How to Convert specific field data to string with PHP

Question

I am trying to convert data from selected cell in SQL table to string so I can check if it's value is NULL. This is how my code looks now:

$columnname="cоntact";
$columnnumber=1;
$column=$columnname . $columnnumber;
$selectcell = mysqli_query($link, "SELECT '{$column}' FROM contacts WHERE id='{$userid}'");
$result = mysqli_fetch_assoc($selectcell);
$cell=(string)$result;

if (is_null($cell)) {
    // do something
}

And this is the error I get:

Notice: Array to string conversion

I am sorry if my mistake is obvious, I am pretty new to php.

Two things to notice right away: '{$column}' wrong quotes, and $result is an array, you can't just convert it to a string. You might want $cell=(string)$result[$column];? — Qirel
– Qirel, Commented Sep 23, 2017 at 15:00
You're already using an API that supports prepared statements with bounded variable input, you should utilize parameterized queries with placeholders (prepared statements) to protect your database against SQL-injection! Get started with mysqli::prepare() and mysqli_stmt::bind_param(). — Qirel
– Qirel, Commented Sep 23, 2017 at 15:05

Jordan · Accepted Answer · 2017-09-23 15:01:46Z

1

Simple one really, you're using the resulting MySQL array in your is_null function. You need to access the column specifically as so:

$columnname="cоntact";
$columnnumber=1;
$column=$columnname . $columnnumber;
$selectcell = mysqli_query($link, "SELECT `{$column}` FROM contacts WHERE 
id='{$userid}'");
$result = mysqli_fetch_assoc($selectcell);
$cell=(string)$result[$column]; //note accessing the [$column] value of the $result array

if (is_null($cell)) {
    // do something
}

edited Sep 23, 2017 at 15:01

answered Sep 23, 2017 at 15:00

Jordan

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4 Comments

Qirel Over a year ago

What about the quotes in the query, any thought on those? ;-)

Qirel Over a year ago

Eeverything returned from MySQL is a string, it doesn't care about the type of column. So the typecasting to string is redundant (its nothing wrong by doing it either), but still, you addressed both issues, +1 :-)

Skibidy Over a year ago

Maybe I didn't describe well what I try to do. The code you edited above is giving me the name of the column (column1). I am trying to get the content of specific cell in the table. Edit: Do I even need to convert it to string? Is there a way to check if it's Null without converting?

Qirel Over a year ago

The original code gives you the column, because you select a string (look at the quotes). But no, you don't need to convert it to string.

Mureinik · Accepted Answer · 2017-09-23 15:01:50Z

0

The easiest approach, IMHO, would be to fetch the data as a non-associative array and take the first field:

$result = mysqli_fetch_row($selectcell);
$cell = $result[0];

answered Sep 23, 2017 at 15:01

Mureinik

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