I executed this code in localhost but it kept looping the table infinitely. How do I fix this? Is there something I should add? The looping never stops until I click the x button to stop loading the page.
//fetching data in descending order (lastest entry first)
//$result = mysql_query("SELECT * FROM users ORDER BY id DESC"); // mysql_query is deprecated
$query=("SELECT * FROM songs ORDER BY songid DESC");
$result = mysql_query($query) or die(mysql_error());
?>
<html>
<head>
<title>View</title>
</head>
<body>
Add New Data<br/><br/>
<table width='80%' border=0>
<tr bgcolor='#CCCCCC'>
<td>Song ID</td>
<td>Title</td>
<td>Artist</td>
<td>Genre</td>
<td>Language</td>
<td>Lyrics</td>
<td>Updated by</td>
</tr>
<?php
$counter == 1;
$res = mysql_fetch_array($result);
//while($res = mysql_fetch_array($result)) { // mysql_fetch_array is deprecated, we need to use mysqli_fetch_array
do {
echo "<tr>";
echo "<td>".$res['songid']."</td>";
echo "<td>".$res['title']."</td>";
echo "<td>".$res['artist']."</td>";
echo "<td>".$res['genre']."</td>";
echo "<td>".$res['language']."</td>";
echo "<td>".$res['lyrics']."</td>";
echo "<td>".$res['update']."</td>";
echo "<td>Edit | Delete</td>";
} while($res)
?>
</table>
You are missing $res = mysql_fetch_array($result); at the end of the loop. Currently, the while statements checks if $res is (still) truthy, but it never changes, so it will always evaluate to TRUE (and so, the loop will always continue to run).
I was just about to write the same thing as Daan Meijer. res never changes and so its always true for the while-statement. I just want to add that mysql_fetch_assoc($result) works a littlebit faster than mysql_fetch_array($result). So if you have a big database, mysql_fetch_assoc($result) is the better choice.
Related
I have implemented a table from data tables
Link [https://datatables.net/]
i would like to use two tables in one site with different columns and datas in the columns after the mysqli connection i insert the data sets with a while mysqli fetch array function the first table works properly
"urlaubstage" -> is correct
but table 2
no matter what i do even var_dump i dint not get any reaction but the table is display correctly on the page but with empty columns
This is the html code
<table id="table_id" class="display">
<thead>
<tr>
<th>Urlaubstage Jahr</th>
<th>Urlaubstage Anspruch</th>
<th>Urlaubstage Beansprucht</th>
<th>Urlaubstage Rest</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysqli_fetch_array($sql)){
$urlaubsTage = $row[4];
echo "<tr>";
echo "<td>{$urlaubsTage}</td>";
echo "<td>Anspruch</td>";
echo "<td>Beansprucht</td>";
echo "<td>Rest</td>";
echo "halllo";
}
echo "</tr>";
?>
</tbody>
</table>
!!!!!!!!!!!!<p>HERE STARTS THE SECOND TABLE</p>!!!!!!!!!!!!!!!!!!!!!!!!
<table id="table_id2" class="display">
<thead>
<tr>
<th>Urlaub Antragsdatum</th>
<th>Urlaub Startdatum</th>
<th>Urlaub Enddatum</th>
<th>Urlaubs Status</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysqli_fetch_array($sql)){
var_dump($row); -> EVEn VAR_DUMP IS NOT SHOWN
echo "<tr>";
echo "<td>Antragsdatum</td>";
echo "<td>Startdatum</td>";
echo "<td>Enddatum</td>";
echo "<td>Status</td>";
echo "halllo";
}
echo "</tr>";
?>
</tbody>
</table>
JQUERY CODE
....
$('#table_id').DataTable();
//FUNKTION FÜR ZWEITE TABELLE
$('#table_id2').DataTable();
....
Picture of Code and Table
ok i got the answer by myself, after the first loop runs the query $sql with mysqli_fetch_array($sql) you cant use it anymore on the second loop why? cause it ran on the first loop and its over solution
rename;
$sql = mysqli_query(same query);
$sql2 = mysqli_query(same query);
first loop while($row = mysqli_fetch_array($sql);
second loop while($row = mysqli_fetch_array($sql2);
You don't need to execute the same query twice. You don't need the while loop either. Try to get the results into an array and then loop the array multiple times.
$result = $conn->query($sql)->fetch_all();
//...
foreach ($result as $row) {
//...
}
// Repeat the same loop again without calling query again
//foreach...
I really can't figure out what I'm doing wrong here. I'm doing a query to check whether there are records in a DB table called 'newCards'.
