mysql_* functions in PHP?
I'm not sure I understand you correctly.
You can restrict the number of rows returned by SELECT query using the LIMIT clause.
Make sure you either put an ORDER BY clause in there or determine that you don't care 'which' 3 rows will get inserted.
See here:
http://dev.mysql.com/doc/refman/5.0/en/select.html

Checking if mysql_query returned anything or not

$query = "SELECT * FROM `table`";
$results = mysql_query($query, $connection);
If 'table' has no rows. whats the easiest way to check for this.?
Jeremy Ruten's answer above is good and executes quickly; on the other hand, it only gives you the number of rows and nothing else (if you want the result data, you have to query the database again). What I use:
// only ask for the columns that interest you (SELECT * can slow down the query)
$query = "SELECT some_column, some_other_column, yet_another_column FROM `table`";
$results = mysql_query($query, $connection);
$numResults = mysql_num_rows($results);
if ($numResults > 0) {
// there are some results, retrieve them normally (e.g. with mysql_fetch_assoc())
} else {
// no data from query, react accordingly
}
You could use mysql_num_rows($results) to check if 0 rows were returned, or use this faster alternative:
$query = "SELECT COUNT(*) AS total FROM table";
$results = mysql_query($query, $connection);
$values = mysql_fetch_assoc($results);
$num_rows = $values['total'];
Alternatively you can simply check if the result of mysql_fetch_assoc is false.
$query = "SELECT * FROM `table`";
$results = mysql_query($query, $connection);
$Row = mysql_fetch_assoc($results);
if ($Row == false)
{
$Msg = 'Table is empty';
}
One thing i noticed that was missed was the fact that the query might not succeed, so you do need to check if the $results variable is set. I'll use the answer given by yjerem as an example.
$query = "SELECT COUNT(*) AS total FROM table";
$results = mysql_query($query, $connection);
if ($results) { // or use isset($results)
$values = mysql_fetch_assoc($results);
$num_rows = $values['total'];
}
If you loop through the results, you can have a counter and check that.
$x = 1;
$query = mysql_query("SELECT * FROM table");
while($row = mysql_fetch_assoc($query))
{
$x++;
}
if($x == 1)
{
//No rows
}

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