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An7hony

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Posts posted by An7hony

    searching a number range using BETWEEN

    in PHP Coding Help

    Posted

    I have read that i can search for a number within a mysql database using BETWEEN

     

    i.e: SELECT  * FROM `table_name` WHERE `field_name` BETWEEN 99 AND 200

     

    How would i do this with the code below?

     

    <?php
$salary = explode(":", $_GET['salary_to']);
	 }
$types[] = (isset($_GET['salary_to']) && strlen(trim($_GET['salary_to'])) > 0) ?
     "`salary_from` LIKE '%". mysql_real_escape_string(trim($salary[0])) ."%' AND `salary_to` LIKE '%". mysql_real_escape_string(trim($salary[1])) ."%'" : ''; 
?>

     

    Thanks for any help

    checkboxes

    in PHP Coding Help

    Posted

    that seems to be working.

     

    Although how do i pull the results i.e

     

    $message = '

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"

    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

    <html xmlns="http://www.w3.org/1999/xhtml">

    <head></head>

    <body>

    <table>

    <tr><td>Categories</td><td>' . $JobCategories . '</td></tr>

    <tr><td>Company</td><td>' . $company . '</td></tr>

    <tr><td>Title</td><td>' . $title . '</td></tr>

    <tr><td>First Name</td><td>' . $firstname . '</td></tr>

     

    If i put $JobCategories[] into this it throws me an error

    checkboxes

    in PHP Coding Help

    Posted

    how would i get checkboxes to work. I have tried:

     

    $JobCategories = ($_GET['JobCategories[]']) ? $_GET['JobCategories[]'] : $_POST['JobCategories[]'];

     

    html looks like:

     

    <li><input name="JobCategories[]" type="checkbox" value="Admin" /> Admin</li>

    <li><input name="JoCategories[]" type="checkbox" value="Campaigning" /> Campaigning</li>

     

    redirect

    in PHP Coding Help

    Posted

    i tried this

     

    $this->url = $_SESSION['url'] = $_SERVER['PHP_SELF'] = $_SERVER['QUERY_STRING'];

     

    but it gave me:

     

    elements/view=view&id=5

     

    rather than:

     

    jobs.php?view=view&id=5

    redirect

    in PHP Coding Help

    Posted

    Hi all,

     

    I have

     

    a page url that looks like this:

     

    jobs.php?view=view&id=5

     

    Their is a login on this page, the form ends with this command:

     

    header("Location: ".$session->referrer);

     

    The fuction for this is:

     

    <?php
function startSession(){
      global $database;  //The database connection
      session_start();   //Tell PHP to start the session

      /* Determine if user is logged in */
      $this->logged_in = $this->checkLogin();

      /**
       * Set guest value to users not logged in, and update
       * active guests table accordingly.
       */
      if(!$this->logged_in){
         $this->username = $_SESSION['username'] = GUEST_NAME;
         $this->userlevel = GUEST_LEVEL;
         $database->addActiveGuest($_SERVER['REMOTE_ADDR'], $this->time);
      }
      /* Update users last active timestamp */
      else{
         $database->addActiveUser($this->username, $this->time);
      }
      
      /* Remove inactive visitors from database */
      $database->removeInactiveUsers();
      $database->removeInactiveGuests();
      
      /* Set referrer page */
      if(isset($_SESSION['url'])){
         $this->referrer = $_SESSION['url'];
      }else{
         $this->referrer = "/";
      }

      /* Set current url */
      $this->url = $_SESSION['url'] = $_SERVER['PHP_SELF'];
   }
?>

     

    Basically

     

    $this->url = $_SESSION['url'] = $_SERVER['PHP_SELF'];

     

    seems to cut my url to:

     

    jobs.php

     

    instead of:

     

    jobs.php?view=view&id=5

     

    Can anyone here tell me how to fix this?

     

    Thanks Guys

    date format

    in PHP Coding Help

    Posted

    Hi Guys

     

    Could somebody tell me how i would put a line break into the code below

     

    i.e:

    1.07 <br />

    2010

     

    <?php echo date('d.m.Y',strtotime($news['created_at']));?>

     

    Thanks for your help

    cant get php email form to work on fasthost

    in PHP Coding Help

    Posted

    I read it on the fasthost knowledge base

    http://www.fasthosts.co.uk/knowledge-base/?article_id=65

     

    Use the PHP mail function and set the mail from using the following line of code - replacing $email_from with the correct domain name.

     

        * ini_set("sendmail_from", " $email_from ");

     

    You need to add a fifth "-f" parameter to the sendmail function. This will set the name of the from address.

     

        * mail($email_to, $email_subject, $email_message, $headers, '-f'.$email_from);

     

     

    cant get php email form to work on fasthost

    in PHP Coding Help

    Posted

    sorry the rest of it goes like this

     

     

    <?php
if($_POST['to']==""){
   $to = "";
   echo '<div class="error">Please select a <strong>Branch</strong></div>';
}else{
   $to = $_POST['to'];
}



if($_POST['firstname']==""){
   $firstname = "";
   echo '<div class="error">Please fill in your <strong>First name</strong></div>';
}else{
   $firstname = $_POST['firstname'];
}

if($_POST['surname']==""){
   $surname = "";
   echo '<div class="error">Please fill in your <strong>Surname</strong></div>';
}else{
   $surname = $_POST['surname'];
}
?>

     

    Then gets inserted into a mysql database

     

    and then runs into the script in the first post.

