Can't find solution for this so i'm posting new question. As thread name say, i want to use while in other while when using fetch_array in PHP. I want to display info about user in every post.
My example code:
<?php
$users_query = "SELECT * FROM users";
$users_result = mysqli_query($db, $users_query);
$posts_query = "SELECT * FROM posts";
$posts_result = mysqli_query($db, $posts_query);
while($row1 = $posts_result->fetch_array()) {
echo '<div class="post"><div class="user-info">';
while($row2 = $users_result->fetch_array()) {
echo $row2["user_name"]." ".$row2["user-posts-count"];
}
echo '</div>'.$row1["message"].'</div>';
}
?>
It's just example but showing what i want. Second while executed only once, so info about user in post is displayed only in first post. Any solution?
Thank you and sorry for my bad english, Adrian.
Related
Noob alert here. I'm just (trying to) teaching myself PHP because I need it on my internship job. Now I am creating a survey and I am trying to use while loops to show the questions and their respective answers onto my XAMPP. I am using this code:
<?php
mysqli_select_db($conn, "surveyordina");
//code hier schrijven
$sql = "SELECT questions_body FROM survey_questions where subthema_id = 1";
$sql2 = "SELECT answer_body FROM survey_answers where answer_id = 1 or answer_id = 2 or answer_id = 3";
$result = mysqli_query($conn, $sql);
$result2 = mysqli_query($conn, $sql2);
if(mysqli_num_rows($result2) > 0){
while($row = mysqli_fetch_assoc($result)) {
echo "<br>" ."Vraag: <br>" . $row["questions_body"]. "<br>";
while($row_answer = mysqli_fetch_assoc($result2)) {
echo $row_answer["answer_body"]. "<br>";
}
}
}
else{
echo "No results";
}
Now the returning part of this code looks like this:
Vraag:
Is the organization aware of where all the personal data is stored? Be it on-site or in the cloud, hosted by the company or by a third party?
Yes
No
I don't know
Vraag:
Is the organization able to locate and find personal data of a particular data subject?
Vraag:
Does the organization have technology in place to return all personal data on a given data subject, given a single search from personnel?
Vraag:
testerino kekkerino
I am trying to get the Yes, No, and I don't know parts under each and every question but only seem to get it under one of the questions.
Is there a method to return the whole answer part under each question? If so, am I on the right path or am I doing something completely wrong?
Thanks in advance.
Once you hit the end of a result-set, mysqli_fetch_assoc() doesn't just keep cycling through. So you're hitting the end of the $result2 set within the first outer loop, and then that's it, there are no more results to fetch.
You should fetch all of the $result2 results outside of the $result loop, assign the results to a variable, and then loop through that array within the $result loop.
Something like:
if(mysqli_num_rows($result2) > 0){
$result2Set = mysqli_fetch_all($result2, MYSQLI_ASSOC);
while($row = mysqli_fetch_assoc($result)) {
echo "<br>" ."Vraag: <br>" . $row["questions_body"]. "<br>";
foreach($result2Set as $row_answer) {
echo $row_answer["answer_body"]. "<br>";
}
}
}
This question already has answers here:
How can I get an unknown username given an ID?
(2 answers)
Closed 1 year ago.
am new here, i have an issue with displaying logged in user profile, hoping you guys can help, here is my code :
<?php session_start(); include 'dpconfig.php'
<?php $run = mysqli_query($conn,"Select * from user Where id = $_SESSION['uid]");
$row = mysqli_fetch_array($run, MYSQLI_BOTH); { }
$showid = $row[0];
$showfirst = $row[1];
$showlast = $row[2];
$showuid = $row[3];
echo $showid; echo $showfirst; echo $showlast; echo $showuid;
Now basically this code gives me the details of the first id in my database even if i login different users, i need help selecting data from table name(user) to display logged in user profile, using sessions. Thanks
$run = mysqli_query($conn,"Select * from user where username='xxx' and pass='xx'");
it will return login user detail.
Try something like this, you need to remember to check that the array holds values as well, then you can respond to it...
$conn = dbconfig;
$id = $_SESSION['id'];
$sql = "SELECT * FROM user WHERE id='$id'";
$check = mysqli_query($conn, $sql) or die ("err $id " . mysqli_error ($conn));
$check2 = mysqli_num_rows($check);
if ($check2 != 0) {
while ($row = mysqli_fetch_assoc($check)) {
$userid = $row['id']; // repeat for all db columns you want
}
}
Sorry if there are any typos, done this on my phone quickly. If you need further help gimme a shout.
Updated the code to show more information with the error message, to help you get to the bottom of why it's not working for you.
