PHP - Perform Work Php Mysql Js Jquery And Yii Framework

Hello,

i want when click next button to show the 4 content ... ive try everything

Code: [Select]

               <div id="MoreBoxBorder"><div id="MoreBoxBorderLine"></div></div>
   <div id="ArrowMoreBox"><a class="next browse right"><div id="ArrowMoreImage"></div></a></div>

<?php
$query = "SELECT *, date_FORMAT(Date, '%H:%i') as Date_Time FROM tblnews WHERE LatestNews = 1 AND Category != 10 ORDER BY `Date` DESC LIMIT 4";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)) {  
?>
               <div id="MoreBox" id="comments">
               <div class="box">
                            <a href="<?php echo BASE_URL;?>index.php?pid=article&artidread=<?php echo $row['Id'];?>"><img src="<?php echo html_entity_decode($row['Thumb']);?>" border="0"  width="110" height="64" /></a>
                            <strong><a href="<?php echo BASE_URL;?>index.php?pid=article&artidread=<?php echo $row['Id'];?>"><strong><?php echo html_entity_decode(substr($row['Title'],0,437));?></strong></a></strong>
                            <p><?php echo html_entity_decode($row['SubTitle']);?></strong>
<div id="MoreIcoBox">
<a href="<?php echo BASE_URL;?>index.php?pid=article&artidread=<?php echo $row['Id'];?>"><img id="MoreIco" src="<?php echo BASE_URL;?>images/news/more.png" border="0" /></a>
</div>

                </div>
</div>

<?php } ?>

Like the image.. when i click next image will display another 4   (& auto scroll every 10seconds)

Any help please

Thank you

Hi,

I am trying to build a jQuery and PHP love it button where the user clicks it and the number increments and saves to a DB.

I have it all working apart from the Saving to the DB part : )

I have this currently to save but it doesnt seem to work.
Code: [Select]
<?php

mysql_connect("localhost", "web183-loveit", "password") or die ("Error.");
mysql_select_db("web183-loveit") or die ("error");

$increase = "UPDATE results SET value=value+1 WHERE id=1;";
$active_rate = mysql_query("SELECT * FROM results WHERE id=1;");
$val = 0;

if($rt = mysql_fetch_assoc($active_rate)) {
$val = $rt['value'];
}

if($_POST['action'] == 'add') {
mysql_query($increase);
print $val++;
}  

?>


Please see it here http://bit.ly/dKx15z

Any ideas??

cheers

Craig

hi,
i made my own page, but when i do an insert using mysql, jquery and php, the arabic text does not show properly
it is a comments page.
here is the DB structure
Code: [Select]

CREATE TABLE IF NOT EXISTS `app_comments` (
  `comment_id` int(20) NOT NULL auto_increment,
  `comment` text NOT NULL,
  `user_id` int(20) NOT NULL,
  `comment_date` date NOT NULL,
  `app_id` int(20) NOT NULL,
  PRIMARY KEY  (`comment_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;
here is the jquery piece that transfers to php code.
Code: [Select]
    function submit_comment(){
        $(".post_comments_button").click(function() {
            var post_comments = $('.post_comments').val();
            var app_id = $('#app_id').val();
            //alert (app_id);
           
            if(confirm('Add Comment?')) {
               
                var string = "task=add_comment&amp;app_id=" + app_id + "&amp;comment="+post_comments;
                            $.ajax({
                url : "appajax.php",
                type : "POST",
                data : string,
                success : function(data) {
                    alert(data);
                    window.location.reload(true);
                   
                }
            });
               
               
               
            }
            /**/
            });
    }and here is the acutal insert code in php
$new_appquery = 'INSERT INTO app_comments (comment, user_id, app_id,comment_date)
                          VALUES ("' . $comment . '",' . $user_id . ',  ' . $app_id . ',NOW());';
                         
                          and the actual problem is, the arabic text shows like this in my application:
                         
                          Quote

بالتوفيق للريال

                         
how can i solve this issue?

I am having a problem making this work. It adds and deletes but it does it by passing the date directly to the PHP files and refreshing the screen. I want it to do it through the jquery Ajax script. This should work according to what I have read but no mater what I try the jquery script does not work.     <html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.11.0/jquery-ui.min.js"></script>

</head>

<body>
<script type="text/javascript">
$(document).ready(function(){
$('MySignform').submit(function(e) {
e.preventDefault();

$.ajax({
type: 'POST',
url: $('MySignform').prop('action'),
data: $('MySignform').serialize(),
success: function() {
alert("It worked!");
}
});
});
});
</script>
<script type="text/javascript">
$(document).ready(function(){
$('DelForm').submit(function(e) {
e.preventDefault();

