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i want to change this into html format
echo " http://somedomain/somepath/usn_handler.php?usn='" . $row['usn'] .
"' >" . $row['usn'] . " ";
This will work
?>
link
<?php
If you are just making a link to the same domain omit the full URL
?>
link
<?php
Assuming you want to make a hyper link. But really you should just google first.
I think you mean something like this :
<div>
http://somedomain/somepath/usn_handler.php?usn=<?php echo $row['usn']; ?>' >" <?php echo $row['usn']; ?>
</div>
or :
<div>
<a src="http://somedomain/somepath/usn_handler.php?usn=<?php echo $row['usn']; ?>" > <?php echo $row['usn']; ?></a>
</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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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 $jsonObj->data->image; ?>" />
<?php
}
?>
You can add else part and do whatever you want.
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I am having multiple images, for that individual image i am having download button
<div>
<?php
for($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 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 have this small piece of code
<?php
foreach ($options['attributes_info'] as $attribute) :
?>
<?php
echo '<br/> ' .
$this->getPdfHelper()->fixEncoding(
$attribute['label'] . ': ' . $attribute['value']
)
?>
<?php
endforeach;
?>
How can I bold (HTML tag <b>) the 'value' that is echoed?
<?php foreach ($options['attributes_info'] as $attribute) : ?>
<?= $this->getPdfHelper()->fixEncoding($attribute['label']) ?> : <b><?= $this->getPdfHelper($attribute['value']) ?></b>
<?php endforeach; ?>
You can wrap the text with the tag.
<p><b>This text is bold</b>.</p>
http://www.w3schools.com/html/html_formatting.asp
You can simply surround it with the tag:
'<b>' . $attribute['value'] . '</b>'