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I have following data in mysql, and I want to echo in this format:
Data:
Name,Image URL,Link
I want to print it dynamically like this : http://screensaver.cf/screensavers.php
I don't know user's screen width, still I want it to appear as much as possible in width.
How can I do this in html and PHP?
My code:
<?php
require("config.php");
?>
<div id="main-wrap">
<div class="container">
<div id="main">
<div id="content"><div id='wsite-content' class='wsite-elements wsite-not-footer'>
<div class="paragraph" style="text-align:left;">
<?php
$sql='SELECT * FROM `games` where 1=1';
$data = mysql_query($sql);
echo '<h4 class="result">Result:</h4>';
while($row = mysql_fetch_row($data)){
$table=WHAT TO DO HERE TO MAKE IT LOOK LIKE THAT???????
echo $table;
echo '<br><br>';
}
?>
</div>
</div>
</div>
</div>
</div>
<?php
include("footer.php");
?>
P.S I know mysql is depreciated and I am constantly working to learn Mysqli, as I am in 8th class, I don't have much time.
<style>.a{float:left;}
</style>
>
in while loop use this
<div class = "a">
<?php <img src='".$row['url']."' width='140' height='140'> ?>
</div>
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I tried this code for displaying an item name, price and a photo from database for searching an item name in a search bar
<?php
try
{
require('connection.php');
$sql2="SELECT * FROM items
where item.item_name like '".$txt1."%'
$result2=$db->query($sql2);
if ($result2->rowCount()!=0) {
foreach ($result2 as $r2) { ?>
<a href="images/<?php echo $r2['item_photo'] ?>" class="fh5co-card-item image-popup">
<figure>
<div class="overlay"><i class="ti-plus"></i></div>
<img src="images/<?php echo $r2['item_photo']?>" alt="Image" class="img-responsive">
</figure>
<div>
<?php echo "<p><span class='price kk'>" ; ?>
<p style="font-family:'georgia';text-align:center;font-size:20px;color:black;"><?php echo ($r2['item_name']); ?></span></p>
<h2 style="font-family:'georgia';text-align:center;font-size:18px;color:grey;"><?php echo ("BD ".$r2['item_price']); ?></h2>
</div>
<?php }
} ?>
The problem is the sql, There is an extra name after where
$sql2="SELECT * FROM items
where item.item_name like '".$txt1."%' ;
There should be only item_name not item.item_name. Change this as:
$sql2="SELECT * FROM items
where item_name like '".$txt1."%';
And your code may be exploited with sql injection. Be carefull for production use this code.
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I am having multiple images, for that individual image i am having download button
<div>
<?php
for($a=0;$a<5;$a++) { ?>
<img src='image_<? echo $a;?>.jpg'/>
<button id='dwn_<? echo $a;?>'>Download</button>
<br>
<? } ?>
</div>
If i click that download button i need to save that corresponding image to my local system.
Kindly give some solution for this.
the easiest way without javascript i think it would ( I edited your code, but i cant try it now)
<div>
<?php
for($a=0;$a<5;$a++) { ?>
<img src='image_<? echo $a;?>.jpg'/>
<form method="get" action='image_<? echo $a;?>.jpg'>
<button id='dwn_<? echo $a;?>' type="submit">Download</button>
</form>
<br>
<? } ?>
</div>
You can delete the id in button if you only use for download
download.php
.........
<?php
if(isset($_REQUEST) && ($_REQUEST['image_name'])){
$filename = $_REQUEST['image_name'].'.jpg';
// Write download code here
}
display_page.php
............
<div>
<?php
for($a=0;$a<5;$a++) { ?>
<img src='image_<? echo $a;?>.jpg'/>
<a href="download.php?image_name=<?php echo $a; ?>" >Download</a>
<br>
<? } ?>
</div>
Most easiest way by using html 5
<div>
<?php
for($a=0;$a<5;$a++) { ?>
<img src='image_<? echo $a;?>.jpg'/>
<a href='image_<? echo $a;?>.jpg' download>
<span id='dwn_<? echo $a;?>'>Download</span >
</a>
<br>
<? } ?>
</div>
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I need the data to have a different style if the sold = 0 in the mysql table
and when i use the code below the website shows a blank white page
(?=$vehicle-> this is the vehicle reference
and sold is the column withing the sql table
<?php
if (?=$vehicle->sold?!= 1)
{
<div class="foo">
<div class="fboverlay"></div>
<a>
<img src="/media.php?productId=<?=$vehicle->vehicle_id?>&file=<?=$vehicle->main_image?>" />
</a>
</div>
}
else {
<img src="/media.php?productId=<?=$vehicle->vehicle_id?>&file=<?=$vehicle->main_image?>" />
}
?>
You need to learn the PHP from the basics. Learn about operators, PHP and HTML secion, etc..
Anyway, i fixed your code. The condition is if $vehicle->sold is not equal to 1. But i think, (in your OP you mentioned it should be 0) you want this: $vehicle->sold == 0
//Use sytnax like this. See php opeartors.
if ($vehicle->sold != 1) {
?> <!-- Close the php -->
<div class = "foo">
<div class = "fboverlay"></div>
<img src = "/media.php?productId=<?= $vehicle->vehicle_id ?>&file=<?= $vehicle->main_image ?>" />
</div>
<?php
//Open the php again
} else {
?> <!-- close the php -->
<img src = "/media.php?productId=<?= $vehicle->vehicle_id ?>&file=<?= $vehicle->main_image ?>" />
<?php //Open again
}
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This is a code snippet I have and I would like to be able to select the out of PHP. When I try to grab the PHP output and store it into a variable, I receive "undefined".
<h1 id=<?php echo '$test; ?> class="list-name--original">
My last attempt I tried this code:$("h1.list-name--original")[0].outerHTML);
Why not just put all your PHP output in a variable
<script>
var data = '<?=$test?>';
</script>
You can then set the contents of an H1 tag like this
<h1 id="myid"></h1>
<script>
$('h1#myid').text(data);
</script>
Try
<h1 id="<?php echo $test; ?>" class="list-name--original">
in jQuery
$(".list-name--original").html();
or
$("#<?= $test ?>").html();
Try this:
<?php $test = "Hello world!"; ?>
<h1 id="example" class="list-name--original"><?php echo $test; ?></h1>
<script language="javascript">
$("#example").html();
</script>
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All of this needs to be inside my loop so I don't know if I can break it (don't think I can) but I need PHP to read it as HTML.
$movieDetails =
<header class="masthead" role="banner">
<div class="logo-container">
<a class="logo" href="">Movie Details</a>
</div>
</header>
<div class="bubble-container">
<div class="speech-bubble speech-bubble-top">
Text that goes into the bubble
</div>
</div>
You can just close your PHP tag to render text that will display as HTML.
<?php
foreach($movies as $movie) {
?>
<div>
<?php echo htmlspecialchars($movie); ?>
</div>
<?php
}
?>
$movieDetails = <<<HTML
<header class="masthead" role="banner">
<div class="logo-container">
<a class="logo" href="">Movie Details</a>
</div>
</header>
<div class="bubble-container">
<div class="speech-bubble speech-bubble-top">
Text that goes into the bubble
</div>
</div>
HTML;
echo $movieDetails;
http://www.php.net/manual/en/language.types.string.php#language.types.string.syntax.heredoc