Multiple prepared statements (SELECT) - php

I'm trying to change my SQL queries with prepared statements.
The idea: I'm getting multiple records out of the database with a while loop and then some additional data from the database in the loop.
This is my old SQL code (simplified):
$qry = "SELECT postId,title,userid from post WHERE id='$id'";
$rst01 = mysqli_query($mysqli, $qry01);
// loop trough mutiple results/records
while (list($postId,$title,$userid) = mysqli_fetch_array($rst01)) {
// second query to select additional data
$query05 = "SELECT name FROM users WHERE id='$userid";
$result05 = mysqli_query($mysqli, $query05);
$row05 = mysqli_fetch_array($result05);
$name = $row05[name ];
echo "Name: ".$name;
// do more stuff
// end of while loop
}
Now I want to rewrite this with prepared statements.
My question: is it possible to run a prepared statement in the fetch of another prepared statement ? I still need the name like in the old SQL code I do for the $name.
This is what've written so far.
$stmt0 = $mysqli->stmt_init();
$stmt0->prepare("SELECT postId,title,userid from post WHERE id=?");
$stmt0->bind_param('i', $id);
$stmt0->execute();
$stmt0->bind_result($postId,$title,$userid);
// prepare second statement
$stmt1 = $mysqli->stmt_init();
$stmt1->prepare("SELECT name FROM users WHERE id= ?");
while($stmt0->fetch()) {
$stmt1->bind_param('i', $userid);
$stmt1->execute();
$res1 = $stmt1->get_result();
$row1 = $res1->fetch_assoc();
echo "Name: ".$row1['name'] ;
}
It returns an error for the second statement in the loop:
Warning: mysqli_stmt::bind_param(): invalid object or resource mysqli_stmt in ...
If I use the old method for the loop and just the prepared statement to fetch the $name it works.

You can actually do this with a single JOINed query:
SELECT p.postId, p.title, p.userid, u.name AS username
FROM post p
JOIN users u ON u.id = p.userid
WHERE p.id = ?
In general, if you are running a query in a loop, there is probably a better way of doing it.

Related

store sql statement in MySQL then run it

Is it possible to store the following SQL statement in MySQL then run it in a prepared statement?
Mysql table:
Table name: mystatements
Columns:id, statements
The following syntax is stored in the statements field:
SELECT id, AES_DECRYPT(secret,'$key') as txtsecret
FROM TABLE_1
Now in php:
first: I do a select query to get my statement
$stmt = $mysqli->prepare("SELECT statements FROM mystatements limit 1");
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
$statement.=$row['txtstatement'];
}
second: using the variable ($statement) from the the query above and add it to query below to run the in the prepared statement:
$key='password123';
$stmt = $mysqli->prepare($statement);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
echo $row['txtsecret'];
}
Also my stored syntax contains AES_DECRYPT(secret,'$key') just to complicate things. is what i'm trying to achieve possible? have I gone about this completely the wrong way?
Ok..
$key='password123';
$sql = str_replace('$key', $key, $statement); //replace $key to correct value
$stmt = $mysqli->prepare($sql);
Result:
SELECT id, AES_DECRYPT(secret,'$key') as txtsecret FROM TABLE_1
to
SELECT id, AES_DECRYPT(secret,'password123') as txtsecret FROM TABLE_1

