I'm new here, so do not be angry if I ask something that is already answered.
I connected sql database:
connect.php
<?php
$connect = mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("mobilni", $connect) or die(mysql_error());
?>
displaydata.php
<?php
include "connection.php";
$sql= "SELECT * FROM imena WHERE Okrug='Beogradski'";
$query=mysql_query($sql) or die (mysql_error());
?>
<div class="beyondheader"></div>
<div class="header">
<div id='cssmenu'>
<ul>
<li class='active'><a href='index.html'>Početna</a></li>
<li><a href='proizvodjaci.html'>Proizvodjači</a></li>
<li><a href='oglasi.html'>Oglasi</a></li>
<li><a href='about.html'>O nama</a></li>
</ul>
</div>
</div>
<div class="middle">
<div class="leftmiddle">ss
</div>
<div class="rightmiddle">
<?php
while ($row = mysql_fetch_array($query)){
?>
<div class="divmobilni">
Ime:<div class="mobilniime"><?php echo $row['Ime'];?></div>
Okrug:<?php echo $row['Okrug'];?>
</div>
<?php } ?>
</div>
</div>
And everything is working fine. Now I want to put every result in the other <div> automatically?
How to do that?
Sorry if I wasn't clear enough. I want to sort results to be like this, in one <div> goes: Ime: Okrug: [that is data for one person]
Now, I want to make that all data from my SQL table display on this page, but every Person separately from this <div>.
To be something like this, with data from table: (This is just example drawn in Paint)
I fixed this with adding only break to the end of PHP. Thanks in any case!
Exactly the same as you would with php open and close <?php ?>
Use this inside the divs you would like or just use echo with the div inside
Related
This one is going to be a bit hard to explain but I am going to try my best.
I have a database with a table called content with 3 columns, I’m trying to get the values as rows and put them in different <div>’s. here is what I wrote so far
<?php
$sql = "SELECT * FROM content";
$query = mysqli_query($db, $sql) or die (mysqli_error());
$contentDisplay = '';
while ($row = mysqli_fetch_array($query)) {
$content_id = $row["id"];
$content_title= $row["title"];
$content_text = $row["text"];
$contentDisplay .= '<h1>'.$content_title.'</h1> <p>'.$content_text.'</p>' ."\n";
}
$row_cnt = $query->num_rows;
printf("Result set has %d rows.\n", $row_cnt);
?>
The row count gives 4 rows which is correct amount of rows.
In my html I have <?php echo $contentDisplay; ?> which puts out all 4 rows after each other, but I need to show the first row in my first <div>, the second row in my second <div> and so forth. thanks in advance
Update: I forgot to say that I have my <div>'s in the html part with different styles. all i need is the first row in one <div> (like on top of the page) and second row in another <div> (like on the bottom of the page) :)
Update2: here is the html code
<div class="content-top">
<section>
<h1></h1>
<p></p>
</section>
</div>
<div class="content-left">
<section>
<h1></h1>
<p></p>
</section>
</div>
<div class="content-center">
<section>
<h1></h1>
<p></p>
</section>
</div>
<div class="content-right">
<section>
<h1></h1>
<p></p>
</section>
</div>
i want the first row to be displayed in <div class="content-top"> the second row in <div class="content-left"> the third row in <div class="content-center"> and the fourth row in <div class="content-right">
<?php
$sql = "SELECT * FROM content";
$query = mysqli_query($db, $sql) or die (mysqli_error());
$rows = [];
while ($row = mysqli_fetch_array($query))
$rows[] = '<h1>'.$row['title'].'</h1> <p>'.$row['text'].'</p>' . "\n";
?>
<div class="content-top">
<section>
<?php echo $rows[0]; ?>
</section>
</div>
<div class="content-left">
<section>
<?php echo $rows[1]; ?>
</section>
</div>
<div class="content-center">
<section>
<?php echo $rows[2]; ?>
</section>
</div>
<div class="content-right">
<section>
<?php echo $rows[3]; ?>
</section>
</div>
$contentDisplay .= '<div><h1>'.$content_title.'</h1> <p>'.$content_text.'</p></div>' ."\n";
Just add divs for each row? :) This way, each row in the database will get a div.
