I am new in php. I make a quiz app and I want to show questions that is not repeated again . here's my code.
Please help me to show require result.
<?php
include('connect.php');
$sql = "SELECT * FROM quiz_question WHERE theme_id= 2 ORDER BY RAND ()";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$id = $row['id'];
echo "
<h2>" . $row["question"]. "</h2>";
break;
}
}
$check_id = array ($row['id']);
echo $check_id['0'];
if(array ($row['id']) == $check_id){
echo "no question ";
}
else{
echo "
<h2>" . $row["question"]. "</h2>";
}
?>
Your question is not clear. But I guess, you can solve it by array_unique($array).
array_unique($array);
<?php
include('connect.php');
$sql = "SELECT * FROM quiz_question WHERE theme_id= 2 ORDER BY RAND ()";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$id = $row['id'];
$question = $row['question'];
echo "
<h2>" . $row["question"]. "</h2>";
$check_id = array($id);
}
}
$check_id_unique=array_unique($check_id);
?>
Related
I'm trying to count two different tables but it's not working, and I cannot find out where the problem is, and how to implement it with PHP.
$sql = "SELECT COUNT(users.user_id) AS totalUsers FROM users, SELECT COUNT(bikes.bike_id) AS totalBikes FROM bikes";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "<td><td>" . $row["totalUsers"] . "</td></tr>";
echo "<td><td>" . $row["totalBikes"] . "</td></tr>";
}
} else {
echo "0 results";
}
$conn->close();
Get the amounts from each table separately. Each SELECT is a unique resultset from your query, listing them with commas won't retrieve all data as you are expecting.
$result = $conn->query("SELECT COUNT(users.user_id) AS totalUsers FROM users");
$totalUsers = 0;
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
$totalUsers = $row["totalUsers"];
}
$result = $conn->query("SELECT COUNT(bikes.bike_id) AS totalBikes FROM bikes");
$totalBikes = 0;
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
$totalBikes = $row["totalBikes"];
}
$conn->close();
Then you can work with the values stored in $totalUsers and $totalBikes:
echo "<tr><td>" . $totalUsers . "</td></tr>";
echo "<tr><td>" . $totalBikes . "</td></tr>";
I have trouble to select a set of specific data using ID from the database. For example, employee one has a unique id of e000000001, when I click the view button in the index will lead to employee detail page which shows the detail of that particular employee instead of all the employees' detail. Thank you.
//from index.php page
<?php
require_once 'db/dbEmpList.php';
$sqlStr = "SELECT * FROM employees;";
$result = $connection->query($sqlStr);
if ($result->num_rows > 0) {
echo "<table class='table table-sm'><thread><tr><th>Full Name</th><th>Employee ID</th><th>Position</th><th>View Employee's Details</th></tr>";
while ($row = $result->fetch_assoc()) {
echo "<tr><td>"
. $row["empName"]. "</td><td>"
. $row["empID"]. "</td><td>"
. $row["position"]. "</td>"
. "<td> <a href='employeedetail.php?id={$row["empID"]}'>View</a>"
. "</td></tr>";
}
}
// from employee page
require_once 'db/dbEmpDetail.php';
$sql = "SELECT * FROM employees where empID = '{$row["empID"]}' ";
$result = mysqli_query($connection, $sql);
if (mysqli_num_rows($result)) {
while ($row = mysqli_fetch_assoc($result)) {
echo '<tr>' .'<td>' .$row["empName"].'</td>'.'<td>'. $row["position"].'</td>' .'<td>'.$row["empNRIC"].'</td>' .'<td>'.$row["empID"].'</td>' .'<td>'.$row["empEmail"].'</td>' .'<td>'.$row["empPwd"].'</td>' . "</tr>";
}
} else {
echo "0 results";
}
mysqli_close($connection);
?>
// FROM EMPLOYEE PAGE
The way you retrieve URL query string is wrong. You should be using $_GET to get the query string from URL. In your case it should be $_GET['id']. See the code below:
require_once 'db/dbEmpDetail.php';
$employeeid = trim(mysqli_real_escape_string($_GET['id']));
$sql = "SELECT * FROM employees where empID = '".$employeeid."' ";
$result = mysqli_query($connection, $sql);
if (mysqli_num_rows($result)) {
while ($row = mysqli_fetch_assoc($result)) {
echo '<tr>' .'<td>' .$row["empName"].'</td>'.'<td>'. $row["position"].'</td>' .'<td>'.$row["empNRIC"].'</td>' .'<td>'.$row["empID"].'</td>' .'<td>'.$row["empEmail"].'</td>' .'<td>'.$row["empPwd"].'</td>' . "</tr>";
}
}
else {
echo "0 results";
}
mysqli_close($connection);
?>
$sql = "SELECT * FROM today WHERE heading='$heading' and day='$day'";
$sql1 = "SELECT * FROM today WHERE day='$day'";
$result = $conn->query($sql);
$result1 = $conn->query($sql1);
if ($result->num_rows > 0) {
echo "<div id='post'><h1>".$row["heading"]."</h1>
<aside class='related-post'>".while($row = $result1->fetch_assoc())
{echo'<img src='".$row["image"]."'>;}
.</aside>}";
I have been using while loops for fetching data from table. My connection is working is perfect but I need another loop in the first that is not working. Isn't it the good way?
