I am trying to loop through a table in my database and show all the details in a table. Firstly, it should loop through my main table, 'TBook', and get the 'date', 'period', roomID', and 'teacherinitials'. Then, using the roomID, it should look in my other table, 'Rooms', to get the 'room' name and the 'description'. After that it should display the 'date', 'period', 'room' & 'description', and 'teacherinitials'.
This is my code:
<?php
// Create connection
$con=mysqli_connect("host","user","pass","database");
// Check connection
if (mysqli_connect_errno($con)) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//Get number of rows
$sql="SELECT * FROM TBook";
$result=mysqli_query($con, $sql);
$rowcount=mysqli_num_rows($result);
$sql2="SELECT room, description FROM Rooms WHERE roomID = $roomID";
$res=mysqli_query($con,$sql2);
//Start table
echo "<table>";
echo "<tr><th>Date</th><th>Period</th><th>Room</th><th>Teacher Initials</th></tr>";
// Loop through database
while ($row = $result->fetch_assoc()) {
$row = mysql_fetch_array($result);
$date = $row['date'];
$period = $row['period'];
$roomID = $row['roomID'];
$teacherinitials = $row['teacherinitials'];
while ($row2 = $res->fetch_assoc()) {
$room = $row2['room'];
$description = $row2['description'];
}
// Show entries
echo "<tr>
<td>".$date."</td>
<td>".$period."</td>
<td>".$room." (".$description.")</td>
<td>".$teacherinitials."</td>
</tr>";
}
echo "</table>";
?>
However, instead I get an error saying "Fatal error: Call to a member function fetch_assoc() on a non-object in /home/user/public_html/my_bookings-results.php on line 56". Line 56 is this line:
while ($row2 = $res->fetch_assoc()) {
It does show the table headers beneath that, but nothing else. What is going wrong?
change first while loop to
while ($row = mysql_fetch_array($result))
remove below line after while loop
$row = mysql_fetch_array($result);
change:
while ($row2 = $res->fetch_assoc()) {
to
while ($row2 = $res->fetch_array($res)) {
Related
$con=mysqli_connect("localhost","root","","database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
error_reporting(E_ALL ^ E_NOTICE);
$query = "SELECT * FROM TT_posts WHERE post_status='publish' AND
ping_status='open'";
$result = $con->query($query);
while($row1 = $result->fetch_assoc())
foreach ($result as $row1){
$image = "SELECT * FROM TT_posts WHERE post_title='$row1[post_name]'";
$result1 = $con->query($image);
while($row2 = $result1->fetch_assoc())
foreach ($result1 as $row2){
echo "<img src='".$row2[guid]."'>";
echo "<p>".$row1[post_title]."</p>";
}}
?>
actually, below query returns 8 results.
$query = "SELECT * FROM TT_posts WHERE post_status='publish' AND
ping_status='open'";
When it executed the loop, it stops at the first result. I don't know what exactly stops the code.
I think you've got your variable names mixed up. Your first while assigns a row to $row1, then your foreach refers to $result, not $row1. The variable $row1 is actually being overwritten every time it loops around the foreach.
You don't actually need either foreach:
while ($row1 = $result->fetch_assoc()) {
$image = "SELECT * FROM TT_posts WHERE post_title='$row1[post_name]'";
$result1 = $con->query($image);
while ($row2 = $result1->fetch_assoc()) {
echo "<img src='".$row2[guid]."'>";
echo "<p>".$row1[post_title]."</p>";
}
}
I have been trying to create a PHP function that can display a MySQL table as a HTML table using PHP. So far I am able to output any table I choice, yet I am encountering a problem when the MySQL table contains empty rows, because empty cells result in the HTML tables. My code is as such:
<?php
function getTABLE(){
$db_host = 'HOST.com';
$db_user = 'USER1';
$db_pwd = 'PASSWORD';
$database = 'testdb';
$table = 'FAQTable';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
//// sending query and only result cell that are not NULL
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h3><center>Table: {$table}</h3>";
echo "<table border='1'><tr>";
//// printing table headers
for($i=0; $i<$fields_num; $i++){
$field = mysql_fetch_field($result);
echo "<td>{$field->name}</td>";
}
//// printing table rows
while($row = mysql_fetch_row($result)){
echo "<tr>";
//// $row is array... foreach( .. ) puts every element
//// of $row to $cell variable
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>";
}
print "</TABLE>";
mysql_close();
}
print getTABLE();
?>
My dilemma is in the "printing table rows" section of the code. I am hoping there is a way in the while($row = mysql_fetch_row($result)) to only accept rows that have values in them. Any ideas?