With $count I check how many results it's returning: it shows me '1'. But the while loop isn't returning ANY thing. The only things I'm seeing are the <th>'s at the top of the table, but no table records are present, while $count is giving '1' as a result. Which is true, cause there is actually 1 record present in DB.
How can I fix this?
<?php
$query = $db->prepare("SELECT * FROM `newCards` WHERE `company` = :companyID");
$query->bindParam(":companyID", $enterprise['id'], PDO::PARAM_INT);
$query->execute();
$count = $query->rowCount();
echo $count;
if(empty($query->fetch())){
echo "Geen gevonden";
} else {
?>
<table>
<tr>
<th>Ontvanger</th>
<th>Saldo</th>
<th></th>
</tr>
<?php
while($result = $query->fetch()){
?>
<tr>
<td><?php echo $result['id']; ?></td>
<td>2</td>
<td>3</td>
</tr>
<?php
}
?>
</table>
<?php
}
?>
$query->fetch() already fetches a record. So next call to fetch() fetches next record or nothing if there're no records. In your case with one record second fetch() fetches nothing, so while never starts.
You can change your code to:
if($count){?>
<table>
<tr>
<th>Ontvanger</th>
<th>Saldo</th>
<th></th>
</tr>
<?php
while($result = $query->fetch()){
// show row
}?>
</table>
} else {
// echo that no rows found
}
I think fetch in first if is executed so that is why second returns empty,
try to assign it to var before conditions or check wit $cont var
I believe you want to return an array indexed by column names with
->fetch(PDO::FETCH_ASSOC)
More information can be found here http://php.net/manual/en/pdostatement.fetch.php
Because fetch() fetches the first row, even when checking in empty(), it will try to fetch the next row when you use while($result = $query->fetch()){. You can either check the value from $count (like shown by u_mulder), but you should beware of the note in the manual for rowCount() (emphasis mine)
If the last SQL statement executed by the associated PDOStatement was a SELECT statement, some databases may return the number of rows returned by that statement. However, this behavior is not guaranteed for all databases and should not be relied on for portable applications.
You can use a do..while structure and check if the fetch was successful or not instead. If you change out if(empty($query->fetch())){ with if (!$row = $query->fetch()) {, you check if there was a row fetched or not (as fetch() returns null on an empty result). Then $row is ready to use, and you can use it before the first loop takes place.
<?php
$query = $db->prepare("SELECT * FROM `newCards` WHERE `company` = :companyID");
$query->bindParam(":companyID", $enterprise['id'], PDO::PARAM_INT);
$query->execute();
$count = $query->rowCount();
echo $count;
if (!$row = $query->fetch()) {
echo "Geen gevonden";
} else {
?>
<table>
<tr>
<th>Ontvanger</th>
<th>Saldo</th>
<th></th>
</tr>
<?php
do {
?>
<tr>
<td><?php echo $result['id']; ?></td>
<td>2</td>
<td>3</td>
</tr>
<?php
} while ($result = $query->fetch());
?>
</table>
<?php
}
PHP.net on PDOStatement::rowCount()
PHP.net on do..while
I know how to produce results one after another but how do you separate them? So in my sql I'm selecting * from table and limiting it to 4
$sql = "SELECT * FROM table limit 4";
$result = $conn->query($sql);
while($row = $result->fetch_assoc())
{$rows['id']=$row;};
$price = $row['price'];
I dont seem to get any result, any suggestions, sorry guys beginner
...<?php echo $id ?></font></span>
<h4><?php echo $price ?></h4></div>
<div class="planFeatures"><ul>
<li><h1><?php echo $id=2 ?></h1></li>//how do I echo the next id?
<li><?php echo $price2 ?></li> //also the next price which id now is also 2
//and so on......
How do I display the next increments results in a different area of the same page, within another div?
I do get results if I sql and re-select all over again (and say id=2) but I'm sure there is a better way of doing it because I've already got my 4 results with my limit.
It seems you are not saving the results from the query result properly. Each iteration of the loop overwrites the same bucket in the $rows array. Instead, you need to add elements to the $rows array; this will produce an indexed array. Then you can iterate over it and generate the HTML content.
<?php
// Perform query.
$sql = "SELECT * FROM table limit 4";
$result = $conn->query($sql);
// Fetch results
while (true) {
$row = $result->fetch_assoc();
if (!$row) {
break;
}
$rows[] = $row;
}
// Generate HTML content using $rows array.
