PHP - Link Section Of An Image

What I'm trying to do is have it match up the image file name wiht the same name as the link. Sometimes it does but sometimes obviously the image is random. Any ideas?


Code: [Select]
function spotlight($dbc){
$query = "
    SELECT
        *
    FROM
        characters
    WHERE
        characters.statusID = 1 AND characters.styleID = 1
    ORDER BY
        RAND()
    LIMIT
        1";
$result = mysqli_query($dbc, $query);
$row = mysqli_fetch_array($result);
    $shortName = $row[ 'shortName' ];
$labels = array('shortName');
    $img = array();
if($handle = opendir('images/spotlight/')) {
$count = 0;
while (false !== ($file = readdir($handle))) {
if(strlen($file) > 2){
$img[$count] = $file;
$count++;
}
}
}
echo "<a href='bio.php?shortName=" . $shortName . "'><img src='/images/spotlight/".$img[rand(0, (count($img)-1))]."' alt=Spotlight border=0 /></a>";
}

It is easy to get image or link by DomDocument, but I did not find a way to get image with its target link. Imagine a html as
Code: [Select]
<div class=image>
<a href='http://site.com'><img src='imagelink.jpg'></a>
</div>How to get both the image link and href?

$dom = new DOMDocument();
@$dom->loadHTML($html);
$xpath = new DOMXPath($dom);
$hrefs = $xpath->evaluate("/html/body//div[@class='image']");
for ($i = 0; $i < $hrefs->length; $i++) {
$href = $hrefs->item($i);

Now to get the image and its href, we need first getElementsByTagName('a') and getElementsByTagName('img') but they do not work inside foreach.

What's your idea?

Hi   I'm trying to change an image when a certain link is clicked. The image is a logo of a company and the links that will be clicked is english, arabic, and Chinese.   So if someone clicks on Chinese then the logo changes to the Chinese version of the normal logo.       Here is my html

<?php
$langArabic='';
    $langChinese='';
    $langSpanish='';
    $langEng='';

    $linkArabic=getSEOLink($page_id);
    $linkChinese=getSEOLink($page_id);
    $linkSpanish=getSEOLink($page_id);
    $linkEng=getSEOLink($page_id);

    $queryLink = mysql_query("SELECT * FROM $_SEO_TABLE WHERE `id_page`='".$page_id."'");
    if(mysql_num_rows($queryLink) > 0){
        $resultLink = mysql_fetch_object($queryLink);
        if($resultLink->url != ''){
            $linkEng = $_HTTP_ADDRESS."".$resultLink->url;
            if($page_id == 1 || $page_id == '1'){
                $linkEng = $_HTTP_ADDRESS;
            }
        }
        if($resultLink->url_arabic != ''){
            $linkArabic = $_HTTP_ADDRESS."".$resultLink->url_arabic;
        }
        if($resultLink->url_chinese != ''){
            $linkChinese = $_HTTP_ADDRESS."".$resultLink->url_chinese;
        }
        if($resultLink->url_spanish != ''){
            $linkSpanish = $_HTTP_ADDRESS."".$resultLink->url_spanish;
        }
    }

    switch (urlExtension()) {
        case 'html':
            $langClass = '';
            $langEng='langActive';
            $langTag='en';
            break;
        case 'ar':
            if($page->title_arabic == ''){
                $langClass = '_arabic'; 
            }else{
                $langClass = '_arabic';
            }
            $langArabic='langActive';
            $langTag='ar';
            break;
        case 'es':
            $langClass = '';
            $langSpanish='langActive';
            $langTag='es';
            break;
        case 'cn':
            $langClass = '';
            $langChinese='langActive';
            $langTag='cn';
            break;
        default:
            $langClass = '';
            $langEng='langActive';
            $langTag='en';
    }
?>

<a class="main-logo<?php echo $langClass;?>" href="javascript:void(0);"><img src="<?php echo $_HTTP_ADDRESS;?>images/logo-letters.png" /></a>


<div class="language-selection-wrapper<?php echo $langClass;?> overlay-bg<?php echo $langClass;?> border-bottom-white<?php echo $langClass;?> padding-level-one<?php echo $langClass;?>">
                <div class="language-inner-wrapper<?php echo $langClass;?>">
                    <a class="<?php echo $langEng;?>" href="<?php echo $linkEng;?>">English</a><span>|</span>
                    <a class="<?php echo $langArabic;?>" href="<?php echo $linkArabic;?>">العربيّة</a><span>|</span>
                    <a class="<?php echo $langSpanish;?>" href="<?php echo $linkSpanish;?>">Español</a><span>|</span>
                    <a class="<?php echo $langChinese;?>" href="<?php echo $linkChinese;?>">中文</a>
                    <div class="clearboth"></div>
                </div>

 

I'm wondering if anyone can help me find/create a script which allows a preview of a link to be generated?

