I'm just struggling to have my SQL output loop within a table. Currently, it displays one result within the table and the remaining out of it.
If I put the entire table within the while loop, it creates a new table for each row...
Please help!
Code below
if (mysqli_num_rows($result) > 0)
{
echo
"<table class='xsmall' align='center'>
<tr><th>Costume ID</th>
<th>Description</th>
<th>Size</th></tr>";
while($row = mysqli_fetch_array($result))
{
//-----------------Echoing Results-----------------------//
echo
"<tr><td>".$row['Fname']."</td>
<td>".$row['Sname']."</td>
<td>".$row['AvgSpend']."</td></tr>
</table><br>";
}
}
echo "<table>";
while($row = mysqli_fetch_array($result)) {
echo "
<tr>
<td>'".$row['Fname']."'</td>
<td>'".$row['Sname']."'</td>
<td>'".$row['AvgSpend']."'</td>
</tr>";
}
echo "</table>";
Related
I have a webpage that displays cars from the first car in the table to the last car with a while loop.
I have the following columns: Make, Model, Price. In my syntax I have an anchor tag around the Make rows that links to the description page of the Make you click on.
I want my <h> tags to change to the Model of the corresponding Make.
I've spent over an hour trying to achieve this but all I could come up with is this:
<?php
$query = "SELECT Model FROM inventory;";
$Vtitle = $conn->query($query);
$Vtitle_ar = mysqli_fetch_assoc($Vtitle);
echo "<h1>".$Vtitle_ar['Model']."</h1>";
?>
This works to an extent.
Every anchor I click replaces the <h> tags with only the first result under the Model column in my database.
Here is my code for the the entire car inventory page
<?php
$query = "SELECT * FROM inventory;";
/* Try to query the database */
if ($result = $conn->query($query)) {
// Don't do anything if successful.
}
else {
echo "Error getting cars from database:" .$conn->error()."<br>";
}
// Create the table headers
echo "<table id='Grid' style='width: 80%'><tr>";
echo "<th style='width: 50px'>Make</th>";
echo "<th style='width: 50px'>Model</th>";
echo "<th style='width: 50px'>Asking Price</th>";
echo "</tr>\n";
$class = "odd"; // keep track of whether a row was even or odd, so we can style it later
// Loop through all the rows returned by the query, creating a table for each row
while ($result_ar = mysqli_fetch_assoc($result)) {
echo "<tr class=\"$class\">";
echo "<td><a href='viewcar.php?VIN=".$result_ar['VIN']."'>".$result_ar['Make']."<a></td>";
echo "<td>".$result_ar['Model']."</td>";
echo "<td>".$result_ar['ASKING_PRICE']."</td>";
echo "</td></tr>\n";
// if the last row was even, make the next one odd and vice-versa
if ($class=="odd") {
$class = "even";
}
else {
$class = "odd";
}
}
echo "</table>";
$conn->close();
?>
Does anyone how I can do this?
I'm new to programming and I'm trying to use this for an actual project I'm working on for a hair salon's website
Add a WHERE clause to the query.
$vin = $_GET['VIN'];
$stmt = $conn->prepare("SELECT Model FROM inventory WHERE VIN = ?");
$stmt->bind_param("s", $vin);
$stmt->execute();
$stmt->bind_result($model);
$stmt->fetch();
echo "<h1>$model</h1>";
Though not a solution resolved with the use of a where clause as given by #Barmar whilst formatting the code I did find an error within the HTML which was not immediately obvious
The line echo "</td></tr>\n"; has an extra </td> which would break the flow of the html and can have detrimental effects. Also, // Don't do anything if successful. makes no sense - if there are results then process the recordset otherwise show the error ;-)
<?php
$query = "SELECT * FROM inventory;";
if ( $result = $conn->query( $query ) ) {
echo "
<table id='Grid' style='width: 80%'>
<tr>
<th style='width: 50px'>Make</th>
<th style='width: 50px'>Model</th>
<th style='width: 50px'>Asking Price</th>
</tr>";
$i=0;
while( $result_ar = mysqli_fetch_assoc( $result ) ) {
$class = $i %2 == 0 ? 'even' : 'odd';
echo "
<tr class='$class'>
<td><a href='viewcar.php?VIN={$result_ar['VIN']}'>{$result_ar['Make']}<a></td>
<td>{$result_ar['Model']}</td>
<td>{$result_ar['ASKING_PRICE']}</td>
</tr>";
$i++;
}
echo "</table>";
} else {
echo "Error getting cars from database:" .$conn->error()."<br>";
}
$conn->close();
?>
For styling alternate table rows ( the above uses a modulus function to calculate odd/even ) you can do it with some simple CSS - such as
tr:nth-child( odd ){/* rules */}
I have a php code to print out my table including its column name. The printing has to be dynamic because it has to print different size/length tables based on a user input. :
<table>
<?php
while ($row = mysqli_fetch_array($results)) {
while ($fieldInfo = mysqli_fetch_field($results)) { ?>
<th> <?php echo $fieldInfo->name; ?> </th>
<td> <?php echo $row[$fieldInfo->name] ?> </td>
<?php }
} ?>
</table>
this is the query for $results:
$tName = $_POST["tableNames"]; //this data is recieved from another page
require_once("conn.php");
$sql = "SELECT * FROM $tName";
$results = mysqli_query($conn, $sql)
or die ('Problem with query' . mysqli_error($conn));
my code correctly prints out the table name as well as the first row data but it is not formatted correctly here is how it looks:
additionally. for some reason it only prints out the first row even though im using a while loop.
