PHP script stops // looping through table - php

I have a very simple script looping through a MySQL-Table (3500 records) and returning the result in an HTML-Table.
The scripts ends suddenly (after 500) without error and without reaching the end of the table.
Any ideas whats wrong? Many thanks for comments.
Cheers
Tim
echo "<h1>This is PHP version " . phpversion() . "</h1>"; // returns 7.1.10
error_reporting(E_ALL);
ini_set('display_errors', 'on');
echo "<h3>Script Runtime: ". ini_get('max_execution_time') ." Sec.</h3>"; // returns 300
echo "<h3>Memory Limit: ". ini_get('memory_limit') ."</h3>"; // retuns 100M
$i = 0;
$sql = "SELECT * FROM tblClients ORDER BY lastname, firstname";
if (! $recClients = mysqli_query($GLOBALS['conn'], $sql)) {die("<p><b>SQL:</b><br>".$sql."</p>");}
echo "<table>";
echo "<tr>";
echo "<th>Nr.</th>";
echo "<th>Firstname</th>";
echo "<th>Lastname</th>";
echo "</tr>";
while ($rowClients = mysqli_fetch_assoc($recClients)) {
echo "<tr>";
echo "<td>" . $i++ . "</td>";
echo "<td>" . $rowClients['firstname'] . "</td>";
echo "<td>" . $rowClients['lastname'] . "</td>";
echo "</tr>";
}
echo "</table>";

Related

images appearing in warning but not in table?

I am working on a website whereby a load of advertisers are stored in the DB and then displayed to the user by there logo. I know storing directly in to the DB for images is not the done thing, however, I am starting out this way, to get the website running and then will refactor to move to a much more suitable approach.
Currently, I have the following PHP code:
<?php
session_start();
require_once "config.php";
// Create connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "SELECT * FROM advertisers";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>advertiser_Name</th>";
echo "<th>advertiser_URL</th>";
echo "<th>advertiser_Category</th>";
echo "<th>advertiser_logo</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['advertiser_id'] . "</td>";
echo "<td>" . $row['advertiser_Name'] . "</td>";
echo "<td>" . $row['advertiser_URL'] . "</td>";
echo "<td>" . $row['advertiser_Category'] . "</td>";
echo "<td>" . $row['<img src="data:image/jpeg;base64,'.base64_encode($row['advertiser_logo']).'"/>'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>
However, the images are displayed when called from the DB but they are displayed in the warning message rather than in the table?
<?php
session_start();
require_once "config.php";
// Create connection
if ($link === false)
{
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "SELECT * FROM advertisers";
if ($result = mysqli_query($link, $sql))
{
if (mysqli_num_rows($result) > 0)
{
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>advertiser_Name</th>";
echo "<th>advertiser_URL</th>";
echo "<th>advertiser_Category</th>";
echo "<th>advertiser_logo</th>";
echo "</tr>";
while ($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['advertiser_id'] . "</td>";
echo "<td>" . $row['advertiser_Name'] . "</td>";
echo "<td>" . $row['advertiser_URL'] . "</td>";
echo "<td>" . $row['advertiser_Category'] . "</td>";
echo "<td><img src='data:image/jpeg;base64," . base64_encode($row['advertiser_logo']) . "'/></td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
}
else
{
echo "No records matching your query were found.";
}
}
else
{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
mysqli_close($link);
?>
The fact that is showing the image in the warning is because you're using a tag with the source as an array key which is not correct.
The array keys, so what is inside the square bracket, is the reference to the array position. If you're familiar with C for example is the 0, 1, ecc.. and not the value itself.
Yes as #NigelRen mentioned this row $row['<img src="data:image/jpeg; looks very bad.
I think you should use:
echo "<td><img src='data:image/jpeg;base64," . base64_encode($row['advertiser_logo']) . "'/></td>";