Using PHP variable in SQL query

I'm having some trouble using a variable declared in PHP with an SQL query. I have used the resources at How to include a PHP variable inside a MySQL insert statement but have had no luck with them. I realize this is prone to SQL injection and if someone wants to show me how to protect against that, I will gladly implement that. (I think by using mysql_real_escape_string but that may be deprecated?)
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'hospital_name' AND value = '$q'";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried switching '$q' with $q and that doesn't work. If I substitute the hospital name directly into the query, the SQL query and PHP output code works so I know that's not the problem unless for some reason it uses different logic with a variable when connecting to the database and executing the query.
Thank you in advance.
Edit: I'll go ahead and post more of my actual code instead of just the problem areas since unfortunately none of the answers provided have worked. I am trying to print out a "Case ID" that is the primary key tied to a patient. I am using a REDCap clinical database and their table structure is a little different than normal relational databases. My code is as follows:
<?php
$q = 'Hospital_Name';
$query = "SELECT * FROM database.table WHERE field_name = 'case_id' AND record in (SELECT distinct record FROM database.table WHERE field_name = 'hospital_name' AND value = '$q')";
$query_result = mysqli_query($conn, $query);
while ($row = mysqli_fetch_assoc($query_result)) {
echo $row['value'];
}
?>
I have tried substituting $q with '$q' and '".$q."' and none of those print out the case_id that I need. I also tried using the mysqli_stmt_* functions but they printed nothing but blank as well. Our server uses PHP version 5.3.3 if that is helpful.
Thanks again.
Do it like so
<?php
$q = 'mercy_west';
$query = "SELECT col1,col2,col3,col4 FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
if($stmt = $db->query($query)){
$stmt->bind_param("s",$q); // s is for string, i for integer, number of these must match your ? marks in query. Then variable you're binding is the $q, Must match number of ? as well
$stmt->execute();
$stmt->bind_result($col1,$col2,$col3,$col4); // Can initialize these above with $col1 = "", but these bind what you're selecting. If you select 5 times, must have 5 variables, and they go in in order. select id,name, bind_result($id,name)
$stmt->store_result();
while($stmt->fetch()){ // fetch the results
echo $col1;
}
$stmt->close();
}
?>
Yes mysql_real_escape_string() is deprecated.
One solution, as hinted by answers like this one in that post you included a link to, is to use prepared statements. MySQLi and PDO both support binding parameters with prepared statements.
To continue using the mysqli_* functions, use:
mysqli_prepare() to get a prepared statement
mysqli_stmt_bind_param() to bind the parameter (e.g. for the WHERE condition value='$q')
mysqli_stmt_execute() to execute the statement
mysqli_stmt_bind_result() to send the output to a variable.
<?php
$q = 'Hospital_Name';
$query = "SELECT value FROM database.table WHERE field_name = 'hospital_name' AND value = ?";
$statement = mysqli_prepare($conn, $query);
//Bind parameter for $q; substituted for first ? in $query
//first parameter: 's' -> string
mysqli_stmt_bind_param($statement, 's', $q);
//execute the statement
mysqli_stmt_execute($statement);
//bind an output variable
mysqli_stmt_bind_result($stmt, $value);
while ( mysqli_stmt_fetch($stmt)) {
echo $value; //print the value from each returned row
}
If you consider using PDO, look at bindparam(). You will need to determine the parameters for the PDO constructor but then can use it to get prepared statements with the prepare() method.

Conditional query with PDO prepare and bind statement

I am converting all my queries from mysql to PDO, and in this process I found a conditional query like a follows
if (isset($parameters['searchTerm'])) {
$where =" And title LIKE '%{$parameters['searchTerm'] }%'";
}
$sql = "Select * from table data Where tableId = 5 {$where} ";
and when I am trying to convert this query in PDO the expected syntax is as follows
if (isset($parameters['searchTerm'])) {
$where =" And title LIKE :searchTerm";
}
$sql = $dbh->prepare("Select * from table data Where tableId = 5 {$where}");
if (isset($parameters['searchTerm'])) {
$sql ->bindParam(':searchTerm', '%{$parameters['searchTerm'] }%');
}
$sql ->execute();
Now as you can See that the if condition if (isset ($parameters ['searchTerm'] )) {...} is repeated twice.
The reason is
I can not prepare the sql query before $where is being set thus $sql variable is initialized after first if statement
I can not bind the parameters until I prepare the sql so it has to be placed after the $sql is being prepared
So there is one if statement before $sql = $dbh->prepare("Select * from table data Where tableId = 5 {$where}"); and one if statement after.
And my question is: Is there a way to remove this redundant if statement or I have to do it this way only.
you can use handy PDO's feature that lets you to send array with parameters straight into execute()
$where = '';
$params = array();
if (isset($parameters['searchTerm'])) {
$where =" And title LIKE :searchTerm";
$params['searchTerm'] = "%$parameters[searchTerm]%";
}
$sql = "Select * from table data Where tableId = 5 $where";
$pdo->prepare($sql)->execute($params);
Note that PHP syntax in your code is also wrong.

