Warning: mysqli_query() expects parameter 2 to be string, object given [duplicate] - php

Question

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Warning: mysqli_query() expects parameter 2 to be string, object given in
(2 answers)
Closed 3 years ago.
Can anyone help me with this error:
Warning: mysqli_query() expects parameter 2 to be string, object given .. on line 25.
<?php
session_start();
include('includes/dbcon.php');
$query = mysqli_query($con, "SELECT * FROM reservation WHERE r_date='".$date."'
if (!mysqli_query($con,$query))
{
$query = mysqli_query($con, "SELECT * FROM combo where combo_id=1");
$row=mysqli_fetch_array($query);
$price=$row['combo_price'];
$payable=$pax*$price;

Answer 1

<?php
session_start();
include('includes/dbcon.php');
// you're missing some syntax here..
// also your $query IS your query so it should be $query = "SELECT * FROM ";
$query = mysqli_query($con, "SELECT * FROM reservation WHERE r_date='".$date."'
// you don't need this above line.. it does it all right here...
if (!mysqli_query($con,$query))
{
$query = mysqli_query($con, "SELECT * FROM combo where combo_id=1");
$row=mysqli_fetch_array($query);
$price=$row['combo_price'];
$payable=$pax*$price;
// missing closing brackets. }

Answer 2

Your code has multiple problems. Missing ;, repeated calls to mysqli_query, SQL injection and no error checking.
Instead of checking whether the query was successful with if enable exceptions at the top of your file. Use prepared statements, preferably in object-oriented way.
session_start();
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); // this line enables exceptions
include 'includes/dbcon.php';
$stmt = $con->prepare('SELECT * FROM reservation WHERE r_date=?'); // ? is a placeholder for binding data
$stmt->bind_param('s', $date); // bind data to SQL statement as a string(s)
$stmt->execute();
$reservations = $stmt->get_result();
// if your SELECT found some record then loop on the result set fetching each row one by one
while ($row = $reservations->fetch_assoc()) {
$combos = $con->query("SELECT * FROM combo where combo_id=1"); // if there is no data to be bound then we can use query
$row = $combos->fetch_assoc(); // fetch the matching combo row
$price = $row['combo_price'];
$payable = $pax * $price;
}

Answer 3

Your variable named query should only be your... query
$result = mysqli_query($con, "SELECT * FROM reservation WHERE r_date='".$date."'";
Also even if you think you will get back a record, function mysqli_fetch_array
will always return an array. So you need to select the first item in the array and then the key or index.
$price = $row[0]['combo_price'];
Some code practices. Don't put everything inside your IF. Because if it fails $payable will be undefined and throw an error. Initialize it on top of your script. Also you need to store the return value of mysqli_query as you need to free the memory used for it.
mysqli_free_result($result);