php query not working without giveing any error

Question

i am having issues with the query below its showing success with no error but not inserting to the database

here is my code

public function uploadImg($table, $imageurl){
        //path to tore the uploaded images
        $target = "../images/" .basename($_FILES['image']['name']);
        $image = $_FILES['image']['name'];
        //echo $target;
        $sql = "INSERT INTO {$table} ($imageurl) values('$target')";
        $q = $this->DBcon->prepare($sql);
        $q->execute();


        if(move_uploaded_file($_FILES['image']['tmp_name'], $target)){
            echo "uploaded successfully";
        }else{
            echo "error uploading";
        }
    }
}

And this will not show any error because you are not showing them — Masivuye Cokile, May 25 '17 at 08:14
what's the value of the `prepare` statement without using the [parameterized queries](https://stackoverflow.com/questions/4712037/what-is-parameterized-query) ?? — hassan, May 25 '17 at 08:15
@MasivuyeCokile "where do u assign value to $imageurl" ,, here `uploadImg($table, $imageurl)` — hassan, May 25 '17 at 08:16
@Imm why not to print your error and see ? `$q->errorInfo()` — hassan, May 25 '17 at 08:17
@hassan isn't he supposed to assign value to the `imageurl` after successfully moved the file to the upload directory? — Masivuye Cokile, May 25 '17 at 08:20
@Imm you not helping if you not gonna take part in the comments, we don't wanna assume we wanna give direct straight answers — Masivuye Cokile, May 25 '17 at 08:21

score -2 · Answer 1 · edited May 25 '17 at 08:27

-2

Try:

$sql = "insert into " . $table . " set " . $imageurl . "=" . $target;

if $tabel and $imageurl exists, a record will be created. And if $target has a string, it will be registered in the field which is represented by $imageurl.

edited May 25 '17 at 08:27

hassan

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answered May 25 '17 at 08:25

Tom Dalgaard Petersen

1

Adeojo Emmanuel IMM · Answer 2 · 2018-04-23T23:52:41.990

-2

fixed

simply change your query from

$sql = "INSERT INTO {$table} ($imageurl) values('$target')";

to

$sql = "INSERT INTO {$table} ('$imageurl') values('$target')";

edited Apr 23 '18 at 23:52

answered May 25 '17 at 08:42

Adeojo Emmanuel IMM

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php query not working without giveing any error

2 Answers2