?>
<table>
<thead>
<tr>
<th>ID</th>
<th>Price</th>
</tr>
</thead>
<tbody>
<?php foreach ($rows as $row):?>
<tr>
<td>ID: <?php print $row['id'];?></td>
<td>Price: <?php print $row['price'];?></td>
</tr>
<?php endforeach;?>
</tbody>
</table>
I took some liberty in the above example and generated a simple HTML table. Of course you can modify this to generate whatever you want.
I hope I've interpreted your question accurately, apologies if not!
So I'm trying to make a HTML table that gets data from a MySQL database and outputs it to the user. I'm doing so with PHP, which I'm extremely new to, so please excuse my messy code!
The code that I'm using is: braces for storm of "your code is awful!"
<table class="table table-striped table-hover ">
<thead>
<tr>
<th>#</th>
<th>Name</th>
<th>Description</th>
<th>Reward</th>
<th>Column heading</th>
</tr>
</thead>
<tbody>
<?php
$con = mysql_connect("localhost", "notarealuser", 'notmypassword');
for ($i = 1; $i <= 20; $i++) {
$items = ($mysqli->query("SELECT id FROM `items` WHERE id = $i"));
echo ("<tr>");
echo ("
<td>
while ($db_field = mysqli_fetch_assoc($items)) {
print $db_field['id'];
}</td>");
$items = ($mysqli->query("SELECT name FROM `items` WHERE id = $i"));
echo ("
<td>
while ($db_field = mysqli_fetch_assoc($items)) {
print $db_field['name'];
}</td>");
$items = ($mysqli->query("SELECT descrip FROM `items` WHERE id = $i"));
echo ("
<td>
while ($db_field = mysqli_fetch_assoc($items)) {
print $db_field['descrip'];
}</td>");
$items = ($mysqli->query("SELECT reward FROM `items` WHERE id = $i"));
echo ("
<td>
while ($db_field = mysqli_fetch_assoc($items)) {
print $db_field['reward'];
}</td>");
$items = ($mysqli->query("SELECT img FROM `items` WHERE id = $i"));
echo ("
<td>
while ($db_field = mysqli_fetch_assoc($items)) {
print $db_field['img'];
}</td>");
echo ("</tr>");
}
?>
</tbody>
</table>
However, this code is not working - it simply causes the page to output an immediate 500 Internal Server Error. IIS logs show it as a 500:0 - generic ISE. Any ideas?
You are mixing mysql and mysqli, not closing php code block and you are not selecting a database. Plus you don't have to run a query for each field
Try this:
<table class="table table-striped table-hover ">
<thead>
<tr>
<th>#</th>
<th>Name</th>
<th>Description</th>
<th>Reward</th>
<th>Column heading</th>
</tr>
</thead>
<tbody>
<?php
$con = new mysqli("host","user", "password", "database");
$execItems = $con->query("SELECT id, name, descrip, reward, img FROM `items` WHERE id BETWEEN 1 AND 20 ");
while($infoItems = $execItems->fetch_array()){
echo "
<tr>
<td>".$infoItems['id']."</td>
<td>".$infoItems['name']."</td>
<td>".$infoItems['descrip']."</td>
<td>".$infoItems['reward']."</td>
<td>".$infoItems['img']."</td>
</tr>
";
}
?>
</tbody>
</table>
<table class="table table-striped table-hover">
<thead>
<tr>
<th>#</th>
<th>Name</th>
<th>Description</th>
<th>Reward</th>
<th>Column heading</th>
</tr>
</thead>
<tbody>
<?php
$con = mysqli_connect("hostname","username",'password');
$sql= "SELECT * FROM `items` WHERE id <20 ";
$items = (mysqli_query($sql));
while ( $db_field = mysqli_fetch_assoc($items) ) {?>
<tr><td><?php echo $db_field['id'];?></td></tr>
<tr><td><?php echo $db_field['name'];?></td></tr>
<tr><td><?php echo $db_field['descrip'];?></td></tr>
<tr><td><?php echo $db_field['reward'];?></td></tr>
<tr><td><?php echo $db_field['img'];?></td></tr>
<?php}
</tbody>
</table>
Try these, not tested
Where is the question?
There's many problems with this code.
First, you are confused between PHP and HTML.
Code between is PHP. It's executed on the server, you can have loops and variables and assignments there. And if you want some HTML there you use "echo".
Code outside is HTML - it's sent to the browser as is.
Second - what you seem to be doing is querying each field separately. This is not how you work with SQL.
