PHP - Clickable Buttons Within Php If Statements

images are uploaded to a directory and the sql database contains the filename. I added 'url' field and am trying to link the two

I have used on of the posts to help me adapt some script however I'm not sure I have the syntax right.  The image in DW live view looks ok and I get the link finger ok but when previewed in browser the image is not shown at all or any link.

Any ideas


<? echo !empty($row['pic']) ?  "<a href=\"{$row['url']}\" <img src=\"js/images/{$row['pic']}\">" : ''; ?>


it echos out to a table

Mike

I want to display 3 clickable images in a single row which repeats as long as there is data in the database, so far it is displaying a single clickable image from the database. below is all the code..

  <table width="362" border="0">
         
         <?php     
$sql=mysql_query("select * from `publication` GROUP BY `catsue`") or die(mysql_error());
$num=mysql_num_rows($sql);

while($rowfor=mysql_fetch_array($sql))
{
   
      
$cat=$rowfor['catsue'];
   $pic=mysql_query("select * from `category` where `catsue`='$cat'") or die(mysql_error());
   $picP=mysql_fetch_array($pic);
$base=basename($picP['title']);
?>
   <tr>
              <td width="352" height="88"><table width="408" border="0">
                <tr>
                  <td width="113" rowspan="5"><a href="archive_detail.php?id=<?php echo $rowfor['id'];?>&category=<?php echo $rowfor['catsue'];?>"><img src="ad/pic/<?php echo $base;?>" width="100" height="100" border="0"/></a></td>
                  <td width="94">Title</td>
                  <td width="179" height="1"><?php echo $rowfor['catsue'];?>&nbsp;</td>
                </tr>
                <tr>
                  <td>&nbsp;</td>
                  <td width="179" height="3">&nbsp;</td>
                </tr>
                <tr>
                  <td>&nbsp;</td>
                  <td width="179" height="8">&nbsp;</td>
                </tr>
                <tr>
                  <td>&nbsp;</td>
                  <td width="179" height="17">&nbsp;</td>
                </tr>
                <tr>
                  <td>&nbsp;</td>
                  <td width="179" height="36">&nbsp;</td>
                </tr>
              </table></td>
              </tr>
              <?php

}?>
           
          </table>

Is it possible for a clickable image to pass to a php page? It can lead to a linked web page, so can it be like a submit button on a form and pass to the $_POST in php?

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=320820.0

I'm trying to make my text and images into clickable links. The images and text are in a database so its getting really tricky for me since I'm new to PHP. I've tried many different ways without success.

Here is a snippet of the code I'm working with. As you can see, I'm also working with an image re-sizer. I would really like to make those images and text
clickable links(i.e. Title). Any help is appreciated!

if ($res) {
while ($newArray = mysqli_fetch_array($res, MYSQLI_ASSOC)) {
$detail = $newArray['id'];
$photo = $newArray['photo'];
$id = $newArray['title'];
$price = $newArray['price'];


list($width) = getimagesize($photo);
// set the maximum width of the image here
$maxWidth = 100;
if ($width > $maxWidth)


echo "<p><img alt=\"Image\" width=\"$maxWidth\" src=\"$photo\" />";

echo "Title:".$id." Price:".$price."<br/";

Hi,

 I want to make the phone number in the below code as clickable. Below is the code php and mysql code I used.

 

    <?php  
    $connect = mysqli_connect('localhost', 'root', '1234', 'contact');  
    $query ='SELECT * FROM tbl_employee ORDER BY ID DESC';  
    $result = mysqli_query($connect, $query); 
    ?>

   

<table id="employee_data" class="table table-striped table-bordered">
                <thead>  
                <tr>  
                <td>Name</td>  
                <td>Location</td>  
                <td>Department</td>  
                <td>Phone Number</td>  
                </tr>  
                </thead>  
                          <?php  
                          while($row = mysqli_fetch_array($result))  
                          {  
                               echo '  
                               <tr>  
                                    <td>'.$row["name"].'</td>  
                                    <td>'.$row["location"].'</td>    
                                    <td>'.$row["department"].'</td>  
                                    <td>'.$row['phone_number'].'</td>
                  
                               </tr>  
                               ';  
                          }  
                          ?>  
</table>

 

I'm not sure if this is the right section, if not, please move it on proper one, i'm still newbie.