Update: I tried to finish echo and again started as follow but still an error
while($row = $result->fetch_assoc()) {
echo "<div id='post'><h1>"
.$row["heading"].
"</h1><div class='post-side'><img class='post-image' src='"
.$row["image"].
"'><div class='post-data'><p><strong>Age: </strong><span>$age</span></p><p><strong>Date of birth: </strong><span>"
.$row["day"].
"-"
.$row["month"].
"-"
.$row["year"].
"</span></p></div></div></div><div class='description'><p>"
.$row["description"].
"</p></div><div class='bottom-related'><aside class='related-post'>";
while($row = $result1->fetch_assoc())
{echo"<img src='"
.$row["image"].
"'>/";}.echo"</aside><aside class='ad2'>".$includead."</aside></div>";
}
echo "</div>";
} else {
echo "No table found";
}
$conn->close();
You're trying to concatenate to a string a WHILE loop; this is wrong.
You should echo your first part, end with it and then do your while loop, and echo the end afterwards:
Your quotes are a bit messed up as well
if ($result->num_rows > 0)
{
echo "<div id='post'><h1>".$row["heading"]."</h1>
<aside class='related-post'>";
while($row = $result1->fetch_assoc())
{
echo'<img src="'.$row["image"].'">';
}
echo '</aside>';
}
You can't concatene while with String, it's a syntaxic error
Also, you have a probleme when trying to echo a String, you can use this syntax:
echo "PHP"; // will evaluate PHP variables and whitespace inside a string
echo 'PHP'; // will evaluate nothing;
But you can not start flushing a string ' and finish by " or vice-versa.
Here the correct code :
<?php
$sql = "SELECT * FROM today WHERE heading='$heading' and day='$day'";
$sql1 = "SELECT * FROM today WHERE day='$day'";
$result = $conn->query($sql);
$result1 = $conn->query($sql1);
if ($result->num_rows > 0) {
echo "<div id='post'><h1>" . $row["heading"] . "</h1><aside class='related-post'>";
while($row = $result1->fetch_assoc()) {
echo'<img src="' . $row["image"] .'">';
}
echo "</aside>";
}
I'm pretty new at PHP/MySQL, so please be patient with me.
I am trying to get a list of members in a table to show up on a page. Right now it's showing the first member about 10 times and not displaying anyone else's name. I DID have it working, but I don't know what happened. I just want it to display everyone's name once. Here is my code:
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching];
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name];
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
I miss look on your code, since it is mess, but disregard the mysqli and mysql thing, you want to show how many student in the teacher's classes.
<?php $select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$row = mysql_fetch_array($select);
$rows = mysql_num_rows($select);
$teaching = $row[teaching]; <--- This only get first row of the course, if you want multiple course under same username, you need to loop it.
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
$row2 = mysql_fetch_array($select2);
$student=$row2[student_name]; <----- This only get the first row of the student name, if you want multiple student under a course, you need to loop it.
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
}
else {
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
$row3 = mysql_fetch_array($select3);
while($row2 = mysql_fetch_array($select2)) {
$house=$row3[house]; <----This only show the first row of $house under same student, so you need to loop it too.
