how to php echo individual column values of a result row from mysql query?

Question

The below code works perfectly and gives me the "id" "firstname" and "lastname". What I want is to use a loop to echo all the field values in the result row without having to quote each column name like

$row["id"]

below is the working code

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

 // Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "<br> id: ". $row["id"]. " - Name: ". $row["firstname"]. " " . 
$row["lastname"] . "<br>";
}
} else {
echo "0 results";
}

$conn->close();
?>

any thoughts please

Zedex7 · Accepted Answer · 2017-08-22 06:26:34Z

1

This should work..

while($row = $result->fetch_array()) {
  $i=0;
  while($i<count($row)){
    echo $row[$i];
    $i++;
  }  
}

edited Aug 22, 2017 at 6:26

answered Aug 22, 2017 at 5:35

Zedex7

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9 Comments

Neo Over a year ago

I got an error for this as below: Internal Server Error The server encountered an internal error or misconfiguration and was unable to complete your request. Please contact the server administrator at webmaster@*****.com to inform them of the time this error occurred, and the actions you performed just before this error. More information about this error may be available in the server error log. Additionally, a 500 Internal Server Error error was encountered while trying to use an ErrorDocument to handle the request.

Zedex7 Over a year ago

Its not related to php code.. answer updated.. check again.. notice fetch_array()

Neo Over a year ago

I edited the code to be like below; while($row = $result->fetch_array()) { $i=0; while($i<count($row)){ echo $row[$i]; $i++; } } but it still gives me only the first value "someuser".

Neo Over a year ago

I think the count($row) equals just one. I tried sizeof($row); and still the same result. Anymore thoughts?

Zedex7 Over a year ago

try print_r($row) after $i=0.. and check if it having array of multiple key values..

|

Passionate Coder · Accepted Answer · 2017-08-22 05:36:21Z

1

just use a foreach loop inside while like this

foreach($row as $key=>$value){
 echo "<br> $key: ". $value. "<br>";
}

answered Aug 22, 2017 at 5:36

Passionate Coder

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2 Comments

Neo Over a year ago

This answer gives me only the first column: " Username: Someuser"

Neo Over a year ago

as @Zaheer Attar suggested, I figured out the mistake in my query. Your code works just fine as is and outputs "Username: Someuser" and so on for all columns. Thanks.

Meloman · Accepted Answer · 2017-08-22 07:07:12Z

0

If I understand what you want to do is automate the output of the name of each column with its value (independent of the number of columns you get).

So do it like this :

if ($result->num_rows > 0) {
    foreach($result->fetch_assoc() as $row) { // browse each records
        foreach($row as $col => $value) { // browse each columns on a record
            echo "<br>$col: $value<br>";
        }
    }
} 
else {
    echo "no result";
}

edited Aug 22, 2017 at 7:07

answered Aug 22, 2017 at 6:41

Meloman

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3 Comments

Neo Over a year ago

Tried it, it produces continuously for the number of columns in the record.

Neo Over a year ago

it prints Array()Array()Array()Array()Array() continuously for the number of columns.

Neo Over a year ago

Thanks, but I got the answer.

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