However, here's my problem, i'm using jquery to replace standard check boxes with images, and here's the code:


<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.0/jquery.min.js"></script>
<script type="text/javascript">

$(document).ready(function() {

/* see if anything is previously checked and reflect that in the view*/
$(".checklist input:checked").parent().addClass("selected");

/* handle the user selections */
$(".checklist .checkbox-select").click(
function(event) {
event.preventDefault();
$(this).parent().addClass("selected");
$(this).parent().find(":checkbox").attr("checked","checked");
}
);

$(".checklist .checkbox-deselect").click(
function(event) {
event.preventDefault();
$(this).parent().removeClass("selected");
$(this).parent().find(":checkbox").removeAttr("checked");
}
);

});

</script>

Here's the css:

<style type="text/css">

body {
font: 12px/1.3 "Lucida Grande","Lucida","Arial",Sans-serif;
}

form {
margin: 0 0 30px 0;
}

legend {
font-size: 17px;
}

fieldset {
border: 1;
}

.checklist {
list-style: none;
margin: 0;
padding: 0;
}

.checklist li {
float: left;
margin-right: 10px;
background: url(i/checkboxbg.gif) no-repeat 0 0;
width: 105px;
height: 80px;
position: relative;
font: normal 11px/1.3 "Lucida Grande","Lucida","Arial",Sans-serif;
}

.checklist li.selected {
background-position: -105px 0;
}

.checklist li.selected .checkbox-select {
display: none;
}

.checkbox-select {
display: block;
float: left;
position: absolute;
top: 55px;
left: 10px;
width: 85px;
height: 23px;
background: url(i/select.gif) no-repeat 0 0;
text-indent: -9999px;
}

.checklist li input {
display: none;
}

a.checkbox-deselect {
display: none;
color: white;
font-weight: bold;
text-decoration: none;
position: absolute;
top: 55px;
right: 10px;
}

.checklist li.selected a.checkbox-deselect {
display: block;
}

.checklist li label {
display: block;
text-align: center;
padding: 8px;
}

.sendit {
display: block;
float: left;
top: 118px;
left: 10px;
width: 115px;
height: 34px;
border: 0;
cursor: pointer;
background: url(i/sendit.gif) no-repeat 0 0;
text-indent: -9999px;
margin: 20px 0;
}

</style>

And finally the form thing

<ul class="checklist"><li><input name="jqdemo" value="value5" type="checkbox" id="choice_c"/><label for="choice_c">Some Text Here to be checked - clicked</label><a class="checkbox-select" href="#">Click</a><a class="checkbox-deselect" href="#">Cancel</a></li></ul>


I would like that my text on top of checkboxes become clickable.

Hi, I have a blog page I am working on and I am trying to make some enhancements that I need some help with. In MySQL I have a table with id, title and post. In my page display I want to enable the user to click on the title and display a post on it's own so the URL would look like www.mysite.com/blog/?id=1 or something like that. Here is the PHP function to display the post from MySQL. all on one page.


function GetBlogPosts($inId=null, $inTagId =null)
{
if (!empty($inId))
{
$query = mysql_query("SELECT * FROM blog_posts WHERE id = " . $inId . " ORDER BY id DESC LIMIT 2"); 
}
else if (!empty($inTagId))
{
$query = mysql_query("SELECT blog_posts.* FROM blog_post_tags LEFT JOIN (blog_posts) ON (blog_post_tags.postID = blog_posts.id) WHERE blog_post_tags.tagID =" . $tagID . " ORDER BY blog_posts.id DESC");
}
else
{
$query = mysql_query("SELECT * FROM blog_posts ORDER BY id DESC");
}

$postArray = array();
while ($row = mysql_fetch_assoc($query))
{
$myPost = new BlogPost($row["id"], $row['title'], $row['post'], $row['postfull'], $row["author_id"], $row['date_posted']);
array_push($postArray, $myPost);
}
return $postArray;
}

The below is constantly loading the password page no matter what is enter in the password input on the form (on another page). I have checked the $login variable by echoing it and it is carrying the value from the field.


