Simple MySQL echo - "Trying to get property of non-object"

Question

I'm trying to figure out why I'm receiving this error when trying to echo data from a database table, but can't exactly see what I'm doing.

$query = "SELECT * from web_projects"; // Select all rows from web_projects table
$result = mysqli_query ($con,$query);
while ($row = mysqli_fetch_array ($result)) {
foreach($row as $web) {
    echo "<p>Name: ".$web->name."</p>";
    echo "<p>Technologies: ".$web->tech."</p>";
    echo "<p>Description: ".$web->description."</p>";
}
}

Theres only one row in the table, but when compiled I'm getting this:

Notice: Trying to get property of non-object in /Users/leecollings/Sites/leecollings.co/index.php on line 31 Name:

Notice: Trying to get property of non-object in /Users/leecollings/Sites/leecollings.co/index.php on line 32

etc

Have I missed something glaringly obviously?

You are fetching an array, not an object. Use the $array['syntax'], not the $object->syntax. You're also already looping through the array, the $web variable simply isn't an object. — deceze
– deceze ♦, Commented Sep 12, 2014 at 8:19
You should remove the mysql tag here. Your problem is just about php. — Logar
– Logar, Commented Sep 12, 2014 at 8:20
var_dump($row), var_dump($web) to see what you're working with. — deceze
– deceze ♦, Commented Sep 12, 2014 at 8:22
@deceze Ok, so I changed to $web['name'] etc but I'm now getting Warning: Illegal string offset 'name', for all three items. Have I done that right? — Lee
– Lee, Commented Sep 12, 2014 at 8:22

lpg · Accepted Answer · 2014-09-12 08:22:28Z

3

I would remove foreach line, and use array accessing:

$query = "SELECT * from web_projects"; // Select all rows from web_projects table
$result = mysqli_query ($con,$query);
while ($row = mysqli_fetch_array ($result)) {
    echo "<p>Name: ".$row['name']."</p>";
    echo "<p>Technologies: ".$row['tech']."</p>";
    echo "<p>Description: ".$row['description']."</p>";
}

answered Sep 12, 2014 at 8:22

lpg

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Comments

Andrey Chul · Accepted Answer · 2014-09-12 08:22:23Z

2

Error message is very clear: $web is not object. Read the manual more thoroughly: mysqli_fetch_array() returns array or NULL, not object.

answered Sep 12, 2014 at 8:22

Andrey Chul

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cwek · Accepted Answer · 2014-09-12 08:23:19Z

1

You are fetching an array, but you are trying to echo an object. Use this instead:

echo "<p>Name: ".$web['name']."</p>";

Full edited code:

$query = "SELECT * from web_projects"; // Select all rows from web_projects table
$result = mysqli_query ($con,$query);
while ($row = mysqli_fetch_array ($result)) {
    echo "<p>Name: ".$row['name']."</p>";
    echo "<p>Technologies: ".$row['tech']."</p>";
    echo "<p>Description: ".$row['description']."</p>";
}

answered Sep 12, 2014 at 8:23

cwek

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2 Comments

Lee Over a year ago

Thanks for this, I've got it working using the exact same method. :)

cwek Over a year ago

I showed him difference between object and array. This is why I didn't change the variable name.

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