My advice to you is to prepare two arrays:
First one: containing column names and second: containing data.
When use two foreach to generate first row with header and second one to display data. You have forgot to add <tr> tags to divide rows.
Use
The mysqli_fetch_field() function returns the next column in the result set as an object. It will only returns all column names not the records of table.
You need to use mysqli_fetch_array() for getting all records:
while ($info = mysqli_fetch_array($results,MYSQLI_ASSOC)) {
{
echo $info['rid'];
echo $info['level'];
....
}
I ended up with using a taras' suggestion of storing the column names in an array:
<table>
<?php
while ($fieldInfo = mysqli_fetch_field($results)) { ?>
<th> <?php echo $fieldInfo->name; ?> </th>
<?php
$colNames[] = $fieldInfo->name;
?>
<?php }
while ($row = mysqli_fetch_array($results)) { ?>
<tr>
<?php for ($i=0; $i<sizeof($colNames); $i++) { ?>
<td><?php echo $row[$colNames[$i]] ?>
<?php } ?>
</tr>
<?php } ?>
</table>
As of my understand, do you want to display all table and their columns?
So you can format like below
$sql = "SHOW TABLES FROM dbname";
$result_tables = mysqli_query($link, $sql);
echo "<table border=1>";
echo "<tr><td>Table name</td><td>Fields name</td></tr>";
while($row = mysqli_fetch_array($result_tables)) {
echo "<tr>";
echo "<td>".$row[0]."</td>";
$sql2 = "SHOW COLUMNS FROM ".$row[0];\\row[0] is used to get table name
$result_fields = mysqli_query($link, $sql2);
echo "<td>";
while($row2 = mysqli_fetch_array($result_fields)) {
echo $row2['Field'].',';
}
echo "</td>";
echo "</tr>";
}
Hi so im trying to put all of the useres in a database, table into a html table this is what i have:
<table class="table table-striped">
<thead>
<tr>
<th>UUID</th>
<th>Full Name</th>
<th>Email</th>
<th>Access Key</th>
<th>Phone Number</th>
<th>Activated</th>
<th>Role</th>
</tr>
</thead>
<tbody>
<?php
include_once('inc/conf/databaseConnect.php');
$query = mysql_query("SELECT * FROM list_users ORDER by id");
while($row = mysql_fetch_array($query)){
echo "<tr>";
echo "<td>".$row['uuid']."</td>";
echo "<td>".$row['firstname'].$rowtwo['lastname']."</td>";
echo "<td>".$row['email']."</td>";
echo "<td>".$row['security_key']."</td>";
echo "<td>".$row['phone_no']."</td>";
echo "<td>".$row['activated']."</td>";
echo "<td>".$row['role']."</td>";
echo "</tr>";
}
?>
</tbody>
</table>
This dosnt return anything, or any errors. It connects to the database correctly ive checked that its just not retrieving the users.
Image of database structure
databaseConnect.php:
<?php
//Create Connnection
$sqlLink = mysqli_connect('localhost', 'root', 'classified', 'user_details');
//If Error Connecting
if(!$sqlLink) {
die('<center><br><h3>Error connecting to servers Database.');
}
?>
After seeing your edit because you did not originally show us which connection method you were using; the almighty answer here (least the most important one) is that you can't mix different MySQL APIs.
Can I mix MySQL APIs in PHP?
You need to use the same one from connection to query.
Plus that $rowtwo should be $row as I stated in comments along with my asking about what was inside your databaseConnect.php file.
Get to work with prepared statements also to help protect against an SQL injection.
https://en.wikipedia.org/wiki/Prepared_statement
Thanks. I have fixed the issue i updated to using mysqli's methods
<?php
include_once('inc/conf/databaseConnect.php');
$query = $sqlLink->query("SELECT * FROM list_users ORDER by id");
while($row = $query->fetch_array()){
echo "<tr>";
echo "<td>".$row['uuid']."</td>";
echo "<td>".$row['firstname'].$row['lastname']."</td>";
echo "<td>".$row['email']."</td>";
echo "<td>".$row['security_key']."</td>";
echo "<td>".$row['phone_no']."</td>";
echo "<td>".$row['activated']."</td>";
echo "<td>".$row['role']."</td>";
echo "</tr>";
}
?>
At least check for errors:
if ($query = mysql_query("SELECT * FROM list_users ORDER by id");) {
...