$.ajax({
type: 'POST',
url: $('DelForm').prop('action'),
data: $('DelForm').serialize(),
success: function() {
alert("It worked!");
}
});
});
});
</script>
<br />
<br />
<br />
<center>
<form name="MySignform" action="addsigningfee.php" method="post">
<label>Signing fee</label>
<input name="signingfee" type="text" id="signingfee"/>
<input name="pid" type="hidden" value="7"/>
<p></p>
<button id="form-submit" type="submit">Save</button>
</form>
<br />
<div id="result"></div>
<br />





<form name="DelForm" action="delsigningfee.php" method="post">
<select name="signingfee">

<?php
require_once("connect.php");
foreach ($db->query("SELECT signingfee,id FROM options WHERE signingfee <> '' AND pid = '7'") as $row)
{
echo "<option value=" . $row['id'] . ">" . $row['signingfee'] . "</option>";
}
?>
</select>


<p></p>
<button type="submit" >Delete</button>
</form>
</center>
</body>
</html>     Adding and Deleteing mysql date using jquery Ajax I am having a problem making this work. It adds and deletes but it does it by passing the date directly to the PHP files and refreshing the screen. I want it to do it through the jquery Ajax script. This should work according to what I have read but no mater what I try the jquery script does not work. Some help getting me In the right direction would be great.


  Code:

1. Create a denomination program. See the example below. Input amount: 3897

1000: 3

500: 1

200: 1

100: 1

50: 1

20: 2

10:0

5:1

1: 2

2. Create a program that will compute the ideal age of a man's wife. See the example below.

Input Man's Age: 27

Formula: Man'sAge/2 + 7

Your ideal Wife's age is: 20 

 

Please send the Output as reply and please include the codes

Edited June 14 by Given-kun
Edit

Hi Guys

Sorry for the lame title but I have absolutely no idea how to word this.

I have the following code which pulls message from a database:


$messages_query = mysql_query("SELECT * FROM messages WHERE id='$message_id' AND recipient='$username'");
$messages_row = mysql_fetch_assoc($messages_query);

$sender = $messages_row['sender'];
$date = $messages_row['date'];
$time = $messages_row['time'];
$subject = $messages_row['subject'];
$message = $messages_row['message'];
$recipient = $message_query['recipient'];

echo $sender . "<br />";
echo $date . "<br />";
echo $time . "<br />";
echo $subject . "<br />";
echo $message . "<br />";


As you can see, it only pulls the data if id='$message_id' AND recipient='$username'.

I wish to perform a die ("You are not authorised to view this message!"); if id='$message_id' or recipient='$username' is different.

Example:
The database reads id: 1 & username: user

The data would be displayed if $message_id = 1 and $username = user.
The die would be performed if $message_id = 5 and $username = user

I hope that makes sense, iv had a problem trying to word it correctly let alone come up with a solution.

Does anybody know what i could do?

Thanks

Hi,   I'm using an example below regarding a whole URL. I'm looking for some conditional PHP that will display certain content depending on if the URL contains the word 'liz' if not, display else.  

$host = $_SERVER['SERVER_NAME'] . $_SERVER['REQUEST_URI'];


if($host == 'liz-4.website.com') 
{
echo $this->__('The domain contains the word liz.');
}
else
{
echo $this->__('The domain does not contain the word liz.');
}

 

Hi every one, Here is the problem..
I have a string which looks like this '(243/433)*100' , i want to calculate its value, but dont know how to do.
if i use type conversion then value becomes zero.
Please help.

This is more of a logic question.  I have built a beta version of my site and I am currently working on getting beta V2 out, but I want to implement location based searching of users (particularly with a google api integration).  Aside from the API stuff, how do I perform location based searches on the backend.  For instance, I have all of my users locations (city, state, country), but how can I say to grab all users by that state when that state is typed in or grab all users by that city when the city is typed in and the same thing for country?  Is there anyway to set this up with a default radius too?  So if I search for a town, I get a 20 Mile radius result?  I really have no idea how to go about the dynamic look up match. Any good tutorials on this?

Hi ,

 Presently we are preparing an excel for download report by querying to data base.
 We are querying to DB and looping through each result row and writing to excel.
for 2lacs data its  too much time.

I found the time taken max in DB query and processing the DB data .
Ant way I can make it faster ?

Any parallel processing I can apply to get it faster......I expecting Huge data report in future.

Hello I have a question. I'm trying to perform multiple queries based off the results from a query.  Is this possible ?