<?php
$off="yes";
$login=$_POST['login'];
if($off=="yes" && (!isset($_SESSION['login']) || $_SESSION['login']!="pass")){
    if($login=="" || $_SESSION['login']!="pass"){
    header('Location: ./index2.html');
    exit();
    }
    elseif(!isset($_SESSION['login'])){
    session_start();
    $_SESSION['login']=$login;
    header('Location: ./index.php');
    exit();
    }
}
elseif($off=="no" || $_SESSION['login']=="pass"){
content goes here.
}
?> 

Anything wrong with this because I got a parse error ONLY after adding to create this.

<?php
                            $visible = array('Yes', 'No');
                            foreach($visible as $visible):
                            $visible2 = array('yes', 'no');
                            foreach($visible2 as $visible2):
                        ?>      

Hey Guys,

I am creating an array and I have a need to put a if statement inside it. Is this even possible?

Here is where I am at so far, its just not working.


 function load_permissions($gid)
    {
     $data = array();
     $this->db->select('`groups`.`name` AS `permission_name`, `permissions`.``');
$query = $this->db->get('`groups`,`permissions`');
if ($query->num_rows() > 0){
foreach ($query->result_array() as $row){
  $data[] = array(
  "name" => $row['permission_name'],
  "read" => if($row['level'] >= 1)
   { 
    img("img/true.gif")
   } else { 
    img("img/false.gif")
   },
 
  );
  }
 }
 
 $query->free_result();
return $data;



Thanks,
Peter

Hi Guys,
Newbie trying to understand PHP conditionals.
In the below Code: [Select]
[[*id]] is a global variable which equates to my current page id so it could have the value of 1 for example. I tested it using echo and the correct values pass to it in the cms I'm using. What I cant achieve is for the conditionals below to work correctly so perhaps you can see a syntax error or show me away of doing the same using an array or something perhaps? THANKS

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

if ($id=="1" || $id=="2")
  echo "we are in 1 or 2";
elseif ($id=="3" || $id=="4")
  echo "we are in 3 or 4";
else
  echo "not in 1 to 4";
?>

Hello, I am learning PDO SQL statements, I have mananged to connect using PDO:


Code: [Select]
try
{
$this->link = $dbh = new PDO('mysql:host='.$this->dbhost.';dbname='.$this->dbname.'', $this->dbuser, $this->dbpass);
}
catch (PDOException $e)
{
echo 'Connection failed: ' . $e->getMessage();
}


I am not getting any error, so I guess that is a good start. I then tried the PDO SQL statement out. My old mySQL query is commented out in the


Code: [Select]
if (isset($_SESSION['id']) && isset($_SESSION['password']))
{
$_SESSION['id'] = ( isset( $_SESSION['id'] ) ) ? $_SESSION['id'] : FALSE;
$_SESSION['password'] = ( isset( $_SESSION['password'] ) ) ? $_SESSION['password'] : FALSE;


//$logged = mysql_query("SELECT * FROM `db_members` WHERE `id`='".$_SESSION['id']."' AND `password` = '".$_SESSION['password']."'");
//$logged = mysql_fetch_array( $logged );

// the new pdo statement
$ff = $dbh->prepare('SELECT * FROM db_members WHERE id = '.$_SESSION['id'].' AND password = '.$_SESSION['password'].'');

$ff->execute();
$logged = $ff->fetchAll();

echo $logged['username'];
}

I am trying to assign the session to logged variable. So all I am asking is go into db_members and check the id and password that is the same as session id and password and collect the rows data such as username.

My login script and everything works perfectly, even session id and password is valid when echo'd but I cannot assign it to the logged variable like my old sql statements.


Code: [Select]
$logged = mysql_query("SELECT * FROM `db_members` WHERE `id`='".$_SESSION['id']."' AND `password` = '".$_SESSION['password']."'");
$logged = mysql_fetch_array( $logged );

I used other PDO statements and it works perfectly but I just don't understand why this is not working.... can I please get some help if you have any solution to this?

Hey guys,

is there any way to have a LIKE condition on an IF statement?
Something like:
Code: [Select]
if $GET_['WTV'] LIKE %LOVE% {
$category=romance
Any clues?
Thanks in advance.

I am trying to write an if statement that checks to make sure the information put in the text box is an imbedded file. Does anyone know how to do that?