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in my opencart I have image in products category page.
If I have set image in category show in category if I haven't set image show onather one.
<?php if ($thumb) { ?>
<div class="category-img" style="background: url('<?php echo $thumb; ?>') no-repeat;"></div>
<?php }else{ ?>
<div class="category-img" style="background: url(image/catalog/category/default-thum.jpg) 50% 50% no-repeat;"></div>
<?php } ?>
I want to modify code.
If I have set image then show image,
else if
No showing.
Can someone tell me how to modify code?
Use ternary operator .Check this code:
<div class="category-img" style="background: url('<?php echo ($thumb ?: 'image/catalog/category/default-thum.jpg'); ?>') 50% 50% no-repeat;"></div>
http://php.net/manual/en/language.operators.comparison.php
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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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If <?php echo $jsonObj->data->image; ?> has a value, like 1234.jpg I want to show this:
<img src="https://www.example.com/img/<?php echo $jsonObj->data->image; ?>" />
If <?php echo $jsonObj->data->image; ?> has no value, then <img src="https://www.example.com/img/<?php echo $jsonObj->data->image; ?>" /> should remain hidden.
Any help please?
Use php if-else statement for that.
<?php
if(!empty($jsonObj->data->image)) {
?>
<img src="https://www.example.com/img/<?php echo $jsonObj->data->image; ?>" />
<?php
}
?>
You can add else part and do whatever you want.
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My CSS and HTML are not working. div2 should render below div1 but it's not working on mobile.
#div1 {
width:100%
}
#div2 {
width:100%;
}
<div id = "main">
<div id = "div1">
</div>
<div id = "div2">
</div>
</div>
div {
display:inline-block
}
#div1 {
width: 100%
}
#div2 {
width: 100%;
}
<div id="main">
<div id="div1">
</div>
<div id="div2">
</div>
</div>
making the divs diplay inline:block should solve the problem
Try this one
#div1,#div2 {
width: 100%
display:inline-block;
vertical-align: top;
}
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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 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
}