Can't convert object into a string

Question

I want to output the results of a query in text on a page so I can see what it looks like. Problem is that I am receiving errors if I use an echo like I did below or even use a print.

The error I am receiving is something about not being able to convert an object into a string

My question is what is best way to output query results on the page?

Below is the code:

$answersql = "INSERT INTO Answer (Answer) 
    VALUES (?)";

      if (!$insertanswer = $mysqli->prepare($answersql)) {
      // Handle errors with prepare operation here
    }  

    $insertanswer->bind_param("s", $_POST['value'][$i]);

        //$insertanswer->execute();

        echo $insertanswer;

        if ($insertanswer->errno) {
          // Handle query error here
        }

        $insertanswer->close();

you're not trying to output the results there, you're trying to echo a pdo object. what exactly were you hoping for with this? — Kai Qing
– Kai Qing, Commented Oct 9, 2012 at 23:59
Here's a PDO guide. You might be looking for a way to check affected rows or something since that query above is an insert. phpro.org/tutorials/Introduction-to-PHP-PDO.html — Kai Qing
– Kai Qing, Commented Oct 10, 2012 at 0:02
PDO is nice, but he is using the MySQL-Improved driver, not PDO. — Geoff Montee
– Geoff Montee, Commented Oct 10, 2012 at 0:07
You can echo and print objects that have a __toString method that returns a string, but I don't think that's what you actually need here. What errors are you getting? — ACJ
– ACJ, Commented Oct 10, 2012 at 0:20

Geoff Montee · Accepted Answer · 2012-10-10 00:06:17Z

1

You aren't outputting the query result. It appears that you are trying to output the query itself.

However, $insertanswer is an instance of mysqli_statement. This class does not appear to offer access to the query.

That is why you cannot print it as a string. It isn't a string.

answered Oct 10, 2012 at 0:06

Geoff Montee

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Morufu Salawu · Accepted Answer · 2012-10-10 00:08:54Z

0

You may try to type cast it using the string. for instance you declare a variable to hold the resultant conversion, like $answersql = "INSERT INTO Answer ((string)Answer)) VALUES (?)"; or you may type cast the assigned variable thus: (string)$answersql = "INSERT INTO Answer (Answer) VALUES (?)"; depending on how or where you need your result.

answered Oct 10, 2012 at 0:08

Morufu Salawu

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rwhite · Accepted Answer · 2012-10-10 00:19:58Z

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Couple items. On your binding, you'll want to assign $_POST to a variable first. I see you have an counter there, but don't see where you are iterating over any thing. I'll assume you can assign the $_POST['value'] to a scale variable.

 $input = $_POST['value'];

Then INSERT returns true or false and an ID if you have set an auto_increment attribute on your db table column. If so, you can then use

 $insertid = $insertanswer->insert_id;

However the way you have the statement written, it wouldn't return any value. Just true or false and an id (if auto_increment attribute is set).

answered Oct 10, 2012 at 0:19

rwhite

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