$result = mysql_query("SELECT id FROM sometable");

for each id returned

$result1 = mysql_query("SELECT * FROM someothertable WHERE id=result from $result query");

Any help would be GREAT

<?php



require("connection.php");




$query = 'SELECT * FROM job_post WHERE id_jobpost= :Id';
$start = $bdd->prepare($query);
$start->execute(array(':Id' => $_GET['id_jobpost'])); 
$result = $start->fetch();

foreach ($result as $results) {
    $jobtitle = $result['jobtitle'];
    $id_jobpost = $result['id_jobpost'];
    $description = $result['description'];
    
}


?> 

https://prnt.sc/uk3913

Whats wrong with this code

Hi everyone,

Would someone be able to help me out here, I cannot get this to work.
Basically I have 3 pages, I with a search box which then run's a search script to display the results.  Then the results are clickable, the idea is when the user clicks the result it takes them to another page which I want to display the info from the database.....if that makes sense, but when the link is clicked nothing is displaying.

So here is what I have done so far:

index.php - This is the page with the search box, code is :
Code: [Select]
<?php
$dbhost = 'localhost';
$dbuser = '*********';
$dbpass = '********';

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die                      ('Error connecting to mysql');

$dbname = '***********';
mysql_select_db($dbname);
?>
<form name="form" action="search.php" method="get">
  <input type="text" name="q" />
  <input type="submit" name="Submit" value="Search" />
</form>

When they hit submit it takes them to the results page, code is:
Code: [Select]
<?php

  // Get the search variable from URL
  $var = @$_GET['q'] ;
  $trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10; 

// check for an empty string and display a message.
if ($trimmed == "")
  {
  echo "<p>Please enter a search...</p>";
  exit;
  }

// check for a search parameter
if (!isset($var))
  {
  echo "<p>We dont seem to have a search parameter!</p>";
  exit;
  }

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","********","********"); //(host, username, password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("**********") or die("Unable to select database"); //select which database we're using

// Build SQL Query  
$query = "select * from Fish where Commonn like \"%$trimmed%\"  
  order by Commonn"; // EDIT HERE and specify your table and field names for the SQL query

 $numresults=mysql_query($query);
 $numrows=mysql_num_rows($numresults);

// If we have no results, offer a google search as an alternative

if ($numrows == 0)
  {
  echo "<h4>Results</h4>";
  echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>";

// google
 echo "<p><a href=\"http://www.google.com/search?q=" 
  . $trimmed . "\" target=\"_blank\" title=\"Look up 
  " . $trimmed . " on Google\">Click here</a> to try the 
  search on google</p>";
  }

// next determine if s has been passed to script, if not use 0
  if (empty($s)) {
  $s=0;
  }

// get results
  $query .= " limit $s,$limit";
  $result = mysql_query($query) or die("Couldn't execute query");

// display what the person searched for
echo "<p>You searched for: &quot;" . $var . "&quot;</p>";

// begin to show results set
echo "Results<br><br>";
$count = 1 + $s ;

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["Commonn"];

  echo "$count.)&nbsp;<a href='info.php?ID=$ID'>$title</a>" ;
  $count++ ;
  }

$currPage = (($s/$limit) + 1);

//break before paging
  echo "<br />";

  // next we need to do the links to other results
  if ($s>=1) { // bypass PREV link if s is 0
  $prevs=($s-$limit);
  print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt; 
  Prev 10</a>&nbsp&nbsp;";
  }

// calculate number of pages needing links
  $pages=intval($numrows/$limit);

// $pages now contains int of pages needed unless there is a remainder from division

  if ($numrows%$limit) {
  // has remainder so add one page
  $pages++;
  }

// check to see if last page
  if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {

  // not last page so give NEXT link
  $news=$s+$limit;

  echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
  }

$a = $s + ($limit) ;
  if ($a > $numrows) { $a = $numrows ; }
  $b = $s + 1 ;
  echo "<p>Showing results $b to $a of $numrows</p>";
  
?>

When they click the results from this page it takes them to another page which should (but doesnt) display the info from the mysql database, code is:

Code: [Select]
<?php


// now you can display the results returned


  echo "$title" ;



?>


Im really sorry if this is obvious or if I have dont something completely wrong but im very new to php and trying to learn on the job.

Many Thanks in advance

Jay

Hi I have the file "pback.php" which displays a mysql table.
One row allows me to update the data.
If I press "update it leads me to update.php, but there I just get white fields instead of the data from the id requested.
If I type anything in the field and press confirm it opens update_ac but just says error.
What do I do wrong?
Thanks!
Quote

pback.php

Code: [Select]
<?php
$host="xxx"; // Host name
$username="xxx"; // Mysql username
$password="xxx"; // Mysql password
$db_name="xxx"; // Database name
$tbl_name="sp_users"; // Table name