I have already tried using the following lines with no luck:
$result = mysql_query("SELECT * FROM {$table} WHERE * IS NOT NULL");
$result = mysql_query("SELECT COUNT(id) FROM {$table} where answer IS NOT NULL or answer <>'' ");
$result = mysql_query('SELECT COUNT(*) FROM {$table} WHERE answer <> ""');
$result = mysql_query("SELECT * FROM {$table} WHERE CHAR_LENGTH>0");
$result = mysql_query("SELECT * FROM {$table} WHERE val1 is <> '' ");
$result = mysql_query("SELECT * FROM {$table} WHERE col1 is <> '' ");
//// Outputs funky count in a separate table, but not the desired table with no empty cells
$result = mysql_query("SELECT COUNT(answer) FROM {$table} WHERE CHAR_LENGTH(answer)>0");
$result = mysql_query("SELECT COUNT(answer) FROM {$table} WHERE LENGTH(answer)>0");
$reslts = mysql_query("SELECT * FROM {$table}");
while($row = mysql_fetch_row($reslts)){
$empty_count = 0;
$count = count($row);
for($i = 0; $i < $count; $i++)
if($row[$i] === '' || $row[$i] === 'NULL')
$empty_count++;
$result = ($count);
}
So Thanks To Paul Spiegal for helping with this PHP function that can output any MySQL Table into a HTML Table to be displayed on a website... The working function is as follows, just change the values for the variables to access any MySQL data:
function getTABLE(){
$db_host = 'www.host.com';
$db_user = 'user1';
$db_pwd = 'password';
$database = 'testdb';
$table = 'MyTable';
if (!mysql_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
//// sending query
$result = mysql_query("SELECT * FROM {$table}");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($result);
echo "<h3><center>Table: {$table}</h3>";
echo "<table border='1'><tr>";
//// printing table headers
for($i=0; $i<$fields_num; $i++){
$field = mysql_fetch_field($result);
echo "<td>{$field->name}</td>";
}
//// printing table rows
while($row = mysql_fetch_row($result)){
echo "<tr>";
if (strlen(implode('', $row )) == 0) {
continue;
}else {
foreach($row as $cell)
echo "<td>$cell</td>";
echo "</tr>";
}
}
print "</TABLE>";
mysql_close();
}
print getTABLE();
Using implode() function, you can combine all cells into one string. If that string is empty, you skip printing that row.
while($row = mysql_fetch_row($result)){
if (strlen(implode('', $row)) == 0) {
continue; // skip this empty row
} else {
// TODO: print this row
}
}
The code below works fine for printing one record from a database table, but what I really want to be able to do is print all the records in the mysql table in a format similar to my code.
I.E.: Field Name as heading for each column in the html table and the entry below the heading. Hope this is making sense to someone ;)
$raw = mysql_query("SELECT * FROM tbl_gas_meters");
$allresults = mysql_fetch_array($raw);
$field = mysql_query("SELECT * FROM tbl_gas_meters");
$num_fields = mysql_num_fields($raw);
$num_rows = mysql_num_rows($raw);
$i = 1;
print "<table border=1>\n";
while ($i < $num_fields)
{
echo "<tr>";
echo "<b><td>" . mysql_field_name($field, $i) . "</td></b>";
//echo ": ";
echo '<td><font color ="red">' . $allresults[$i] . '</font></td>';
$i++;
echo "</tr>";
//echo "<br>";
}
print "</table>";
Just as an additional piece of information you should probably be using PDO. It has more features and is helpful in learning how to prepare SQL statements. It will also serve you much better if you ever write more complicated code.
http://www.php.net/manual/en/intro.pdo.php
This example uses objects rather then arrays. Doesn't necessarily matter, but it uses less characters so I like it. Difference do present themselves when you get deeper into objects, but not in this example.
//connection information
$user = "your_mysql_user";
$pass = "your_mysql_user_pass";
$dbh = new PDO('mysql:host=your_hostname;dbname=your_db;charset=UTF-8', $user, $pass);
//prepare statement to query table
$sth = $dbh->prepare("SELECT name, colour FROM fruit");
$sth->execute();
//loop over all table rows and fetch them as an object
while($result = $sth->fetch(PDO::FETCH_OBJ))
{
//print out the fruits name in this case.
print $result->name;
print("\n");
print $result->colour;
print("\n");
}
You probably also want to look into prepared statements. This helps against injection. Injection is bad for security reasons. Here is the page for that.
http://www.php.net/manual/en/pdostatement.bindparam.php
You probably should look into sanitizing your user input as well. Just a heads up and unrelated to your current situation.
Also to get all the field names with PDO try this
$q = $dbh->prepare("DESCRIBE tablename");
$q->execute();
$table_fields = $q->fetchAll(PDO::FETCH_COLUMN);
Once you have all the table fields it would be pretty easy using <div> or even a <table> to arrange them as you like using a <th>
Happy learning PHP. It is fun.
Thanks guys, got it.