} else {
echo '<b>MySQL error:</b><br>' . mysql_error() . '<br />';
}
You must use mysql_fetch_assoc() instead of mysql_fetch_array or fetch like this mysql_fetch_array($result, MYSQL_ASSOC).
I have a site where members can mark other members as 'a favourite'. User are able to search the members table in various ways and I want to show from any of the results that are returned whether or not the users returned are favourites of the current user.
This is some very simplified code I have been using to try and get this query to work but I just can't figure it out. Whenever I add 'GROUP BY' to avoid duplicate results from my LEFT JOIN the 'if' statement does not work. The 'if' statment does work however, if I omit the 'GROUP BY' but I get all rows from members table and the favourites table. Thanks.
$result = mysqli_query($db_conx, "SELECT members.*, user_favourites.* FROM members LEFT JOIN user_favourites ON members.id = user_favourites.fav_id GROUP BY members.id");
echo "<table border=''>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
<th>A favourite of User</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
if ($visitor == $userid ){
$msgs = "x";
}
else { $msgs = "0";
}
echo "<td>". $msgs. "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
I suggest to you, tu use 2 query's insted one. Simple query is a fast querys. If your site scale up, your query will be slow query and it cause performance problems.
Idea:
For one hand: SELECT FROM members, get all and put it inside array, key=member.id
$members[$row['member_id']]=$row;
On the other hand: SELECT * FROM user_favourites, and put it inside the previous array, refereced by the key fav_id use distinc if you have duplicates.
$members[$row['fav_id']]['favourites']=$row;
Perfect now you have all you need, an array with all information, iterate it.
JilianJ something like this:
<?
//Prepare
$all_users=array()
$query_users='SELECT * from user';
$query_favourites='SELECT * FROM user_favourites';
//Now I find all members information
$users=mysqli_query($db_conx,$query_users);
while($user = mysqli_fetch_array($users)) {
$all_users[$user['id']]=$user;
}
//Now I add favourite information to the users information
$favourites=mysqli_query($db_conx,$query_favourites);
while($favourite = mysqli_fetch_array($favourites)) {
$all_users[$favourite['fav_id']]['fovourite'][]=$favourite['fav_id'];
}
?>
<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Email</th>
<th>A favourite of User</th>
</tr>
<? foreach ($information as $key=>$item) {?>
<tr>
<td><?=$item['firstname'];?></td>
<td><?=$row['lname'];?></td>
<td><?=$row['email'];?></td>
<td>
<? foreach($item['favourites'] as $key2=>$item2) { ?>
<p><?=$all_members[$item2]['name];?></p>
<? } ?>
</td>
</tr>
<? } ?>
</table>
Good luck
I've been doing a lot of basic stuff with PHP, but have come up with a new twist on how I want to display some tabular data. I looked at the basic rules for constructing pivot/cross tab displays, but what I'm looking for seems to be a bit different. The primary data is stored in 1 table as (along with other fields):
the station
ID of ambulance at that station
current status of ambulance (available, unavailable, out of service)
My present code outputs this...but looking for this:
http://www.countryhicksystems.com/upload/stack.jpg
I know we're to use PDO or MySQLi, but I set this up as a quick prototype.
//Log in, etc. code before this
$query = "SELECT * FROM $usertable WHERE $ems = $editID
Order by equip_station, equip_descrp";
$result = mysql_query($query);
if($result)
{
echo "<table width=\"90%\" border='1' align=\"left\" height=\"Auto\" cellpadding=\"3\" cellspacing=\"3\">
<tr>
<th width=\"65\"><div align=\"center\">Station</div></th>
<th >Status</th>
<th >As of Date</th>
<th >Description</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td class=\"columnfont\">" .$row["$name"]."</div></td>";
if ($row['equip_status'] == 'Available')
{
echo "<td class=\"green\">" .$row["$id"]."</div></td>";
}
elseif ($row['equip_status'] == 'Unavailable')
{
echo "<td class=\"red\">" .$row["$id"]."</div></td>";
}
else
{
echo "<td class=\"black\">" .$row["$id"]."</div></td>";
}
echo "<td class=\"cellfont\">" .$row["$availdate"]."</div></td>";
echo "<td class=\"cellfont\">" .$row["$descr"]."</div></td>";
echo "</tr>";
}
}
?>
</p>
</div>
How would I change my present code to display Station Ambulances across a row instead of down a column?
Thanks in advance everyone.