Left join using PDO

I am using the following PDO query:
<?php
$cadena = $_SESSION[Region];
// We Will prepare SQL Query
$STM = $dbh->prepare("SELECT `id_mesero`, `nombre_mesero`,`alias_mesero`, `rest_mesero` FROM tbmeseros WHERE cadena_mesero='$cadena'");
// For Executing prepared statement we will use below function
$STM->execute();
// we will fetch records like this and use foreach loop to show multiple Results
$STMrecords = $STM->fetchAll();
foreach($STMrecords as $row)
{
The value from the 'rest_mesero' field is the index from the table 'tbrestaurantes'.
I would need to join some fields values from 'tbrestaurantes' to the PDO query, but I don't know how to do it using PDO.
Any help is welcome.
UPDATED QUESTION TEXT
This is my proposal for the query :
$dbh->prepare("SELECT * FROM tbmeseros LEFT JOIN tbrestaurantes ON tbmeseros.rest_mesero = tbrestaurantes.id_restaurante WHERE tbmeseros.cad_mesero = ?");
But is show an error:
Warning: PDOStatement::execute() [pdostatement.execute]: SQLSTATE[HY093]: Invalid parameter number: no parameters were bound in /.../AdminMeseros.php on line 80
Line 80 is
$STM->execute();
This is my updated query:
<?php
$cadena = $_SESSION[Region];
$STM =$dbh->prepare("SELECT * FROM tbmeseros LEFT JOIN tbrestaurantes ON tbmeseros.rest_mesero = tbrestaurantes.id_restaurante WHERE tbmeseros.cad_mesero = ?");
$STM->bindParam(1, $cadena);
// For Executing prepared statement we will use below function
$STM->execute(array($cadena));
// we will fetch records like this and use foreach loop to show multiple Results
$STMrecords = $STM->fetchAll();
foreach($STMrecords as $row)
{
And here table's screenshots:
For tbmeseros:
For tbrestaurantes:
The value of $cadena is 'HQ3'.
When you put a parameter in the SQL, you need to supply the value for the parameter. There are two ways to do that:
1) Call bindParam():
$STM->bindParam(1, $cadana);
2) Provide the values when calling execute():
$STM->execute(array($cadana));
You need to fill the ? in the query:
$q = $dbh->prepare("SELECT * FROM tbmeseros LEFT JOIN tbrestaurantes ON tbmeseros.rest_mesero = tbrestaurantes.id_restaurante WHERE tbmeseros.cad_mesero = ?");
$q->bindValue( 1, 'x' );
$q->execute();
print_r( $q->fetchAll( PDO::FETCH_ASSOC );
In prepare you can use '?' and then bindValue to attach an escaped value to the query. Your query doesn't appear to have this and that is the cause of the error.

mysqli prepared statement without bind_param

I have this code for selecting fname from the latest record on the user table.
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$sdt=$mysqli->('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$sdt->bind_result($code);
$sdt->fetch();
echo $code ;
I used prepared statement with bind_param earlier, but for now in the above code for first time I want to use prepared statement without binding parameters and I do not know how to select from table without using bind_param(). How to do that?
If, like in your case, there is nothing to bind, then just use query()
$res = $mysqli->query('SELECT fname FROM user ORDER BY id DESC LIMIT 1');
$fname = $res->fetch_row()[0] ?? false;
But if even a single variable is going to be used in the query, then you must substitute it with a placeholder and therefore prepare your query.
However, in 2022 and beyond, (starting PHP 8.1) you can indeed skip bind_param even for a prepared query, sending variables directly to execute(), in the form of array:
$query = "SELECT * FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->execute([$_POST['ID']]);
$result = $stmt->get_result();
$row = $result->fetch_assoc();
The answer ticked is open to SQL injection. What is the point of using a prepared statement and not correctly preparing the data. You should never just put a string in the query line. The point of a prepared statement is that it is prepared. Here is one example
$query = "SELECT `Customer_ID`,`CompanyName` FROM `customers` WHERE `Customer_ID`=?";
$stmt = $db->prepare($query);
$stmt->bind_param('i',$_POST['ID']);
$stmt->execute();
$stmt->bind_result($id,$CompanyName);
In Raffi's code you should do this
$bla = $_POST['something'];
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
$stmt = $mysqli->prepare("SELECT `fname` FROM `user` WHERE `bla` = ? ORDER BY `id` DESC LIMIT 1");
$stmt->bind_param('s',$_POST['something']);
$stmt->execute();
$stmt->bind_result($code);
$stmt->fetch();
echo $code;
Please be aware I don't know if your post data is a string or an integer. If it was an integer you would put
$stmt->bind_param('i',$_POST['something']);
instead. I know you were saying without bind param, but trust me that is really really bad if you are taking in input from a page, and not preparing it correctly first.

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