Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in...

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Hey, I'm new to mysql and am having trouble with adding information to a database. Error "Data Not Inserted Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in... on line 56. How can I fix this? Thanks.

PHP

<?php

// php code to Insert data into mysql database from input text
if(isset($_POST['submit']))
{
    $hostname = "localhost";
    $username = "";
    $password = "";
    $databaseName = "";
    
    // get values form input text and number

   $title = $_POST['title'];
   $first_name = $_POST['first_name'];
   $surname = $_POST ['surname'];
   $gender = $_POST ['gender'];
   $dob = $_POST ['dob'];
   $street_address = $_POST ['street_address'];
   $suburb = $_POST ['suburb'];
   $state = $_POST ['state'];
   $home_phone = $_POST ['home_phone'];
   $mobile_phone = $_POST ['mobile_phone'];
   $work_phone = $_POST ['work_phone'];
   $email = $_POST ['email'];
   $martial_status = $_POST ['martial_status'];
   $occupation = $_POST ['occupation'];
   $parent_name = $_POST ['parent_name'];
   $preferred_dr = $_POST ['preferred_dr'];
   $medicare_no = $_POST ['medicare_no'];
   
   //$allergies = $_POST ['allergies'];
   //$postcode = $_POST ['postcode'];
    // connect to mysql database using mysqli

    $connect = mysqli_connect($hostname, $username, $password, $databaseName);
    
    // mysql query to insert data

   $query = "INSERT INTO `Patient Details Table` (`Title`, `First Name`, `Surname`, `Gender`, `D.O.B`, `Street Address`, `Suburb`, `State`, `Home Phone`, `Mobile Phone`, `Work Phone`, `Email`, `Martial Status`, `Occupation`, `Parent Name`, `Preferred Dr`, `Medicare No.`) VALUES ('$title', '$first_name', '$surname', '$gender', '$dob', '$street_address', '$suburb' '$state', '$home_phone', '$mobile_phone', '$work_phone', '$email', '$martial_status', '$occupation', '$parent_name', '$preferred_dr', '$medicare_no')";
   //$query = "INSERT INTO `Postcode Table` (`Suburb`, `Postcode`) VALUES ('$suburb', '$postcode')";
   //$query = "INSERT INTO `Allergy Table` (`Allergy`) VALUES ('$allergies')";

    $result = mysqli_query($connect,$query);
    
    // check if mysql query successful

    if($result)
    {
        echo 'Data Inserted';
    }
    
    else{
        echo 'Data Not Inserted';
    }
    
    mysqli_free_result($result);
    mysqli_close($connect);
}

?>

What I have tried:

The code was originally off a video on how to add inputted information into databases. I have made a few changes in order for it to suit my needs. However, I cannot get it working. Any help you be great. Thanks.

Posted 31-Aug-17 2:44am

Member 13387451

Updated 21-Sep-21 10:33am

Jochen Arndt

v2

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Mohibur Rashid 31-Aug-17 23:21pm

Are those spaces in your table name and field name?

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Jochen Arndt · Answer 1 · 2017-08-31T04:23:00

Read the documentation:

PHP: mysqli::query - Manual[^]:
Return Values

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

You are executing an INSERT statement for which the return value is always a boolean. So there is not even no need to call mysqli_free_result() but it is wrong because you (should) know that $result can't be a mysqli_result.

The original source on which your code is based was probably executing a SQL commmand that will return a mysqli_result. But even then your code would be wrong because it should be only freed upon success.

As you can see it is better to write your own code from scratch than modifying existing code (especially if you are a beginner). If the code you have used as base also calls mysqli_free_result() outside a conditional block as in your posted code, it is even worse: You have copied code doing it wrong.

If you need example code for PHP database operations, refer to the official documentation. See for example my above link. It contains a lot of example code.

Patrice T · Answer 2 · 2017-08-31T13:37:00

Not a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]

Member 15366151 · Answer 3 · 2021-09-21T10:33:00

Solution 4

Simple

$title=addslashes($title);

Returns a string with backslashes added before characters that need to be escaped. These characters are:

single quote (')
double quote (")
backslash (\)
NUL (the NUL byte)

Posted 21-Sep-21 10:33am

Member 15366151

Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in...

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