$table = 'tbl_gas_meters';
$result = MYSQL_QUERY("SELECT * FROM {$table}");
$fields_num = MYSQL_NUM_FIELDS($result);
ECHO "<h1>Table: {$table}</h1>";
ECHO "<table border='1'><tr>";
// printing table headers
FOR($i=0; $i<$fields_num; $i++)
{
$field = MYSQL_FETCH_FIELD($result);
ECHO "<td>{$field->name}</td>";
}
ECHO "</tr>\n";
// printing table rows
WHILE($row = MYSQL_FETCH_ROW($result))
{
ECHO "<tr>";
// $row is array... foreach( .. ) puts every element
// of $row to $cell variable
FOREACH($row AS $cell)
ECHO "<td>$cell</td>";
ECHO "</tr>\n";
}
while ( $row = mysql_fetch_array($field) ) {
echo $row['fieldname'];
//stuff
}
Try this :
$raw = mysql_query("SELECT * FROM tbl_gas_meters");
$allresults = mysql_fetch_array($raw);
$field = mysql_query("SELECT * FROM tbl_gas_meters");
while($row = mysql_fetch_assoc($field)){
echo $row['your field name here'];
}
Please note that, mysql_* functions are deprecated in new php version , so use mysqli or PDO instead.
Thanks! I adapted some of these answers to draw a table from all records from any table, without having to specify the field names. Just paste this into a .php file and change the connection info:
<?php
// Authentication detail for connection
$servername = "localhost";
$username = "xxxxxxxxxx";
$password = "xxxxxxxxxx";
$dbname = "xxxxxxxxxx";
$tablename = "xxxxxxxxxx";
$orderby = "1 DESC LIMIT 500"; // column # to sort & max # of records to display
// Create & check connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error); // quit
}
// Run query & verify success
$sql = "SELECT * FROM {$tablename} ORDER BY {$orderby}";
if ($result = $conn->query($sql)) {
$conn->close(); // Close table
$fields_num = $result->field_count;
$count_rows = $result->num_rows;
if ($count_rows == 0) {
die ("No data found in table: [" . $tablename . "]" ); //quit
}
} else {
$conn->close(); // Close table
die ("Error running SQL:<br>" . $sql ); //quit
}
// Start drawing table
echo "<!DOCTYPE html><html><head><title>{$tablename}</title>";
echo "<style> table, th, td { border: 1px solid black; border-collapse: collapse; }</style></head>";
echo "<body><span style='font-size:18px'>Table: <strong>{$tablename}</strong></span><br>";
echo "<span style='font-size:10px'>({$count_rows} records, {$fields_num} fields)</span><br>";
echo "<br><span style='font-size:10px'><table><tr>";
// Print table Field Names
while ($finfo = $result->fetch_field()) {
echo "<td><center><strong>{$finfo->name}</strong></center></td>";
}
echo "</tr>"; // Finished Field Names
/* Loop through records in object array */
while ($row = $result->fetch_row()) {
echo "<tr>"; // start data row
for( $i = 0; $i<$fields_num; $i++ ) {
echo "<td>{$row[$i]}</td>";
}
echo "</tr>"; // end data row
}
echo "</table>"; // End table
$result->close(); // Free result set
?>
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>";
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 );
You misspelled $numberOfFields in your loop, which means you're using a different variable for your loop control. Your loop won't work.
I recommend turning on the error reporting so PHP can catch this stuff for you.
Use this... just while loop
<?php
// Array
while($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();
?>
I have a large database of venues - and I would like to display this data in one page that would only change in some sort of an attribute to the id: (Ex: venues.php?id=1, which would get all the data from row #1.)
Edit: Okay, I updated the code and this is what it looks like now:
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>
And upon going to venues.php?id=1 I get this error:
Parse error: syntax error, unexpected T_LOGICAL_OR in
/home/nightl7/public_html/demos/venues/venues.php on line 8
Do you mean something like:
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
In order to "pass" the id into your url "venues.php?id=1"
You need to use a hybrid html/php form with method=get.
You can see an example html form here: w3schools html forms
This is what I would do:
print '<form name="input" action="venues.php" method="get">';
print 'Venue: <select name = "id">';
$con = mysql_connect("","","");
mysql_select_db($dataBase);
if (!$con){die('Could not connect: ' . mysql_error());}
else {
$opt = array();
$optVal = array();
$i = 0;
$sql = "Select * from venues";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
$opt[$i] = $row['VenueName'];
$optVal[$i] = $row['VenueID'];
print "<option value='$optVal[$i]'>$opt[$i]</option>";
$i++;
}
}
mysql_close($con);
print '</select><br />';
print '<input type="submit" value="Submit" />';
print '</form>'
This will give you a form that will give you a drop down list of all your venues and once a venue is selected will direct you to the venues.php page with the respective id.
at the top of your venues,php page just use
$id = $_GET['id'];
This assigns the id number to the variable $id and then you can use this "select"
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) ;
To collect your venue name from your database using the id supplied in the form.
Good Luck :)
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>