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>"; }
?>
So what you really want to do is
<?php
$select = mysql_query("SELECT * FROM `member_staff` WHERE `username`='$_SESSION[USR_LOGIN]' AND `status`='Active'");
$rows = mysql_num_rows($select);
if ($rows==0){echo "Sorry, you don't appear to be a professor.";}
else { ?>
<?php }
while( $row = mysqli_fetch_array( $select ) ) {
$teaching = $row[teaching];
$select2 = mysql_query("SELECT * FROM `classes_enrolled` WHERE `course`='" . $teaching . "' ORDER BY `student_name`") or die(mysql_error());
$count = mysql_num_rows($select2);
if($count==NULL) {
echo "<table width=\"80%\">\n";
echo "<tr><td><center>Nobody has registered for your class yet!</center></td></tr>\n";
echo "</table>\n";
echo "<br /><br />\n\n";
} else {
while( $row2 = mysql_fetch_array($select2) ) {
$student=$row2[student_name];
echo "<center><font size=\"3\"><b>YEAR 1, TERM 2</b></font></center>";
echo "<table width=\"80%\" class=\"table-stripes\">\n";
echo "<tr><td width=\"50%\"><b>STUDENT</b></td></tr>\n";
$select3 = mysql_query("SELECT * FROM `members` WHERE `username`='" . $student . "'") or die(mysql_error());
while($row3 = mysql_fetch_array($select3)) {
$house=$row3[house];
echo "<tr><td><strong class=\"$house\">$student</strong></td></tr>";
}
echo "</table>";
}
} // END ELSE
}
} // END ELSE
?>
So I try to learn php and decided to make one site where I add images, save them in folder and id, name,type, path in mysql. Then show on page. So far I have upload form and I can upload and save images. Also I showing them successfully on the page.
Now I'm trying to make categories like - Nature, Funny ... etc. So I added one field in my main table -> img_category.
Also I madded second table - cats whit cat_id and cat_name fields. Using this to show the categories on the page:
<?php
$q = mysqli_query($con,"select * from cats");
while ($res = mysqli_fetch_assoc($q))
{
echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
}
So now how can I make when I click on some category link to load images only from this category?
I have managed to make something like this but it doesn't work like is expected
<?php
$q = mysqli_query($con,"select * from cats");
while ($res = mysqli_fetch_assoc($q))
{
echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
}
?>
<hr>
<?php
$cat_id = $_GET['cat_id'];
$query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
$line = mysqli_fetch_array($result, MYSQL_BOTH);
if (!$line) echo '';
$previd = -1;
$currid = $line[0];
if (isset($_GET['id'])) {
do {
$currid = $line[0];
if ($currid == $_GET['id']) break;
$previd = $currid;
$line = mysqli_fetch_array($result, MYSQL_BOTH);
} while ($line);
}
if ($line) {
echo "<div id=\"picture\">";
echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /></a><br />";
echo "<div id=\"caption\">".$line['caption']."</div><br />";
}
else echo "There is no images!\n";
if ($previd > -1) echo '<span>Prev</span>';
echo str_repeat(' ', 5);
$line = mysqli_fetch_array($result, MYSQL_BOTH);
$query = "select * from images order by RAND() LIMIT 1";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
while ($row = mysqli_fetch_array($result, MYSQL_BOTH)){
echo 'Random';
}
echo str_repeat(' ', 5);
if ($line) echo '<span>Next</span> <br /><br />';
echo "</div>";
?>
The results are:
When there is image in the category is showed but and if I click on 'Next' button I get the same image.
If there is no image in the category I get all echoes like link whit the ID of last category for exam: There is no image like link and if I click it I get last category ID loaded. In my case I have 8 categories so ID=8.
Any help is appreciate!
Thank's
EDIT:
Ok this line:
echo '<span>Следваща</span>
Where is pic.php?cat_id=... i think is wrong. Here I must take next image ID not next category ID. But how to change it for image? If i make it pic.php?id=... I get empty page.
I don't understand it. I know that is messy code but is best I can do for now.
EDIT 2:
I've made something like this. Now can you help me how to make query's for next image because now didn't get next image and stay the same.