I am trying to show it all with while loop on a dynamic page, but it wont show anything at all..
if (isset($_GET['id'])) {
$genre = $_GET['id'];
$sql = "SELECT * FROM movstream_genre WHERE ID = '$genre'";
$result = $db->query($sql);
$row = $result->fetch_assoc();
}else{
$sql = "SELECT ID, genre FROM movstream_genre";
$result = $db->query($sql);
}
<html>
<body>
<ul>
<?php while ( $row = $result->fetch_assoc() ): ?>
<li><?=$row['genre'];?></li>
<?php endwhile; ?>
</ul>
</body>
</html>
Anyone know why it wont show anything on the dynamic page, but if the page is not dynamic it works fine :)
Thanks.
If dynamic page means that your sending an id in Url an you are not getting result.
I think that the problem is that you are using $row = $result->fetch_assoc(); in if condition and also in while.
Please use fetching in one place. Better be in while loop.
Hope it helps.
I figured it out
if (isset($_GET['id'])) {
$genre = $_GET['id'];
$sql = "SELECT ID, genre FROM movstream_genre";
$result = $db->query($sql);
}
This it how it needs to look when its a dynamic page :)
well i think it is, it works now.
please echo the result ...
<li><?php echo $row['genre'];?></li>
And in href, it is only genre OR genre.php
In the first query..
$sql = "SELECT * FROM movstream_genre WHERE ID = '$genre'";
$result = $db->query($sql);
$row = $result->fetch_assoc();// this line not required
comment this line
$row = $result->fetch_assoc();
It looks like you haven't enclosed the first part of your PHP code in <?php ?> tags
I have the following code that works by outputting as a link ( the link comes from a field in my database) I wish to do the same for the code below, however i cannot get it work, here is the example of what I have that works, and the code that i wish to make output as a link:
Working Code what I want it to look like
if (!empty($_REQUEST['term'])) {
$term = mysql_real_escape_string($_REQUEST['term']);
$sql = "SELECT * FROM adrenaline WHERE title LIKE '%".$term."%'";
$r_query = mysql_query($sql);
while ($row = mysql_fetch_array($r_query)){
echo '<br> '. $row['title'] .'';
}
}
?>
And the code that i have at the moment, it works by be manually typing in the hyper link, however I wish to make it take the link from the database like the example above
//query the database
$query = mysql_query("SELECT * FROM hobby WHERE id = '1' ");
//ferch the results / convert results into an array
WHILE($rows = mysql_fetch_array($query)):
$title = $rows['title'];
echo "<a href='shard.php'>$title</a>";
endwhile;
?>
Many thanks!
I am not 100% certain if this is what you meant to ask... let me know in comments:
<?PHP
$query = mysql_query("SELECT * FROM hobby WHERE id = '1' ");
if(mysql_num_rows($query) >= 1) {
while($rows = mysql_fetch_array($query)) {
echo sprintf("%s", $rows["description"], $rows["title"]);
}
} else { echo "No hobbies found."; }
?>
I believe you might have faced some syntax issues while dealing with quotes parsing a variable in <a html tag. Consider using sprintf something like in my example.
I have also added a mysql_num_rows() just in case and you can see its a good fail-safe method incase there are no rews found on any select query.
IMPORTANT: STOP using mysql_ functions because its deprecated from new PHP versions. Use PDO or mysqli instead.
I'm trying out my hand at php at the moment - I'm very new to it!
I was wondering how you would go about selecting all items from a mySQL table (Using a SELECT * FROM .... query) to put all data into an array but then not displaying the data in a table form. Instead, using the extracted data in different areas of a web page.
For example:
I would like the name, DOB and favorite fruit to appear in one area where there is already say 'SAINSBURYS' section hardcoded into the page. Then further down the next row that is applicable to 'ASDA' to appear below that.
I searched both here and google and cant seem to find an answer to my strange questions! Would this involve running the query multiple times filtering out the sainsburies data and the asda data where ever I wanted to place the relevant
echo $row['name']." ";
echo $row['DOB']." "; etc etc
next to where it should go?
I have got php to include data into an array (I think?!)
$query = "SELECT * FROM people";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
echo $row['name']." ";
echo $row['DOB']." ";
echo $row['Fruit']." ";
}
?>
Just place this (or whatever your trying to display):
echo $row['name']." ";
Anywhere you want the info to appear. You can place it within HTML if you want, just open new php tags.
<h1>This is a the name <?php echo $row['name']." ";?></h1>
If you want to access your data later outside the while-loop, you have to store it elsewhere.
You could for example create a class + array and store the data in there.
class User {
public $name, $DOB, $Fruit;
}
$users = new array();
$query = "SELECT * FROM people";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)) {
$user = new User;
$user->name = $row["name"];
$user->DOB = $row["DOB"];
$user->Fruit = $row["Fruit"];
$users[$row["name"]] = $user;
}
Now you can access the user-data this way:
$users["USERNAME"]->name
$users["USERNAME"]->DOB
$users["USERNAME"]->Fruit