Basically I need a user to be able to simply enter a url and click submit and then a preview of that link will be shown in the form of some text and an image. Very similar to what Facebook does when a user shares a link. I think they use the description tag of the site in general.

Can anyone point me in the right direction in doing this? I need to be able to do it in php/javascript alone without the use of jquery or plugins.

Hello guys, I have created a mini image hosting website. Well, I have successfully coded the file upload, including security to allow certain image extensions and size as a beginner in PHP.

However, only one thing remains is the image link. You can view the website on this address http://mini-image-hosting.99k.org/ where it is currently hosting on a free web hosting account with a free sub-domain.

Right now, only the image can be uploaded and is being stored in a directory. Nevertheless, I want that when the person uploads an image, he gets also the link, for example: http://mini-image-hosting.99k.org/xxx.jpg something like that.

Can you help me for this?

I have an image which is a complete image that has  4 titles(aquatic weeds,green algae,water quality,bottom muck)
Can i write a code somehow that when i hover over those images or title i can link it to different hyperlinks?

Greetings!
I have a website www.lanceronlinejobs.com/our_franchises.php I have a franchise images. I want to display each franchise record from database whenever a user click on franchise link. Here is my code.
ourfranchises.php  code:

Code: [Select]
<?php 
include('fconnection.php');

$sql = mysql_query("SELECT * FROM tbl_franchise ORDER BY id DESC") or die(mysql_error());
$row3 = mysql_fetch_array($sql); 
?>
<a href="franchiseDetails.php?city=<?php echo $row3['city'];?>"><img src="images/lbatkhela.jpg" width="150" height="150" alt="Batkhela" /></a>----------------------------------------------------------------------

franchiseDetails.php code:

Code: [Select]
<?php
$id = $_GET['id'];

$query = "SELECT * FROM tbl_franchise WHERE id = '$id' ORDER BY id DESC LIMIT 1";

$result = mysql_query($query);

if (!$result) {
echo "NO RECORD FOUND";
} else {

while($row3 = mysql_fetch_array($result)):
?>

Manager Name: <?php echo $row3['manager_name'];?>
Please help me. Any help would be appreciated.

Thanks.

I basically have a PHP Search Form, and when a user fills in a form it outputs the results. Each result displays a image of a property, how could i make them images have their own unique link which will take them directly to the page of the property being shown?

Im using PHP and mySQL tables

Any help is appreciated,

Thank You.

Heres the PHP that outputs the results:
Code: [Select]

<?php
require_once 'mstr_ref.php';
function san($input){
if(get_magic_quotes_gpc()){
$input=stripcslashes($input);
}
$output = mysql_real_escape_string($input);
return $output;
}
if(isset($_POST['submit'])){
$pVars = array('area'=>$_POST['areas'], 'propType'=>$_POST['prop_type'], 'saleType'=>$_POST['ptype'], 'minB'=>$_POST['min_bedrooms'], 'maxB'=>$_POST['max_bedrooms'], 'minP'=>$_POST['min_price'], 'maxP'=>$_POST['max_price']);
foreach ($pVars as $k=>$v){
$v = san($v);
}
$sql = new makeQuery();
$sql->manAdd('location_id', $pVars['area']);
if($pVars['propType'] != 'Any'){
$sql->manAdd('catagory_id', $pVars['propType']);
}
if ($pVars['maxB'] > 0){
$sql->manAdd('bedrooms', $pVars['maxB'], '<=');
}
if($pVars['minB'] > 0){
$sql->manAdd('bedrooms',$pVars['minB'],'>=');
}
if($pVars['saleType'] != 'Any'){
if($pVars['saleType'] == "forsale"){
$sql->manAdd('market_type', 'sale');
if($pVars['minP'] != 0){
$pVars['minP'] = $pVars['minP'] * 1000;
}
if($pVars['maxP'] != 0){
$pVars['maxP'] = $pVars['maxP'] * 1000;
}
}
if($pVars['saleType'] == 'forrent'){
$sql->manAdd('market_type', 'rent');
}
}

