getting MYSQL error code while trying to get json data from database

Question

I am trying to get an AJAX search working, i am very close to this. Here is the php that i am using.

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Products";

try {
  $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
  $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

  $searchValue = $_GET['search'];

  if(isset($searchValue) && $searchValue != ''){
    $search = addslashes($searchValue);
    $statement = $conn->prepare("SELECT ProductName FROM Product WHERE ProductName LIKE('" . $search . "%') ORDER BY ProductName");
    $statement->execute();


    $all = $statement->fetchAll(PDO::FETCH_ASSOC);
    for($i=0; $i<count($all);$i++){
      echo json_encode($all[$i]).ProductName;
    }
  }

}
catch(PDOException $e)
{
  echo "Error: " . $e->getMessage();
}

$conn = null;
?>

The responseText i get is this:

Notice: Use of undefined constant ProductName - assumed 'ProductName' in F:\xampp\htdocs\searchSuggest.php on line 23

{"ProductName":"iMac"}ProductName

Notice: Use of undefined constant ProductName - assumed 'ProductName' in F:\xampp\htdocs\searchSuggest.php on line 23

The only thing i want to display is the "iMac" part of the json object

Why was this question getting down voted (it's heading back up but was at -2)? It's an inexperienced user with a simple error but what's wrong with that? He provides a code sample, an example of the error and clearly states his problem. There is nothing wrong with that. — rg88
– rg88, Commented Dec 21, 2014 at 14:36

GolezTrol · Accepted Answer · 2014-12-21 14:45:14Z

2

The fix

This line is incorrect:

echo json_encode($all[$i]).ProductName;

It looks like you are trying to get the productName as a property, but the operator for that is ->:

echo json_encode($all[$i])->ProductName;

That line is still incorrect though. The result of json_encode is not an object but a string. The right way to fix it, is to use the array result of fetchAll, and get the product name by the array key:

echo $all[$i]['ProductName'];

The error message

The . operator is for string concatination, so you are trying to concatinate to the json string, the constant ProductName, which is not defined. And that's exactly what the warning says: You are using the undefined constant ProductName, so PHP assumes you meant the constant string 'ProductName' instead.

With an -> it still won't work, though, since json_encode returns a string, not an object. You could json_decode it again, but that's a waste of processing time.

Possible solutions

You seem to be trying to treat the result as an object. Which would be done like this:

$all = $statement->fetchAll(PDO::FETCH_OBJ);
for($i=0; $i<count($all);$i++){
  echo $all[$i]->ProductName;
}

or this:

while ($row = $statement->fetch(PDO::FETCH_OBJ)) {
  echo $row->ProductName;
}

edited Dec 21, 2014 at 14:45

answered Dec 21, 2014 at 14:33

GolezTrol

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3 Comments

axiac Over a year ago

It doesn't work this way. json_encode() produces a string not an object.

GolezTrol Over a year ago

@axiac Yeah, the description was a bit unclear. I hope you find it better now?

Adam Piper Over a year ago

it is telling me that ProductName is an undefined index

axiac · Accepted Answer · 2014-12-21 14:35:44Z

0

You probably want:

echo($all[$i]['ProductName']);

answered Dec 21, 2014 at 14:35

axiac

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