Need help displaying mysql database contents to webpage - php

Question

I need help displaying data from mysql to a webpage, I am coding in php.
My database consists of products which are cars(same type e.g Chevy), right now I have 2 rows (I can add more if I want to), each cars contains the image path, and description.
I can show one row (car) but I am unable to show all rows. I know I have to go through a loop to get all the data from the cars database but I am not sure how to implement it.
This is what I have so far. Assuming I already connected to my database
note: the image path I would like to show the picture in my website.
This is how i would like it to display in my webpage:
$query = "SELECT * FROM cars where cars.carType = 'Chevy' AND \
cars.active = 1";
$numberOfFieds = mysqli_num_fields($result);
$numberOfRows = mysqli_num_rows($result);
/* Gets the contents */
$row = mysqli_fetch_row($result);
$rows = mysqli_fetch_assoc($result);
$fieldcarssontable = array_keys($row);
echo "<table>";
while($row = mysqli_fetch_assoc($result)){
echo "<th>" . $fieldcarssontable[imgPath] . "</th>";
echo "<th>" . $fieldcarssontable[description] . "</th>";
}
echo "</tr>";
echo "</table>";

Answer 1

Just add a while loop. mysqli_fetch_assoc returns a row and moves the internal pointer to the next row until all rows are fetched, then it returns false and the while loop will stop
Pseudo syntax to understand while
while ( this is true ) {
execute this
}
So on your case you can say
while ( $row = mysqli_fetch_assoc( $result ) ) {
// process/output $row
}
mysqli_fetch_assoc and mysqli_fetch_row literally do the same, assoc gives you the array with your result field names as index so this is simpler to access ( $row['name'] rather than $row[0] when using fetch_row )
Have fun! :)
EDIT
// connect to your database server
$link = mysqli_connect('localhost', 'my_user', 'my_password', 'my_db');
// an error occured
if (!$link) {
die('Connect Error (' . mysqli_connect_errno() . ') '
. mysqli_connect_error());
}
// build your query
$query = "SELECT
* # select actual fields instead of *
FROM
cars
WHERE
cars.carType = 'Chevy'
AND
cars.active = 1";
// execute query
$result = mysqli_query($link, $query );
if ( !$result ) {
die( 'no result' );
}
// number of fields
$numberOfFields = mysqli_num_fields($result);
// the field names
$fieldNames = mysqli_fetch_fields($result);
// number of result rows
$numberOfRows = mysqli_num_rows($result);
// watch the content of fieldName and compare it with the table header in the output
print_r( $fieldName );
echo "<table>\n";
// table header, not neccessary to put this into a loop if the query isn't dynamic
// so you actually know your field names - you can echo the header without any variable.
// for the sake of learning about loops I added this
foreach( $fieldNames as $index => $fieldName ) {
echo "\t<th>field #" $index . ", name:" . $fieldName . "</th>\n";
}
// now it's time to walk through your result rows, since we only need to check for "true" a while loop does best
while ( $row = mysqli_fetch_assoc( $result ) ) {
echo "\t<tr><td>" . $row['imgPath'] . "</td><td>" . $row['description'] . "</td></tr>\n";
}
echo "</table>\n";
// remove the result from memory
mysqli_free_result( $result );
mysqli_close( $link );

Answer 2

You misspelled $numberOfFields in your loop, which means you're using a different variable for your loop control. Your loop won't work.
I recommend turning on the error reporting so PHP can catch this stuff for you.

Answer 3

Use this... just while loop
<?php
// Array
while($result = mysql_fetch_assoc($result)) {
//show you fields
echo $result["FieldName"];
}
?>
Or use this proper
<?php
// Edit it as per your query
$query = "SELECT * FROM cars";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while($row = $result->fetch_assoc()) {
//show you fields
echo $row["Name"];
}
/* free result set */
$result->free();
}
/* close connection */
$mysqli->close();
?>