MySQL Query not working with PHP Variable? [closed] - php

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Why is this working:
$sql_query = "SELECT * FROM Content WHERE id IN (1,5,7,9)";
But this isn't:
$array_values = "1,5,7,9";
$sql_query = "SELECT * FROM Content WHERE id IN ('$array_values')";
I want to select data from the database based on the integers in the $array_values string.
How can I do it?


because there are ` s in your code here
$sql_query = "SELECT * FROM Content WHERE id IN ('$array_values')";
use :
$sql_query = "SELECT * FROM Content WHERE id IN ( $array_values )";
or
$sql_query = "SELECT * FROM Content WHERE id IN (".$array_values.")";

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SQL not providing proper result. (Using PHP) [closed]

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Closed 2 years ago.
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So here I have a table of Products which contains ID, Name, Type & Price
Now I'm running a function to give me the minimal price of specific product types.
The datatype of Price is Price int(5,2)
function MinPrice($connection) {
echo "Minimum price: <br>";
$sql = "SELECT Type, min(Price)
FROM Products
GROUP BY Type;";
$result = mysqli_query($connection,$sql);
while ($row = mysqli_fetch_assoc($result)) {
//Printing the result
echo $row['Type'].' || '.$row['Price'].'<br>';
}
}
The code works but not properly. I get the names of the product types but for the price, it gives me an error.
Undefined index: Price
Can anyone please help me out with this?

Try this:
$sql = "SELECT Type, min(Price) as Price
FROM Products
GROUP BY Type;";

Why isn't `INSERT ROW` working? [closed]

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Closed 7 years ago.
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I have an SQL piece of code that every time a user inserts a name creates a new column with the name (and this part works fine). I also want it to add a row to that column with the respective comment given by the user. I don't see why this add row statement won't work.
$value = mysql_real_escape_string($_POST['name']);
$comment = mysql_real_escape_string($_POST['comment']);
$add = mysql_query("ALTER TABLE Names ADD $value Text NOT NULL");
$sql = mysql_query("INSERT INTO Names VALUES $comment");

Try with
$sql = mysql_query("INSERT INTO Names (`$value`) VALUES ('$comment')") or die(mysql_error());
You need to specify the column list.
Ideally you should start using PDO or an ORM to be able to work and debug easier

Try this:
$value = mysqli_real_escape_string($db, $_POST['name']);
$comment = mysqli_real_escape_string($db, $_POST['comment']);
$add = mysqli_query($db, "ALTER TABLE Names ADD '{$value}' Text NOT NULL");
$sql = mysqli_query($db, "INSERT INTO Names({$value}) VALUES('{$comment}')");

PDO multiple queries in php /sql [closed]

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Closed 7 years ago.
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I have the following code. I am grabbing values from a form, and using those values to try to update "customers" and "workorder". The "name" value will be duplicate across the customers and workorder table. I keep on getting the error "number of bound variables does not match number of tokens". I am totally new to using PDO, and am unsure on how to proceed. Any ideas?
$pdo = Database::connect();
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "UPDATE customers set name = ?, email = ?, mobile =? WHERE id = ?; UPDATE workorder set name = ?;";
$q = $pdo->prepare($sql);
$q->execute(array($name,$email,$mobile,$id));
Database::disconnect();
header("Location: index.php");

You have 4 variables here:
$q->execute(array($name,$email,$mobile,$id));
Should be 5:
$q->execute(array($name,$email,$mobile,$id, $name));

How to store json string in mysql using php [closed]

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Closed 8 years ago.
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I want to store my JSON string into database but when i give json string it is not working but when i give simple value it work
this is what i am doing
// $data contain json string
// info is a TEXT type in mysql
$q = "Update user set info = $data where userid = $id";
$sql= $this->db->query($q);

You should really check your error log or enable display_errors to see where the query is failing, but my guess is you probably just need to wrap the data field in quotes. Try this:
// $data contain json string
// info is a TEXT type in mysql
$q = "Update user set info = '$data' where userid = $id";
$sql= $this->db->query($q)

You are missing the quotes around $data and $id. And you need to close the string before concatenating variables, like this:
$q = "Update user set info = '". $data."' where userid = '". $id."'";
$sql= $this->db->query($q);

selcet from mysql using php variable [closed]

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This question does not appear to be about programming within the scope defined in the help center.
Closed 9 years ago.
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How can I use a variable inside a mysql select from statment? like so:
$db = $_POST['db'];
$query = "SELECT * FROM $db..";

To literally answer the question:
$db = $_POST['db'];
$query = "SELECT * FROM {$db}..";
OR
$query = "SELECT * FROM {$_POST['db']}..";
OR
$query = "SELECT * FROM ".$_POST['db']."..";
As others have said, accepting unsanitized input from the POST is a very bad idea indeed
at the very least you should do the following
$db = $mysqli->real_escape_string($_POST['db']);
which will atleast ensure that other commands will not be inserted such as INSERTS, UPDATES, or GRANT

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