Each row will be appended to the $contentDisplay variable, but since all you are doing is printing it, you can print the html directly.
I'm not sure why you are being downvoted, since you are asking a valid question with code that shows what you have tried so far.
I have the structure of a Tab component, but I will like that the head of each tab is the name of a person that is in the database stored.
For example, normally this is a tabs structure
<div id="tabs">
<ul>
<li>1</li>
<li>2</li>
<li>3</li>
</ul>
<div class="panels">
<div id="tab-1">
</div>
<div id="tab-2">
</div>
<div id="tab-3">
</div>
</div>
</div>
But instead of Tab 1,2 and 3 as head text for each tab, I will like to have Marc, Josef, Luis and these are name from the DB. This also includes that everytime that a person is added I will have a new tab. Has anybody a idea how I can do this?
I tryed to call the result of the PHP query in the text of the head, but was not working
**********EDITED WITH CODE******************
<?php
$connection = mysql_connect(/*DAta connectionns*/);
if (!$connection)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("", $connection);
$result = mysql_query("SELECT * FROM People");
while($row = mysql_fetch_array($result))
{
$r = $row["name"];
<div id="tabs">
<ul>
<li><?php $r ?></li>
</ul>
<div class="panels">
</div>
</div>
}
?>
I know that there is something wrong, but I can not see
Assume you have a JavaScript Object (from your ajax?) you can do:
var mytabs = [{
name: "Marc"
}, {
name: "Josef"
}, {
name: "Luis"
}];
$('#tabs').tabs();
$.each(mytabs, function(indexInArray, myobj) {
$('#tabs').find('li> a').eq(indexInArray).text(myobj.name);
});
Here is a fiddle you can play with: http://jsfiddle.net/MarkSchultheiss/p4mzy3hb/
<li class="list-group-item">
<a href="/user/Michael" class="thumb-sm pull-left m-r-sm">
<img src="http://www.gravatar.com/avatar/8b7a9ba3cbf958009080f6da12a55029?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a>
<a href="user/Michael" class="clear">
<strong class="block">
<?php include '/includes/connection.php';?>
<?php echo $products['Title'] ; ?>
</strong>
<?php include '/includes/connection.php';?>
<small><?php echo $products['Followers'] ; ?> Followers </small>
</a>
</li>
<li class="list-group-item">
<a href="/user/Steven" class="thumb-sm pull-left m-r-sm">
<img src="http://www.gravatar.com/avatar/a5fb2decd550cdf33cbb8ce7566ba772?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a>
<a href="/user/Steven" class="clear">
<strong class="block">
<?php include '/includes/connection.php';?>
<?php echo $products['Title'] ; ?>
</strong>
<small><?php echo $products['Followers'] ; ?> Followers</small>
</a>
</li>
do i need to manually insert for every content?
I have over 100 contents how can i automatically insert in every line like content 1 display row 1 followers and content box 2 display row 2 followers and so on
The trick here is to make a loop within the code. This means you have to generate all content from out of the database or it's not gonna work. Let me show you an example:
<ul>
<?php
$sql = "SELECT ProjectId, ProjectTitel, ProjectExpertise
FROM project";
$stmt1 = mysqli_prepare($con, $sql);
mysqli_stmt_execute($stmt1);
mysqli_stmt_bind_result($stmt1,$ProjectId,$ProjectTitel,$ProjectExpertise);
while (mysqli_stmt_fetch($stmt1)){
?>
<li class="wow fadeInLeft" data-wow-offset="30" data-wow-duration="1.5s" data-wow-delay="0.15s">
<a href="inc/elements/project.php?id=<?php echo $ProjectId; ?>" class="meer">
<img src="img/portfolio/<?php echo $ProjectId; ?>/thumbnail/1.jpg" alt="<?php echo $ProjectTitel; ?> project">
<div class="project-info">
<div class="project-details">
<h5 class="witte-text blauwe-streep-onder">
<?php echo $ProjectTitel; ?>
</h5>
<div class="details witte-text">
<?php echo $ProjectExpertise; ?>
</div>
</div>
</div>
</a>
</li>
<?php
}
?>
</ul>
I start an UL outside of the PHP code and then I start my PHP query, as you see, i select the items i need from the database. Here i create a while loop and run through my setup of the list element. As you see, i use my items within the project id link (the link is an ajax call to another page), the thumbnail needed to show the picture, the title and the expertise i used.
