Uncaught mysqli_sql_exception

[ivanAdolfo]

insertImg: $sql="`user`, `id`, `imageName`, `imageType`,  `imageSize`, `imgeContent`)  VALUES ($wuser,$id, $imgName, $imgType, $imgSize,$imgContent)";

Values:   XY(user varchar)=1(id int)=noticiacallampa.jpg(imageName varchar)=image/jpeg(imageType varchar)=31 kB(imageSize varchar)=imgContent (longlob it comes from file_get_contents($image)
Fatal error: Uncaught mysqli_sql_exception: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'kB,????' at line 5 in C:\xampp\htdocs\_website\liceoUtil\php\liceoFuncs.php:79 Stack trace: #0 C:\xampp\htdocs\_website\liceoUtil\php\liceoFuncs.php(79): mysqli_query(Object(mysqli), 'INSERT INTO `li...') #1 C:\xampp\htdocs\_website\liceoIUpload.php(32): insertImg(Object(mysqli)) #2 {main} thrown in C:\xampp\htdocs\_website\liceoUtil\php\liceoFuncs.php on line 79

line 79: $result = mysqli_query($conn, $sql);

thanks

[Barand]

An insert query works better if your sql informs the server

1. it's an INSERT query
2. which table to insert into

so needs to begin with "INSERT INTO tablename..."

Also, string literals like $imageName need to be inside single quotes. However you should be using a prepared statement.

I would also advise storing your image files in the filesystem and putting the metadata (such as filename) in the table.

[ivanAdolfo]

21 hours ago, Barand said:

An insert query works better if your sql informs the server

1. it's an INSERT query
2. which table to insert into

so needs to begin with "INSERT INTO tablename..."

Also, string literals like $imageName need to be inside single quotes. However you should be using a prepared statement.

I would also advise storing your image files in the filesystem and putting the metadata (such as filename) in the table.

I try it using quotes as well and the INTO 'tablename' and the error still the same.

I will do what you suggest... storing your image files in the filesystem and putting the metadata (such as filename) in the table.

Thanks.

[Barand]

Post the output from...

echo $sql;