I don't know why but I'm getting error at this line: $row = mysql_fetch_array($run_games)){
Here is the code:
<div class='topnav'> <!--Start of the Top Navigation-->
<?php
include("includes/connect.php");
$select_games = "SELECT * FROM games";
$run_games = mysql_query($select_games);
$row = mysql_fetch_array($run_games)){
$game_category = $row['game_category'];
?>
<!--some links here with the variable of "game_category"-->
<?php } ?>
</div> <!--End of the Top Navigation-->
Please help me :(
It should be ..
while($row = mysql_fetch_array($run_games))
{
echo $game_category = $row['game_category'];
}
You need to loop through the resultset !
Loop through the result set using while.
$row = mysql_fetch_array($run_games)){
Should be
while($row = mysql_fetch_array($run_games)) {
// do stuff ...
}
This is probably a cut and paste error. The first part of your while loop is missing:
$row = mysql_fetch_array($run_games)){
should be
while ($row = mysql_fetch_array($run_games)){
These functions are deprecated,so use latest ones
$row = mysql_fetch_array($run_games));//if only one row
//other wise should use while
while($row = mysql_fetch_array($run_games))
{
}
Related
Hello I was trying to put the results of the query to an array but it's not working intead it only displays i
<?php
$sqlo = mysqli_query($conn, "SELECT * FROM users");
$i=1;
while ($h=mysqli_fetch_assoc($sqlo)) {
echo "<br>counter[i] : ".$counter[$i] = $h['username'];
echo "<br>i++ : ".$i++;
}
?>
As Barmar suggestion is recommended separating the assignment from echo.
Try this code:
<?php
$sqlo = mysqli_query($conn, "SELECT * FROM users");
$i=1;
$counter = [];
while ($h=mysqli_fetch_assoc($sqlo)) {
$counter[$i] = $h['username']; //if you need to store username inside an array
echo "<br>counter[i] : ".$counter[$i];
echo "<br>i++ : ".$i++;
}
create array variable before loop.
$i=1;
$counter[];
while ($h=mysqli_fetch_assoc($sqlo)) {
echo "<br>counter[i] : ".$counter[$i] = $h['username'];
e
echo "<br>i++ : ".$i++;
}
i am making a profile.php page and i would like it to show the user all his projects, this is my first time doing something like this, and i cant find a solution for it
code to show the projects :
$username = $_SESSION['username'];
if ($_SESSION['type'] = "developer"){
$q = "SELECT * FROM `projects` WHERE `developer` = '$username'";
$result = mysqli_query($con,$q);
$row = mysqli_fetch_array($result);
$numrows = mysqli_num_rows($result);
if(empty($numrows)){
echo'
<div class="row">
<div class="col-lg-12 newp">
<p><span class="glyphicon glyphicon-plus plus"></span>Add a new project</p>
</div>
</div>';
}else{
$p_id = $row['project_id'];
$p_name = $row['project_name'];
$p_owner = $row['owner'];
$p_developer = $row['developer'];
$p_price = $row['price'];
$p_date_started = $row['date_started'];
$p_date_end = $row['date_end'];
$p_paid = $row['paid'];
//foreach project the user has do this :
echo"
<div class=\"row\">
<div class=\"col-lg-12\">
<p>$p_name </br>owner : $p_owner, developer : $p_developer, price : $p_price$</br>started : $p_date_started, ends :$p_date_end, paid :$p_paid</p>
</div>
</div>";
}
}
} else {
while($row = mysqli_fetch_array($result)) {
$p_id = $row['project_id'];
...
Besides the other answer given:
You're presently assigning instead of comparing with
if ($_SESSION['type'] = "developer"){...}
^
which the above will fail and everything inside that conditional statement and should read as
if ($_SESSION['type'] == "developer"){...}
^^
with 2 equal signs.
Make sure the session has also been started, it's required when using sessions.
session_start();
You're also open to an SQL injection. Use a prepared statement:
https://en.wikipedia.org/wiki/Prepared_statement
Helo evry1
i want to display data from phpmyadmin table to my php page but the problem is tht it not displaying here is my code
<?php
include('config.php');
$res = mysql_query("SELECT * FROM accord");
while($row = mysql_fetch_assoc($res))
{
$id=$row['id'];
$rupes=$row['rs'];
$mob=$row['mobilen'];
$carmodel=$row['modelcar'];
}
?>
PKR:<?php echo $rupes;?>
here is my code plz help
There might be two problems
1) You have issue with config.php
2) Your query has some issue try putting die like this
mysql_query("SELECT * FROM accord") or die(mysql_error());
You must echo your variables:
<?php
include('config.php');
$res = mysql_query("SELECT * FROM accord");
while($row = mysql_fetch_array($res))
{
echo $id=$row['id'];
echo $rupes=$row['rs'];
echo $mob=$row['mobilen'];
echo $carmodel=$row['modelcar'];
}
?>
Hello and Good Morning,
I am still learning PHP and for some reason my script will not post any data in my foreach loop. Any Idea why? The emailRow Echos out fine but I am going to remove My code is below:
<?php
include 'includes/header.php';
$accountUser = array();
$upgradeEmail = $_GET['currEmail'];
$emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";
<?php echo $emailRow['fbID']; ?>
<?php echo $emailRow['firstName']; ?>
<?php echo $emailRow['lastName']; ?>
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
?>
<table>
<?php foreach($accountUser as $emailData) { ?>
<tr><td> <?php emailData['fbID']; ?> </td><td><?php emailData['firstName']; ?></td><td><?php emailData['lastName']; ?></td></tr>
<?php } ?>
</table>
You have constructed your SQL query in $emailQuery, but never executed it. Call mysql_query(), and pass its result resource to mysql_fetch_assoc().
$emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";
$result = mysql_query($emailQuery);
if ($result)
{
while($emailRow = mysql_fetch_assoc($result, $conn))
{
$accountUser[]=$emailRow;
}
}
else // your query failed
{
// handle the failure
}
Please also be sure to protect your database from SQL injection by calling mysql_real_escape_string() on $upgradeEmail since you're receiving it from $_GET.
$upgradeEmail = mysql_real_escape_string($_GET['currEmail']);
You don't actually echo anything.
as well as not running the query.
and there are some other methods to do things, much cleaner than usual uglyPHP.
a function
function sqlArr($sql){
$ret = array();
$res = mysql_query($sql) or trigger_error(mysql_error()." ".$sql);
if ($res) {
while($row = mysql_fetch_array($res)){
$ret[] = $row;
}
}
return $ret;
}
a code
$email = mysql_real_escape_string($_GET['currEmail']);
$data = sqlArr("SELECT * FROM users WHERE emailOne='$email' AND authLevel=0");
include 'template.php';
a template
<? include 'includes/header.php' ?>
<table>
<? foreach($data as $row) { ?>
<tr>
<td><?=$row['fbID']?></td>
<td><?=$row['firstName']?></td>
<td><?=$row['lastName']?></td>
</tr>
<? } ?>
</table>
YOU HAVE A SYNTAX ERROR. You can't open a new php tag within an existing php tag. You have to close the already open tag first.
As far as getting the query to work,
First you have to fetch data before printing it or echoing it...
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
then you may write statements..
echo $emailRow['fbID']; etc. code.
Secondly you have not fired a query, just written the query statement. Use mysql_query to fire it.
Your code would be something like this..
<?php include 'includes/header.php';
$accountUser = array();
$upgradeEmail = $_GET['currEmail'];
$emailQuery = mysql_query("SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0") or die (mysql_error());
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
echo $emailRow['fbID'];
echo $emailRow['firstName'];
echo $emailRow['lastName'];
print '<table>';
foreach($accountUser as $emailData) {
print '<tr><td>'$.emailData['fbID'].'</td><td>'.$emailData['firstName'].'</td><td>'.$emailData['lastName'].'</td></tr>';
}
print '</table';
?>
Feel free to use this code, modifying it to fit your needs.
Why is this not working?
<?php
$select = "select * from messages where user='$u'";
$query = mysqli_query($connect,$select) or die(mysqli_error($connect));
$row = mysqli_num_rows($query);
$result = mysqli_fetch_assoc($query);
$title = mysqli_real_escape_string($connect,trim($result['title']));
$message = mysqli_real_escape_string($connect,trim($result['message']));
while(($result = mysqli_fetch_assoc($query))){
echo $title;
echo '<br/>';
echo '<br/>';
echo $message;
}
?>
where as this works -
<?php
echo $title;
?>
SORRY TO SAY, BUT NONE OF THE ANSWERS WORK. ANY OTHER IDEAS?
If your mysqli query is returning zero rows then you will never see anything printed in your while loop. If $title and $message are not set (because you would want reference them by $result['title'] & $result['message'] if that are the field names in the database) then you will only see two <br /> tags in your pages source code.
If the while loop conditional is not true then the contents of the while loop will never execute.
So if there is nothing to fetch from the query, then you won't see any output.
Does you code display anything, or skip the output entirely?
If it skips entirely, then your query has returned 0 rows.
If it outputs the <br /> s, then you need to check your variables. I could be wrong, not knowing te entire code, but generally in this case you would have something like
echo $result['title'] instead of echo $title
If $title and $message come from your mysql query then you have to access them through the $result array returned by mysqli_fetch_assoc.
echo $result['title'];
echo $result['message'];
Also if your using mysqli you'd be doing something like this:
$mysqli = new mysqli("localhost", "user", "password", "db");
if ($result = $mysqli->query($query)) {
while ($row = $result->fetch_assoc()) {
print $row['title'];
}
$result->close();
}
Try this:
<?php
$result = mysql_query($query);
while($row = mysqli_fetch_assoc($result)){
echo $title.'<br/><br/>'.$message;
}
?>
Does this work;
<?php
$select = "select * from messages where user='$u'";
$query = mysqli_query($connect,$select) or die(mysqli_error($connect));
$row = mysqli_num_rows($query);
while(($result = mysqli_fetch_assoc($query))){
echo $result['title'];
echo '<br/>';
echo '<br/>';
echo $result['message'];
}
?>
Basically I've made sure that it's not picking the first result from the query & then relying on there being more results to loop through in order to print the same message repeatedly.