Here's more or less what you need to do:
//Query all rows from 1 to 20:
$items = $mysqli->query("SELECT id,name,descrip,reward,img FROM `items` WHERE id between 1 and 20");
//Go through rows
while ( $row = mysqli_fetch_assoc($items) )
{
echo "<tr><td>{$db_field['id']}</td>";
//echo the rest of the fields the same way
});
I'm going to go ahead and assume that the code isn't working and that's because there's several basic errors. I'd strongly suggest doing some hard reading around the topic of PHP, especially since you're using databases, which, if accessed with insecure code can pose major security risks.
Firstly, you've set-up your connection using the procedural mysql_connect function but then just a few lines down you've switched to object-orientation by trying to call the method mysqli::query on a non object as it was never instantiated during your connection.
http://php.net/manual/en/mysqli.construct.php
Secondly, PHP echo() doesn't require the parentheses. PHP sometimes describes it as a function but it's a language construct and the parentheses will cause problems if you try to parse multiple parameters.
http://php.net/manual/en/function.echo.php
Thirdly, you can't simply switch from HTML and PHP and vice-versa with informing the server/browser. If you wish to do this, you need to either concatenate...
echo "<td>".while($db_filed = mysqli_fetch_assoc($item)) {
print $db_field['id'];
}."</td>;
Or preferably (in my opinion it looks cleaner)
<td>
<?php
while($db_filed = mysqli_fetch_assoc($item)) {
print $db_field['id'];
}
?>
</td>
However, those examples are based on your code which is outputting each ID into the same cell which I don't think is your goal so you should be inserting the cells into the loop as well so that each ID belongs to its own cell. Furthermore, I'd recommend using echo over print (it's faster).
Something else that may not be a problem now but could evolve into one is that you've used a constant for you FOR loop. If you need to ever pull more than 20 rows from your table then you will have to manually increase this figure and if you're table has less than 20 rows you will receive an error because the loop will be trying to access table rows that don't exist.
I'm no PHP expert so some of my terminology might be incorrect but hopefully what knowledge I do have will be of use. Again, I'd strongly recommend getting a good knowledge of the language before using it.
Im not sure if a similar question has been asked before, but here goes anyway. (I did do a search and found nothing relating to my question).
I am developing a website in which videos are played using the HTML5 video player. I have a connection to my database, a "watch" page that pulls all the correct data using a variable linked to the id (watch.php?v=1). I would like to have an index page where the most recent videos are pulled. They are ordered by the column "id" and everything works when I try and pull one result from the query. How would I go about getting multiple values? Here is my php code (server details hidden):
<?php
$mysqli = new mysqli("HIDDEN", "HIDDEN", "HIDDEN", "HIDDEN");
$sql = "
SELECT id, title, imgsrc, description
FROM videos
ORDER BY id DESC
LIMIT 2
";
$result = $mysqli->query($sql);
$video = mysqli_fetch_array($result, MYSQLI_ASSOC);
mysqli_close($mysqli);
?>
And here is my HTML code for the table.
<table>
<tr>
<td><h2><? echo $video['title']; ?></h2></td>
</tr>
</table>
That isn't the full code, but once I know the procedure I can apply it where needed!
I'm quite new to php and mysql, I can connect to databases but that's about it so a full walkthrough about what does what would be great!
Many Thanks,
James Wassall
You can iterate in for or while loop by calling mysqli_fetch_array($result, MYSQLI_ASSOC):
<table>
<?php
$mysqli = new mysqli("HIDDEN", "HIDDEN", "HIDDEN", "HIDDEN");
$sql = "
SELECT id, title, imgsrc, description
FROM videos
ORDER BY id DESC
LIMIT 2
";
$result = $mysqli->query($sql);
while($video = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
?>
<tr>
<td><h2><? echo $video['title']; ?></h2></td>
</tr>
<?php
}
mysqli_close($mysqli);
?>
</table>
Note that you should consider checking error statements, null controls etc.
try this
while($video = $result->fetch_array(MYSQLI_ASSOC))
{
?>
<tr>
<td><h2><?php echo $video['title']; ?></h2></td>
</tr>
<?php
}
?>
Each time you call mysqli_fetch_array, only one row is fetched.
You need to do something like
while ($video = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
// output $video['title']
}
(according to: http://www.php.net/manual/en/mysqli-result.fetch-array.php)
Try this,
<table>
<?php while($row = $result->fetch_array(MYSQLI_ASSOC)){ ?>
<tr>
<td><h2><? echo $row['title']; ?></h2></td>
<td><h2><? echo $row['description']; ?></h2></td>
</tr>
<?php } ?>
</table>