在PHP中显示两个值

Question

Hi have a database with several values in one table but when i try to display the values there is nothing displayed. The connection to the database is OK but the code does not display any values.

The table name is beraknings_varden and the values i want to display on the screen is stigning and diameter .

Here is the code that should display the values

$result = mysqli_query("SELECT * FROM beraknings_varden") or die(mysqli-error());

while ($row = mysqli_fetch_array ( $result)){ 
   echo $row[stigning].", ";
   echo $row[diameter];
   echo "<br />";
}

mysqli_close($opendb);

?>

Answer 1

You have some issues in your code:

• You need to add connection link identifier in mysqli_query function.
• Don't know what is this mysqli-error() this should be mysqli_error() .

Modified code :

$result = mysqli_query($opendb ,"SELECT * FROM beraknings_varden") or die(mysqli_error($opendb)); 

while ($row = mysqli_fetch_array($result))
{ 
    echo $row['stigning'].", "; 
    echo $row['diameter']; echo "<br />"; 
} 
mysqli_close($opendb);

Side note :

You must need to read this manual ( mysqli_query ) .

Answer 2

Here we go, first you have to more careful when doing some database query.

You miss to give the connection mysqli_query($opendb, "your Qry") where its mandatory.

Skip over spaces when use the function of mysql. Look at there, mysqli_fetch_array($result) , what you write mysqli_fetch_array ( $result) .

Your new Query: Try this, may be its helpful.

<?php 
    $result = mysqli_query($opendb, "SELECT `stigning`, `diameter` FROM beraknings_varden");

    while ($row = mysqli_fetch_array($result)){ 
       echo $row['stigning'].", ";
       echo $row['diameter'];
       echo "<br />";
    }
    mysqli_close($opendb);
?>

Simply this is also Okey,

mysqli_fetch_array($result);

But you may use some specification.

// Associative array
$row=mysqli_fetch_array($result,MYSQLI_ASSOC);

// Numeric array
$row=mysqli_fetch_array($result,MYSQLI_NUM);

you can also free the resource.

// Free result set
mysqli_free_result($result);

Answer 3

you are missing qutation inside the square brackets.

echo $row['stigning'].", ";

echo $row['diameter'];