In your case, the while loop you need requires more than just the the title and followers. I think you need as well an userid / username so you can take in account which user it is. Your loop would now proces all on the same user, since its staticly defined in your code. Also the avatar picture is staticly defined. Let me try to resolve some for you.
<?php
include '/includes/connection.php';
$sql = "SELECT products.title, products.followers
FROM products"; // as example query
$stmt1 = mysqli_prepare($con, $sql);
mysqli_stmt_execute($stmt1);
mysqli_stmt_bind_result($stmt1,$title, $followers);
while (mysqli_stmt_fetch($stmt1)){
?>
<li class="list-group-item">
<a href="/user/Michael" class="thumb-sm pull-left m-r-sm"> <!-- make the username also to be pulled out of a database. -->
<img src="http://www.gravatar.com/avatar/8b7a9ba3cbf958009080f6da12a55029?&d=mm&r=g?&d=mm&r=g&s=215" class="img-circle">
</a> <!-- you should save the avatar link into a database as well -->
<a href="user/Michael" class="clear"> <!-- make the username also to be pulled out of a database. -->
<strong class="block">
<?php echo $title; ?>
</strong>
<small>
<?php echo $followers; ?>
Followers
</small>
</a>
</li>
<?php
}
?>
This as example. As you see, i added notitions behind some code, to explain this should also be dynamic, since it will help you Remember. being a programmer is to find shortcuts on how you display your information. Code that repeats itself constantly with a changing variable should always be looped, to make sure you dont type unneccesairy code.
Since I dont know how your database is made, i can only guess, Michaels name is probably linked to an ID in the same table your products are, which you can use then in to pull out of the database, by searching in your user table. I hope I make sense here. For instance, your products.userid should be as well in your user table as user.userid. Most likely the userid will have a name linked to it in the user table.
$sql = "SELECT products.title, products.followers, products.userid, user.userid, user.username
FROM products, user
WHERE user.userid = product.userid";
So now you have in each row as well the name of the person of who's title it is. And that you can echo out again in the code i put up. (make sure you bind the result in the same order as you pulled them up)
Writing code is all about making it easier to display your information. Loops is and stays the keyword here, as NadirDev explains.
I hope I helped you getting on the right track.
please someone help me to find the solution for how to retrieve data from mysql database and populate in list view in jquery mobile.
am having the php code as follows
<? php
include('libraries/config.php');
$result = mysql_query("SELECT * from category") or die(mysql_error());
while ($obj = mysql_fetch_object($result)) {
$arr[] = $obj;
}
echo json_encode($data);
echo '{"Deals":'.json_encode($arr).'}';
?>
here am getting the data from mysql in json format but i was not known how to populate this in listview, my html page is as follows
<div id="content-area" style="height:auto;">
<br/>
<ul data-role="listview">
<li>
<a href="comfort_list.html">
<div class="content-home-tab1">
<div class="img-content">
<img id="ic_home" src="images/next_btn.png" style="margin-top:37px; margin-left:448px;" width="22" height="28" />
</div>
<div class="content-home-p">
<b>Comfort</b>
</div>
</div>
</a>
</li>
<li>
<a href="#">
<div class="content-home-tab1">
<div class="img-content">
<img id="ic_home" src="images/next_btn.png" style="margin-top:37px; margin-left:448px;" width="22" height="28" />
</div>
<div class="content-home-p">
<b>Handling Your Lenses</b>
</div>
</div>
</a>
</li>
</ul>
</div>
Here in my code am giving static data to list that is comfort and handling your lens, am having this data in db and i need to get that and to be placed here.