Multiple Entries - Count Value in PHP

Thanks, friends, for the help and feedback. I know i am not good with PHP but still trying to learn and playing with it :D -- My table contains duplicate entries against evaid as Open Close or In Process --- with below code I get the last entered status from DB against each status by using query and if statement to show the data but i want to get the count of it as well. Anyone can help me out --- For example ---
$sql = "SELECT * FROM (SELECT * FROM disagreements ORDER BY addeddate DESC) disagreements GROUP BY evaid";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) { // Here with this query I got last entered status of each row against evaid – as 2 Open – 5 in Process and 10 Closed --- with below if statement – I can echo the rows with status but I want to have count of it that how many are open, in process or closed
if($row["status"]=='Open') { // I want to count this value as 2
echo "<tr>";
echo "<td>" . $row["evaid"]. "</td>";
echo "<td>" . $row["status"]. "</td>";
echo "</tr>";
}
}
} else {
echo "Nothing to Display";
}
mysqli_close($conn);
<?php
$count['open'] = 0;
$count['close'] = 0;
$count['process'] = 0;
while($row = mysqli_fetch_assoc($result)) {
if($row["status"]=='Open')
{
$count['open']++;
echo "<tr>";
echo "<td>" . $row["evaid"]. "</td>";
echo "<td>" . $row["status"]. "</td>";
echo "</tr>";
}
if($row["status"]=='Close')
{ // I want to count this value as 2
$count['close']++;
echo "<tr>";
echo "<td>" . $row["evaid"]. "</td>";
echo "<td>" . $row["status"]. "</td>";
echo "</tr>";
}
if($row["status"]=='Process')
{ // I want to count this value as 2
$count['process']++;
echo "<tr>";
echo "<td>" . $row["evaid"]. "</td>";
echo "<td>" . $row["status"]. "</td>";
echo "</tr>";
}
}
print_r($count);
?>
// Hii.. You can get a count from your SQL query itself, try this
$sql = "SELECT *, COUNT(ID) AS COUNT FROM (SELECT * FROM disagreements ORDER BY addeddate DESC) disagreements GROUP BY evaid";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result))
{
if($row["status"]=='Open')
{
echo "<tr>";
echo "<td>" . $row['count'] . "</td>"; // here you will get a count
echo "<td>" . $row["evaid"]. "</td>";
echo "<td>" . $row["status"]. "</td>";
echo "</tr>";
}
}
}
else
{
echo "Nothing to Display";
}
mysqli_close($conn);

Multiply a databasevalue with a result of a loop

Ive currently got this code where I have two loops: one which displays all my ToDos and one which counts the amount of hours ive worked on a certain ToDo.
What I want to do is to multiply the ['pris'] (cost pr hour) with the amount of hours ['timer'] from the loop.
echo "<div id='innhold'>";
$db = kobleTil();
$sql = "SELECT * FROM oppdrag";
$resultat = $db->query($sql);
while($nesteRad = $resultat->fetch_assoc()) {
echo "<hr />";
echo "<table id='resultat'>";
echo "<tr><th colspan='7'>" . $nesteRad['navn'] . "</th></tr>";
echo "<tr><td><b>Type</b></td><td><b>Startdato</b></td><td><b>Sluttdato</b></td><td><b>Pris pr. time</b></td><td><b>Aktiv</b></td></tr>";
echo "<tr>";
echo "<td>" . $nesteRad['type'] . "</td>";
echo "<td>" . $nesteRad['startDato'] . "</td>";
echo "<td>" . $nesteRad['sluttDato'] . "</td>";
echo "<td>" . $nesteRad['pris'] . "</td>";
echo "<td>" . $nesteRad['aktiv'] . "</td>";
echo '<td>Se på</td>';
echo '<td>Legg til timer</td>';
echo '<td>Slett</td>';
echo "</tr></table>";
echo "<hr />";
$prisx = $nesteRad['pris'];
}
echo "<a href='lioppdrag.php'><h5>Legg inn nytt oppdrag</h5></a>";
echo "<br />";
echo "<h3>Statistikk - Oppgaver arbeidet mest med<br /></h3>";
$statistikk = "SELECT oppdrag.*, (SELECT SUM(timer) FROM timeregistrering WHERE timeregistrering.oppdrID = oppdrag.oppdrID) AS Timer FROM oppdrag ORDER BY Timer DESC";
$print = $db->query($statistikk);
while($rad = $print->fetch_assoc()) {
$timer = $rad['Timer'] . '00px';
echo "<b>Oppdrag: " . $rad['navn'] ." Antal timer: " . $rad['Timer'] . " </b><br/>";
echo "<img src='firkant.png' id='firkant' width='"; echo $timer; echo "'></img><br/><br/>";
$timerx = $rad['Timer'];
}
echo "</div>";
} else {
echo "<div id='innhold'>";
echo "<a href='index.php'>Logg inn for å se innholdet</a>";
echo "</div>";
}
?>
I'll try to set you in the right direction but without the exact tables I can't give you the exact answer. first I'd do this in SQL before you ever get to PHOP and loops. Think in terms of set's not loops, so group your todo's and sum(timer) for each group. Then you can do another column with the sum(timer)*pris calling it Total. Psuedocode would be something like
select ToDo.Name, sum(working.timer), sum(timer)*ToDo.Pris as Total
from ToDo join working on todo.id = working.todo_id
group by ToDo.Name
(and a where clause to restrict the dates worked probably unless you just want everything)
edit: sorry I'm not sure why the code didn't show up when I first looked at the question and I only saw it after I answered but I still think this might help so take a look and let me know

how to check if mysql query return no result(record not found) using php?