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

//$sql="SELECT * FROM $tbl_name ORDER BY user_id";
$sql="SELECT * from sp_users,sp_schools where sp_users.user_id=sp_schools.school_id ORDER BY school_name";
$result=mysql_query($sql);
$num_rows=mysql_num_rows($result);
?>
<style type="text/css">
<!--
.style2 {font-weight: bold}
.style3 {
font-family: Arial, Helvetica, sans-serif;
color: #000000;
}
.style10 {font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 10px; color: #000000; }
#Layer1 {
position:absolute;
left:803px;
top:36px;
width:65px;
height:28px;
z-index:1;
}
#Layer2 {
position:absolute;
left:707px;
top:19px;
width:143px;
height:39px;
z-index:1;
}
-->
</style>

<title>User overview</title>
</html>
<div id="Layer2">
  <form action="<?= $_SERVER['PHP_SELF'] ?>" method="post">
<input type="hidden" name="foo" value="<?= $foo ?>" />
<input type="submit" name="submit" value="Refresh" />
</form></div>
<table width="779" border="0" align="left" cellpadding="0" cellspacing="1" bgcolor="#996600">
<tr>
<td width="777">
 
    <div align="left">
      <table width="779" border="1" cellspacing="0" cellpadding="3">
        <tr>
          <td colspan="6" align="center"><div align="center" class="style1 style3"><strong>SchoolPorta.com Users / Total: <?php echo $num_rows ?></strong></div></td>
          </tr>
        <tr>
          <td width="342" align="center"><span class="style2">school</span></td>
          <td width="62" align="center"><span class="style2">Name</span></td>
          <td width="104" align="center"><span class="style2">Lastname</span></td>
          <td width="130" align="center"><span class="style2">Email</span></td>
      <td width="64" align="center"><span class="style2">Update</span></td>
        </tr>
        <?php
while($rows=mysql_fetch_array($result)){
?>
        <tr>
          <td><span class="style10"><? echo $rows['school_name']; ?></span></td>
  <td><span class="style10"><? echo $rows['user_first_name']; ?></span></td>
          <td><span class="style10"><? echo $rows['user_surname']; ?></span></td>
          <td><span class="style10"><a href="mailto:<?php echo $rows['user_login']; ?>"><?php echo $rows['user_login']; ?></a></span></td>
          <td align="center"><a href="update.php?id=<? echo $rows['user_id']; ?>" class="style10">update</a></td>
        </tr>
        <?php
}
?>
      </table>
    </div></td>
</tr>
</table>

<div align="left">
  <p>&nbsp;   </p>
  <p>&nbsp;</p>
  <p>&nbsp;</p>
  <p>
    <?php
mysql_close();
?>
  </p>
</div>
Quote

update.php

Code: [Select]
<?php

$host="xxx"; // Host name
$username="xxx"; // Mysql username
$password="xxx"; // Mysql password
$db_name="xxx"; // Database name
//$tbl_name="sp_users"; // Table name

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// get value of id that sent from address bar
$id=$_GET['user_id'];

?>
<?php
$sql="SELECT * from sp_users,sp_schools where sp_users.user_id=sp_schools.school_id";
$result=mysql_query($sql) or die("Problem with the query: $sql<br>" . mysql_query());
?>
<table width="400" border="0" cellspacing="1" cellpadding="0">
<tr>
<form name="form1" method="post" action="update_ac.php">
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>&nbsp;</td>
<td colspan="3"><div align="center" class="style1">Update data in mysql </div></td>
</tr>
<tr>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
<td align="center">&nbsp;</td>
</tr>
<tr>
<td align="center">&nbsp;</td>

<td align="center"><span class="style7">Name</span></td>
<td align="center"><span class="style7">Lastname</span></td>
<td align="center"><span class="style7">Email</span></td>
</tr>
<tr>
<td>&nbsp;</td>
<td align="center"><input name="name" type="text" id="name" value="<? echo $rows['user_first_name']; ?>"></td>
<td align="center"><input name="lastname" type="text" id="lastname" value="<? echo $rows['user_surname']; ?>" size="15"></td>
<td><input name="email" type="text" id="email" value="<? echo $rows['user_login']; ?>" size="15"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input name="id" type="hidden" id="id" value="<? echo $rows['user_id']; ?>"></td>
<td align="center"><input type="submit" name="Submit" value="Submit"></td>
<td>&nbsp;</td>
</tr>
</table>
</td>
</form>
</tr>
</table>

<?

// close connection
mysql_close();

?>
</body>
</html>
Quote

update_ac.php

Code: [Select]
<?php
$host="xxx"; // Host name
$username="xxx"; // Mysql username
$password="xxx"; // Mysql password
$db_name="xxx"; // Database name
$tbl_name="sp_users"; // Table name

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// update data in mysql database
$sql="UPDATE $tbl_name SET name='$user_name', lastname='$user_lastname', email='$email' WHERE id='$user_id'";
$result=mysql_query($sql);

// if successfully updated.
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='pback.php'>View result</a>";
}

else {
echo "ERROR";
}

?>