PDO:: Confusion [duplicate] - php

This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 7 years ago.
I have no idea why this is not returning anything. I'll show the code and talk through the steps I've taken.
if (isset($_GET['observation'])) {
require_once("../func/connect.php");
$query = "SELECT * FROM observations WHERE option = ?";
$stmt = $db->prepare($query);
$stmt->bindValue(1, $_GET['observation']);
$stmt->execute();
$row = $stmt->fetch(PDO::FETCH_ASSOC);
echo $row['question'];
} else {
echo 'nope';
}
$row dumps a false boolean, $row['question'] is null.
I've wrote about a million queries and don't have a clue why this doesn't work.
Database table observations consists of id, question & option and the bindValue is correct to match a string in the database.
However, it returns null.

option is a reserved word in mysql so you need to quote it with backticks:
$query = "SELECT * FROM observations WHERE `option` = ?";

Related

How to get a column value using MySQLi? [duplicate]

This question already has answers here:
Single result from database using mysqli
(6 answers)
Closed 5 months ago.
I am trying to get a value from column "odznak" in "users" tab for user "user01" and store it in variable $odznak (for searching in another tab.
$stmt = $conn->prepare("SELECT odznak FROM users WHERE username = 'user01'");
$stmt->execute();
$result = $stmt;
$odznak;
You need to fetch the data (say into an associative array)
On the other hand, as a good practice, please use parameterized prepared statement in your select query
So, change to:
$stmt = $conn->prepare("SELECT odznak FROM users WHERE username = ?");
$stmt->bind_param("s", 'user01');
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
$odznak=$row["odznak"];
Now, $odznak is the retrieved data

PHP PDO mysql query to work with both column = '$var' and column IS NULL [duplicate]

This question already has an answer here:
Write a prepared statement with nullable values in conditions
(1 answer)
Closed 2 years ago.
I have a quite long mysql query, selecting data according to status field. I'm calling it for different statuses and it works well, but I have a scenario when I should get all records where status is null ONLY. Is there a way to do this without having to write 2 different sql queries?
Looks like I can't insert 'IS NULL' or '=' without it being rendered as a string.
I want to achieve this:
$sql = "SELECT name, surname FROM ...
...
WHERE status ?;";
$stmt = $pdo->prepare($sql);
$stmt->execute(array($status === 'undefined' ? 'IS NULL' : " = '$status'"));
After all, here's what I did:
$sql = "SELECT name, surname FROM ...
...
WHERE status <=> ?;";
$stmt = $pdo->prepare($sql);
$stmt->execute(array($status === 'unfinished' ? null : $status));
Using parameterised queries, as indicated, is a safer way of introducing user input into your SQL statements. However, it has the effect of treating all input as a parameter, and therefore will surroung any string literals with quotes - giving rise to the problem you have.
To deal with this issue, why not just modify the logic of the code:
$sql = "SELECT name, surname FROM ...
...
WHERE status";
if ($status === 'undefined') {
$sql .= " IS NULL";
$stmt = $pdo->prepare($sql);
$stmt->execute();
} else {
$sql .= " = ?"
$stmt = $pdo->prepare($sql);
$stmt->execute(array($status));
}
Edit
Updated to move the execution into the relevant part of the if statement becuase the parameters must not be specified if there is no placeholder in the SQL statement.

Select query not showing expected result. print_ only shows the first array correctly then a single string of other rows email field [duplicate]

This question already has an answer here:
PDO fetch returns only first row
(1 answer)
Closed 2 years ago.
A table called "checks" has fields ID;email;pass;entered;firstname;lastname;trading.
The only code in this test is the DB connection and the new PDO connection made prior to these snippets. The following snippet reports correctly that there are 5 users in the table "checks".
$sql = "SELECT COUNT(*) AS num FROM checks";
$stmt = $conn->prepare($sql);
$stmt->execute();
$row = $stmt->fetch(PDO::FETCH_ASSOC);
echo '<br>' . $row['num'] . ' users exist.';
This snippet which follows immediately after the above doesn't show the expected result.
$sql = "SELECT * FROM checks";
$stmt = $conn->prepare($sql);
$stmt->execute($id);
$users = $stmt->fetch(PDO::FETCH_ASSOC);
print_r($users);
The print_r statement results in the first array being printed correctly with all fields/contents correctly displayed.
On the next line it prints only the email field contents from each row as a single string!
I'm probably missing something obvious but I just can't spot it. Help please?
PDO::fetch() returns a single row from the result set. You need PDO::fetchAll() instead.
$sql = "SELECT * FROM checks";
$stmt = $conn->prepare($sql);
$stmt->execute();
$users = $stmt->fetchAll(PDO::FETCH_ASSOC); // <--- here

Select from table, where column is foreign key in PHP [duplicate]

This question already has answers here:
mysqli prepared statement num_rows returns 0 while query returns greater than 0 [duplicate]
(3 answers)
Closed 3 years ago.
Even though there is an entry in the database, with this query, I always get 0 entries back
$sql = "SELECT * FROM saved_food WHERE user_id = ? AND favorite_food LIKE ?";
$stmt = $conn->prepare($sql);
$stmt->bind_param("is", $me['id'], $favFood);
$stmt->execute();
var_dump($stmt->num_rows);
the dump is 0
The user_id colum is a foreign key, and shows to the id of the table "user".
I can't see the error here.
Is there a special method for foreignkey values?
I got the error... facepalm
I forgot to call ->get_result();
$sql = "SELECT * FROM saved_food WHERE user_id = ? AND favorite_food LIKE ?";
$stmt = $conn->prepare($sql);
$stmt->bind_param("is", $me['id'], $favFood);
$stmt->execute();
$result = $stmt->get_result();
var_dump($result->num_rows);

Fetch COUNT DISTINCT data with prepared statements [duplicate]

This question already has answers here:
Can PHP PDO Statements accept the table or column name as parameter?
(8 answers)
Closed 5 years ago.
I have this code to get a COUNT DISTINCT data:
$param = 'email';
$stmt = $conn->stmt_init();
$stmt = $conn->prepare("SELECT COUNT(DISTINCT(?)) FROM contatos");
$stmt->bind_param('s',$param);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($count);
while ($stmt->fetch()) {
echo $count;
}
But echo $count always returns 1, but i have dozens of records...
What is wrong?
Thanks
Binding is not allowed for column names (or table names). Your query is not executing correctly. You need to directly pass the name of the field.
$stmt = $conn->prepare("SELECT COUNT(DISTINCT(email)) FROM contatos");

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