in the page i am posting in a form like this
<form method="post" action="help.html">
<ul data-role="listview" >
<?php while ($row = $stmt->fetch()){
?>
<li>
<a href="help1.php?id=<?php echo $row['categoryID']; ?>">
<div class="content-home-tab1">
<div class="img-content">
<img id="ic_home" src="images/next_btn.png" style="margin-top:37px; margin-left:448px;" width="22" height="28" />
</div>
<div class="content-home-p">
<b><?php echo $row['title'];?>
</b>
</div></div></a>
</li>
<?php }?>
</ul>
</form>
and in another page am trying to get the id as follows
<?php
include('libraries/config.php');
$getID = $_GET['id'];
echo $getID;
$stmt = $db->prepare("SELECT * from category where categoryID ='$_GET['id']'");
$stmt->execute();
?>
but when going to that page it is showing nothing and when refreshed then it is showing so that only am asking how to get it to fetch data and print it when the page is loaded. thanks.
As C.S. says, you don't need to use JSON, just loop through your array and add your list items.
Edited to show basic use of PDO, and demo how to access query results.
The mysql_query extension is deprecated as of PHP 5.5.0, so you I would suggest you use PDO_MySQL
First set up your database connection:
$dbName = "your_database_name_here";
$dbUserName = "your_username_here";
$dbPassword = "your_password_here";
$db = new PDO( 'mysql:host=localhost;dbname='.$dbName, $dbUserName, $dbPassword );
Then run your query, and echo your list (or whatever you need from the database by accessing the $row array within the while loop):
$stmt = $db->prepare("SELECT * from category");
$stmt->execute();
while ($row = $stmt->fetch()){
echo "<li>". $row['title'] ."</li>";
}
p.s. you should get rid of that inline css, it's a bad habit to get into.
Will someone PLEASE show me what I'm doing wrong? I'm new to using phpMyAdmin and db.
I have a db table with 4 items. Looks fine on a plain php page - I get all four items in correct order.
But if I use coding <ul class="column"><li><div class="imgblock"> like shown below, I get 4 duplicates of my last entry in the db table. I think I have to rearrange how I use ", ', and ), etc.? Not sure how though...
<?php
// Make a MySQL Connection
mysql_connect("localhost", "....", "....") or die(mysql_error());
mysql_select_db("....") or die(mysql_error());
$show = "SELECT pn, pgname, img, name, price FROM prodshort";
$result = mysql_query ($show);
while ($show = mysql_fetch_array ($result))
{
$field2= $show['pn'];
$field3= $show['pgname'];
$field4= $show['img'];
$field5= $show['name'];
$field6= $show['price'];
$field7= $show['specs'];
}
?>
<!-- start -->
<ul class="column"><li><div class="imgblock">
<a href="<?echo "$field3";?>">
<img src="<?echo "$field4";?>" width="320" height="240" alt="<?echo "$field5";?>" /></a></div><br />
<?echo "$field5";?>
<ul class="specs">
<?echo "$field7";?></ul>
<div class="price">
#<?echo "$field2";?> $<?echo "$field6";?>
</div></li></ul>
<!-- end -->
The above is probably not the code from the example, as the while loop is outside your items.
It should be more like this:
while ($show = mysql_fetch_array ($result)) {
$field2= $show['pn'];
$field3= $show['pgname'];
$field4= $show['img'];
$field5= $show['name'];
$field6= $show['price'];
$field7= $show['specs'];
?>
<!-- start -->
<ul class="column">
<li>
<div class="imgblock">
<a href="<?echo "$field3";?>">
<img src="<?echo "$field4";?>" width="320" height="240" alt="<?echo "$field5";?>" />
</a>
</div>
<br />
<?echo "$field5";?>
<ul class="specs">
<?echo "$field7";?>
</ul>
<div class="price">
#<?echo "$field2";?> $<?echo "$field6";?>
</div>
</li>
</ul>
<!-- end -->
<?
// end of while loop
}
?>
(You might find it easier to debug if you indent your code correctly, most editors do that for free ;-)