Issue with php/mysql prepared insert statement

Question

I might just be too dumb to see it but here is my issue:

Code:

$stmt = $mysql->prepare("INSERT INTO projects_files (mid,filename,type) VALUES ('?','?','?')");
$stmt->bind_param('isi',$this->id,$File->filename,$File->type);

Error:

Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

I have checked that all vars are assigned and have a value and the datatypes are all right.

Thanks in advance for help

Cully · Accepted Answer · 2014-05-20 08:28:41Z

1

Don't put the question marks in quotes. The database will quote everything appropriately.

$stmt = $mysql->prepare("INSERT INTO projects_files (mid,filename,type) VALUES (?,?,?)");
$stmt->bind_param('isi',$this->id,$File->filename,$File->type);

answered May 20, 2014 at 8:28

Cully

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3 Comments

JensV Over a year ago

My friend just showed up and told me the same :) Thanks for the help tho.

Cully Over a year ago

@Jenjen1324 Mark the question as answered if you've solved your problem.

JensV Over a year ago

I was going to but you answered so fast I had to wait another 12 minutes ;)

Ranjith · Accepted Answer · 2014-05-20 08:31:34Z

1

Remove the bindparam single quotes(') around the question mark(?)

$stmt = $mysql->prepare("INSERT INTO 
                     projects_files (mid,filename,type) 
                     VALUES (?,?,?)");
$stmt->bind_param('isi',$this->id,$File->filename,$File->type);

answered May 20, 2014 at 8:31

Ranjith

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2 Comments

Cully Over a year ago

"isi" is the types parameter. See us2.php.net/manual/en/mysqli-stmt.bind-param.php

Ranjith Over a year ago

@CullyLarson : Oho, I forgot that sorry. Thank for your advice. I updated my answer.

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