$qry = $sql->sqlStart.$sql->stmt.'Group By property.id';
$results = mysql_query($qry) or die (mysql_error()."<br />|-|-|-|-|-|-|-|-|-|-|-|-<br />$qry");
if(mysql_num_rows($results) < 1){
die ("Sorry, No Results Match Your Search.");
}
while($row = mysql_fetch_assoc($results)){
echo '<div class="container" style="float:left;">';
echo '<div class="imageholder" style="float:left;">';
echo "<img class='image1' src='{$row['image_path']}' alt='{$row['summary']}'> <br />";
echo '</div>';
echo '<div class="textholder" style="font-family:helvetica; font-size:14px; float:left; padding-top:10px;">';
echo "{$row['summary']}";
echo "<span style=\"color:#63be21;\"><br><br><b>{$row['bedrooms']} bedroom(s) {$row['bathrooms']} bathroom(s) {$row['receptions']} reception room(s)</b></span>";
if($row['parking'] != null){
echo "<span style=\"color:#63be21;\"><b> {$row['parking']} parking space(s)</b></span>";
}

echo '</div>';
echo '<div style="clear:both"></div>';
}

}
else{
echo "There was a problem, please click<a href='index.php'> Here </a>to return to the main page and try again";
}
?>

Hey everyone,

I could use some help with this one..
I need to create some php code to switch images when a certain page loads.

For example:

These are the pages, test1, test2,test3, so when you click on the test1 link in the navigation a certain image would show up :

something like this:
If test1 exists then show image1.jpg
If test2 exists then show image2.jpg
If test3 exists then show image2.jpg.

I am doing this in wordpress and I can't figure out the if then statement..

Does anyone have any ideas??

thanks

Hello i kinda new to programing mostly learning myself stuf there so i need some help to figure out.

so i have lets say static link who always same and i write something in input field and press search button. so i get new link and immediately going to dat link, which means i leaving my site. in my case i working with image

so my problem, is it somehow possible to display image in same page while writing someting in input field when presing button and image just change down below not going directly to link ? 

P.S. sory for my english

 

i have this code 

<html>
    <head>
    <meta charset="utf-8">
        <title>Finder</title>
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
    </head>

<body>

    <center>
    <h1>  <strong>Finder<br>
      <br>
      </strong></font></h1>

    <form action="#" method="POST">
        <input type="hidden" name="link" value="src1">
        <input type="text" name="inputpart" placeholder="inputpart" size="40">
        <input type="submit" name="submit" value="Search" >

    </form>
        </center>

    <?php require_once 'linkgenerator.php' ?>
</body>
</html>

 

PHP link generator code 

<?php
    if (isset($_POST['link'])) {
        if ('src1' === $_POST['link']) {
            header("Location: https://staticlink" . $_POST["inputpart"] . ".png");
        }
    }
    ?>

 

 

Edited March 18, 2020 by eagle101

Hey guys so I am making an App website and somehow need to make a developer section. So like developers can come and setup their app name,description etc. Then be able to program there own app and use like an API service to get the user username etc. So what would I need to let developers do a such thing like let them make an API key and all that is there some tutorial to make such service. (This will be an important feature to my website so please help, thanks)

Hi php freaks

I've been learning php, but what confuses me are the arrays.
I have ini file, i want to get the data from file and add it to the database. Every section makes a new table and all they keys are added with their values. I figured out how to get the keys and how to see the values, but how can i get the name of the sections. Furthermore, how do i know what keys belong to what section.

I am really grateful for person who helps me to realize how to be a good friend with those complicated arrays.

Hi.. I am relatively new. 

I am trying to show an image link in my code below:

$image = "<img src=<imgpg/" . $part_no . ".JPG />";

<td  class = "guide_body" width = "66%"></td><td align = "center"><?php echo $image;?>

There is more code but these are the problem lines I believe.  Both are within the php environment.

This is producing:
http://localhost/%3Cimgpg/AF00016B.JPG

It is supposed to look like this:
http://localhost/imgpg/AF00016B.JPG

So it is adding %3C for some reason between localhost and imgpg directory.  I have no idea why this is happening.

Is there something I can do to prevent this from happening?  Or is there something I can do to erase the extra characters.  (I'm not sure if they might be different though at some point.  I have the same problem both on my local computer and the remote server.)

Thanks very much for any help!!