php mysql bind param parameters

Question

I'm trying out using prepared statements for the first time and running into the following issue with the below code

Error :

Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, boolean given

Code :

$stmt = mysqli_prepare($db, "INSERT INTO fragrances(name, description, essentialoils, topnotes, middlenotes, basenotes, reference, year, type, price, fragrancehouse, triangle, extractname, extractreference, extractprice, extractfragrancehouse, disccolour, collarcolour, actuatorcolour)
    VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"); 
mysqli_stmt_bind_param($stmt, 'sssssssssssssssssss', $name, $description, $essentialoils, $topnotes, $middlenotes, $basenotes, $reference, $year, $type, $price, $fragrancehouse, $triangle, $extractname, $extractreference, $extractprice, $extractfragrancehouse, $disccolour, $collarcolour, $actuatorcolour);

mysqli_stmt_execute($stmt);

I've looked at many different questions on here and none of their solutions seem to apply for my problem, does anyone know what the issue is?

prepare is failing. you need to get the output of mysqli_error. You also have no placeholders in your query. Use ? or named parameters if mysqli supports it (i'm not sure) — Cfreak
– Cfreak, Commented Oct 10, 2014 at 20:06
Possible duplicate of mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it? — Dharman
– Dharman ♦, Commented Oct 15, 2019 at 23:03

meda · Accepted Answer · 2014-10-10 20:41:55Z

4

$stmt becomes a boolean only when mysqli_prepare returns false.

When this happens it means it failed to prepare the query therefore you need to check for errors:

$stmt = mysqli_stmt_init($db);
if (mysqli_stmt_prepare($stmt, 'INSERT INTO fragrances VALUES...')) {
    //it's all good bind and execute here
}else{
   //we have a problem
   printf("Errormessage: %s\n", mysqli_error($db));
}

edited Oct 10, 2014 at 20:41

answered Oct 10, 2014 at 20:30

meda

45.5k14 gold badges96 silver badges123 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Xand94 Over a year ago

Turned out I had mixed my americanisms up and had a field named collarcolor by mistake! this lead me to the error so is getting the tick, Thanks

Asenar · Accepted Answer · 2014-10-10 20:34:15Z

2

The error message means your mysqli_prepare returned a boolean (and for your case, it returned false).

You need to replace all your field name by the character ? to make your prepared statement. This is how it works.

See example in the official documentation

EDIT See also mysqli_error , which will detail your error. In fact, you should always check a variable before using it:

$stmt = mysqli_prepare($db, "....");
if(!$stmt)
  echo mysqli_error($db); // display error only for debug. Avoid this in production

edited Oct 10, 2014 at 20:34

answered Oct 10, 2014 at 20:06

Asenar

7,0863 gold badges44 silver badges53 bronze badges

2 Comments

Xand94 Over a year ago

Unfortunately that does not change my error message

Asenar Over a year ago

So you have an other error. Maybe related to $db. I edited my answer to help you debugging it.

worldofjr · Accepted Answer · 2014-10-10 20:34:26Z

2

It means your SQL was invalid because the prepare is returning false;

Your SQL should be;

$stmt = mysqli_prepare($db, "INSERT INTO fragrances VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ? )");

Each ? is to show where each parameter needs to be bound respectively.

answered Oct 10, 2014 at 20:34

worldofjr

3,8968 gold badges41 silver badges53 bronze badges

Comments

Rimas · Accepted Answer · 2014-10-29 06:27:09Z

Your INSERT statement is invalid: VALUES clause must be with ? in parantheses (and after field names in parentheses). Also good practice is to check $stmt after assigning:

$stmt = mysqli_prepare($db, 
   "INSERT INTO fragrances (name, description, essentialoils, topnotes, middlenotes, basenotes, reference, year, type, price, fragrancehouse, triangle, extractname, extractreference, extractprice, extractfragrancehouse, disccolour, collarcolour, actuatorcolour)
    VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)");
if ($stmt) {
    mysqli_stmt_bind_param($stmt, 'sssssssssssssssssss', $name, $description, $essentialoils, $topnotes, $middlenotes, $basenotes, $reference, $year, $type, $price, $fragrancehouse, $triangle, $extractname, $extractreference, $extractprice, $extractfragrancehouse, $disccolour, $collarcolour, $actuatorcolour);

    mysqli_stmt_execute($stmt);
    // ...
} else
    printf("Error: %s\n", mysqli_error($db));

Unfortunately that does still create the same error message (removing the if statement)

Collectives™ on Stack Overflow

php mysql bind param parameters

4 Answers 4

1 Comment

2 Comments

Comments

1 Comment

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

2 Comments

Comments

1 Comment

Linked

Related