PHP Select statement need quick correction

Question

My echo statement at the end is certainly wrong, I've put it in just to show what I am trying to do, I'd simply like to to call these three variables as set previously. What would be the right way to do that?

$table_id = 'mynewtable';
$query = "SELECT id,name,price FROM $table_id";
$result_set= mysql_query($query);

$id = "SELECT id FROM $table_id";
$name = "SELECT name FROM $table_id";
$price = "SELECT price FROM $table_id";

if ($result_set){

$record= mysql_fetch_row($result_set);
$idrecord= mysql_fetch_row($id);
$namerecord= mysql_fetch_row($name);
$pricerecord= mysql_fetch_row($price);

        foreach ( $record as $value){
        echo "<td><tr> $idrecord $namerecord $pricerecord </tr></td>";
    }
}

Answer 1

There's no point to fetch each of the elements separately. Use mysql_fetch_assoc to get an array with the columns specified in SELECT statement, as keys. this way you can easily get the echo you're after, without querying the database unnecessarily

$table_id   = 'mynewtable';
$query      = "SELECT id,name,price FROM $table_id";
$result_set = mysql_query($query);

if($result_set)
{
    while ($row = mysql_fetch_assoc($result_set)) {
        echo "<td><tr>" .$row["id"]
        . " ". $row["name"]
        . " " .$row["price"] . "</tr></td>";
    }

}

Answer 2

Your $resultset is coming only from the first query ( $query = "SELECT id,name,price FROM $table_id"; ) and not the others?

This section:

$id = "SELECT id FROM $table_id";
$name = "SELECT name FROM $table_id";
$price = "SELECT price FROM $table_id";

is actually useless, as it is just assigning strings to $id , $name and $price .

Did you mean?:

$result_set= mysql_query($query);
while ($row = mysql_fetch_assoc($result_set)) 
{
    echo "<td><tr> ".$row['id']." ".$row['name']." ".$row['price']." </tr></td>";
}

Answer 3

$table_id = 'mynewtable';
$query = "SELECT id,name,price FROM $table_id";
$result_set= mysql_query($query);

if (mysql_num_rows() > 0){

    while($row = mysql_fetch_array($result_set)) {
        echo $row['id'] .", ". $row['name'] .", ". $row['price'];
    }

}

Please start using mysqli_* instead of mysql_*

Answer 4

why not

$table_id = 'mynewtable';
$query = "SELECT id, name, price FROM $table_id";
$result_set= mysql_query($query);

if ($result_set){
        while( $row = mysql_fetch_array(result_set)){ ?>
        <td>
          <tr> <?php echo $row['id'] . ' ' $row['name'] . ' ' . $row['price']; ?> </tr>
        </td>
<?php
    } // end of while
} // end of if

Answer 5

You do not need to have three separate selects for id, name and price. Just use:

if ($result_set) {
    $row = $result_set -> fetch_assoc();
    $id = $row['id'];
    $name = $row['name'];
    $price = $row['price'];
}

Then you can input it into your table or do whatever you would like with the info. The fetch_assoc() will only work if you are using MySQLi . mysql_fetch_assoc is deprecated and you should switch to either PDO or MySQLi

Answer 6

<?php
$query = "SELECT id,name,price FROM mynewtable";
$result_set = mysql_query($query);

if ($result_set!==false){
    while( $row = mysql_fetch_assoc( $result_set ) ) {
     echo '<td>' $row['id'] . ' ' . $row['name'] . ' ' . $row['price'] . '</td>';
    }
}
else {
    echo "Something went wrong with the query:";
    echo mysql_error( );
}

That seems more like it should work :)

EDIT : by the way, use of mysql_* is highly discouraged, as they will disappear in upcoming versions of PHP. Start using PDO or MySQLi instead.