$cat_id = $_GET['cat_id'];
$cat_id = mysqli_real_escape_string($con, $cat_id);
$query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
$prevSQL = mysqli_query($con,"SELECT cat_id FROM cats WHERE cat_id < $cat_id ORDER BY cat_id DESC LIMIT 1") or die (mysqli_error($con));
$nextSQL = mysqli_query($con, "SELECT cat_id FROM cats WHERE cat_id > $cat_id ORDER BY cat_id ASC LIMIT 1") or die (mysqli_error($con));
$prevobj=mysqli_fetch_object($prevSQL);
$nextobj=mysqli_fetch_object($nextSQL);
$pc = mysqli_fetch_object(mysqli_query($con, "SELECT COUNT(cat_id) as pid FROM cats WHERE cat_id<$cat_id ORDER BY cat_id DESC")) or die (mysqli_error($con));
$nc = mysqli_fetch_object(mysqli_query($con, "SELECT COUNT(cat_id) as nid FROM cats WHERE cat_id>$cat_id ORDER BY cat_id ASC")) or die (mysqli_error($con));
$prev=$pc->pid>0 ? 'Prev |' : '';
$next=$nc->nid>0 ? 'Next' : '';
$row = mysqli_fetch_array($result);
echo "<div id=\"picture\">";
echo "<img src=\"upload/" . $row['name'] . "\" alt=\"\" /><br />";
echo $row['caption'] . "<br />";
echo "</p>";
echo $prev;
echo $next;
As you stated, I guess the error is with the line:
echo '<span>Следваща</span>
I think it should be:
echo '<span>Следваща</span>
EDIT:
Your code should look like:
<?php
$q = mysqli_query($con,"select * from cats");
while ($res = mysqli_fetch_assoc($q))
{
echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
}
?>
<hr>
<?php
$cat_id = $_GET['cat_id'];
$query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
$line = mysqli_fetch_array($result, MYSQL_BOTH);
if (!$line) echo '';
$previd = -1;
$currid = $line[0];
if (isset($_GET['id'])) {
do {
$currid = $line[0];
if ($currid == $_GET['id']) break;
$previd = $currid;
$line = mysqli_fetch_array($result, MYSQL_BOTH);
} while ($line);
}
if ($line) {
echo "<div id=\"picture\">";
echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /></a><br />";
echo "<div id=\"caption\">".$line['caption']."</div><br />";
}
else echo "There is no images!\n";
if ($previd > -1) echo '<span>Prev</span>';
echo str_repeat(' ', 5);
$line = mysqli_fetch_array($result, MYSQL_BOTH);
$query = "select * from images order by RAND() LIMIT 1";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
while ($row = mysqli_fetch_array($result, MYSQL_BOTH)){
echo 'Random';
}
echo str_repeat(' ', 5);
if ($line) echo '<span>Next</span> <br /><br />';
echo "</div>";
?>
Try this
<?php
if(isset($_GET['cat_id'])){
$cat_id = $_GET['cat_id'];
$query = "SELECT * FROM images WHERE img_category = '$cat_id'";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
$line = mysqli_fetch_array($result, MYSQL_BOTH);
if (!$line) echo '';
$previd = -1;
$currid = $line[0];
if (isset($_GET['id'])) {
$previous_ids = array();
do {
$previous_ids[] = $line[0];
$currid = $line[0];
if ($currid == $_GET['id']) break;
$previd = end($previous_ids);
$line = mysqli_fetch_array($result, MYSQL_BOTH);
} while ($line);
}
if ($line) {
echo "<div id=\"picture\">";
echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /><br />\r";
echo "<div id=\"caption\">".$line['caption']."</div><br />";
}
else echo "There is no images!\n";
if ($previd > -1)
echo '<span>Prev</span>';
echo str_repeat(' ', 5);
$line = mysqli_fetch_array($result, MYSQL_BOTH);
$query = "select * from images WHERE img_category = '$cat_id' order by RAND() LIMIT 1";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
while ($row = mysqli_fetch_array($result, MYSQL_BOTH)){
echo 'Random';
}
echo str_repeat(' ', 5);
if ($line) echo '<span>Next</span> <br /><br />';
echo "</div>\r";
}
?>