i am passing images file names via textarea to php script to find information about each image in mysql db .The problem is i am trying to output those image file names that not found in mysql db and inform the user which image file names not found in mysql. my current code fails to output those missing records in db but it correctly outputs information about those images found in db. could any one tell me what i am doing wrong ?
foreach ($lines as $line) {
$line = rtrim($line);
$result = mysqli_query($con,"SELECT ID,name,imgUrl,imgPURL FROM testdb WHERE imgUrl like '%$line'");
if (!$result) {
die('Invalid query: ' . mysql_error());
}
//echo $result;
if($result == 0)
{
// image not found, do stuff..
echo "Not Found Image:".$line;
}
while($row = mysqli_fetch_array($result))
{
$totalRows++;
echo "<tr>";
echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['imgPURL'] . "</td>";
echo "<td>" . $row['imgUrl'] . "</td>"; echo "</tr>";
}
};
echo "</table>";
echo "<br>totalRows:".$totalRows;
You can use mysqli_num_rows() in mysqli
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result))
{
$totalRows++;
echo "<tr>";
echo "<td>" . $row['ID'] ."(".$totalRows. ")</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['imgPURL'] . "</td>";
echo "<td>" . $row['imgUrl'] . "</td>";
echo "</tr>";
}
} else {
echo "<tr><td colspan='4'>Not Found Image:".$line.'</td></tr>';
}
You want to use mysqli_num_rows
if(mysqli_num_rows($result)) {
// Do your while loop here
}
Use mysqli_num_rows to compare the number of rows in the result set.

Filtering mysql results via select dropdown

I'm a new member of StackOverflow, and although I've been using the website for a long time, it's my first time posting a question, in a hope that someone will be able to help me. I'll start by saying that my knowledge of PHP and MySQL is basic, but what I'm trying to do isn't too complex in my opinion, so hopefully I won't be asking for much. I've done a lot of prior research, but I just couldn't find the right answer.
In short, this is what I'm trying to do:
I've got an html form, which upon submission writes data to a database, and then publishes a table on a separate html page. With each successful submission a new table gets generated and published, while the old one gets pushed underneath. This all works fine, and I've also implemented pagination so that only 5 tables are visible per page.
What I'd like to be able to do is allow people to ONLY view/display results (tables) based on a specific criteria, in this case "rating", by selecting a rating from a drop-down on the page where tables are published. Rating is one of the fields in my form which gets submitted to a database and then published in one of the rows in a table.
Below is the code which publishes tables. Thanks in advance for your help!
<?php
include('dbconnect.php');
mysql_select_db("vtracker", $con);
$result = mysql_query("SELECT * FROM userdata");
$age = "Age:";
$rating = "Rating:";
$country = "From:";
$name = "Name:";
while($row = mysql_fetch_array($result))
{
echo "<table id='mft_table' cellspacing='0'>";
echo "<tbody>";
echo "<tr>";
echo "<td class='row1'>" .$name . " " . $row['personsname'] . "</td>";
echo "<td rowspan='4'>";
echo "<div class='mft_column'>" . $row['mft'] . "</div>";
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row2'>" . $country . " " . $row['nationality'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row3'>" . $age . " " . $row['personsage'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row4'>" . $rating . " " . $row['rating'] . "</td>";
echo "</tr>";
echo "</tbody>";
echo "<br>";
echo "</table>";
}
?>
for both true and false use can add thid in your code:
if($_POST['rating_dropdown']!='')
{
$temp_rating = $_POST['rating_dropdown'];
$query=mysql_query("SELECT * FROM userdata WHERE rating = '$temp_rating'");
}
else
{
$query=mysql_query("SELECT * FROM userdata");
}
Dunno if this works, it's just a hinch. haha.
It will see if the rating is true(not null), if it's true it will echo the results.
while($row = mysql_fetch_array($result))
{
if ($rating)
echo "<table id='mft_table' cellspacing='0'>";
echo "<tbody>";
echo "<tr>";
echo "<td class='row1'>" .$name . " " . $row['personsname'] . "</td>";
echo "<td rowspan='4'>";
echo "<div class='mft_column'>" . $row['mft'] . "</div>";
echo "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row2'>" . $country . " " . $row['nationality'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row3'>" . $age . " " . $row['personsage'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td class='row4'>" . $rating . " " . $row['rating'] . "</td>";
echo "</tr>";
echo "</tbody>";
echo "<br>";
echo "</table>";
}
}
Once the dropdown gets selected and posted to your display page, use this code:
$temp_rating = $_POST['rating_dropdown'];
mysql_query("SELECT * FROM userdata WHERE rating = '$temp_rating'");
Keep in mind, however, that you should be using PDO or mysqli extension, not the mysql extension. According to PHP's website:
This extension is deprecated as of PHP 5.5.0, and will be removed in
the future. Instead, the MySQLi or PDO_MySQL extension should be used.
See also MySQL: choosing an API guide and related FAQ for more
information.

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