     

     

    cant get php email form to work on fasthost

    in PHP Coding Help

    Posted

    Hi this is killing me, the email form works on all hosts except fasthost.

     

    I have checked that the from email is hosted with fasthost and it is.

     

    I have changed the script to:

     

      ini_set("sendmail_from", $from);

      mail( "$to", "candidate application", $message, "From: \"-f\".$from");

     

    But the emails dont send?

    Any help is really really appreciated

     

    Thanks

     

    <?php

  $from = "our@fasthosteddomain.com";

  $message .= "You have an General job application from: ";
  $message .= $firstname;
  $message .= " ";
  $message .= $surname;
  $message .= "\n";
  $message .= "\n";
  $message .= "Looking for: ";
  $message .= $typeofwork;
  $message .= "\n";
  $message .= "\n";
  $message .= "Job Type: ";
  $message .= $jobtype;
  $message .= "\n";
  $message .= "\n";
  $message .= "Transport: ";
  $message .= $transport;
  $message .= "\n";
  $message .= "\n";
  $message .= "Please login to http://www.ourdomain.com/admin to view their details";
  $message .= "\n";
  $message .= "\n";
  $message .= "Their Telephone number is: ";
  $message .= $telephone;
  $message .= "\n";
  $message .= "\n";
  $message .= "Their email address is: ";
  $message .= $email;
  $message .= "\n";
  $message .= "\n";
  $message .= "Cover Letter:";
  $message .= "\n";
  $message .= $coverletter;
  $message .= "\n";
  $message .= "\n";
  $message .= "If CV attached: http://www.ourdomain.com/cvs/".$fileName;

  ini_set("sendmail_from", $from); 
  mail( "$to", "candidate application", $message, "From: \"-f\".$from");
  
  }
  }
?>

    EOF

    in PHP Coding Help

    Posted

    How would i call safe_output()

    when using <<< EOF

     

    Heres my code:

     

    <?php

$job_list = <<< EOH
<div class="joblisting">
                      <div id="jobtools">
                      <h1>Job Tools</h1>
                      <div class="content">
                      <ul>
                      <li><a href="#" title="Apply for this Job" class="apply">Apply for job</a></li>
                      <li><a href="#" title="Save this Job" class="savejob">Save this job</a></li>
                      </ul>
                      </div>
                      </div>
                    	
                      <div class="jobtop"> 
                      <div class="jobimage"><img src="skin/images/logo.gif" /></div>
                      <div class="headers">
                      <h1><a href="jobs.php?view=show&id=$id; ?>"> safe_output($title); </a></h1>

?>

    sticky jump menu

    in PHP Coding Help

    Posted

    How would i make a jump menu sticky? I have tried to do it the same as a drop down menu but have had no success

     

    <select name="select" onChange="MM_jumpMenu('parent',this,1)" class="basic-combo" >
              <option value="#" selected>Locations. . .</option>
		  <option value="<?php echo 'http://'.$_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI']. '&location=0'; ?>">All Locations</option>
</select>

    uploading a file

    in PHP Coding Help

    Posted

    Here is some of my form:

     

    <form method="POST" enctype="multipart/form-data" action="" onsubmit="return false;" id="myid">


<label for="txtmessage" class="labsmall">Upload<span>*</span></label> <input type="hidden" name="MAX_FILE_SIZE" value="20000000000"><input name="userfile" type="file" class="browse" id="userfile"><br />

<input type="image" src="images/applybut.gif" class="applybutton" onclick="ajax_form('myid','validate.php','receiver');"  />

     

    Here is some of my validate.php:

    <?

//file upload
$uploadDir = '/blabla/';	
$fileName = $_FILES['userfile']['name'];
$tmpName  = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];

    // get the file extension first
$ext      = substr(strrchr($fileName, "."), 1); 

// generate the random file name


// and now we have the unique file name for the upload file
    $filePath = $uploadDir . $fileName;

    // move the files to the specified directory
// if the upload directory is not writable or
// something else went wrong $result will be false
    $result    = move_uploaded_file($tmpName, $filePath);
if (!$result) {
	echo 'Error uploading the <strong>file</strong>';

}

$SQL = " INSERT INTO table ";
$SQL = $SQL . " (name, size, type, path) VALUES ";
$SQL = $SQL . " ('$fileName', '$fileSize', '$fileType', '$filePath') ";
$result = mysql_db_query($db,"$SQL",$cid);
if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); }

mysql_close($cid);
?>

     

    Will this work using: onclick="ajax_form('myid','validate.php','receiver');"

     

    The rest of my data gets posted except the file upload

     

    Thanks for the help guys

    sort by

    in PHP Coding Help

    Posted

    Hi I have 2 jump menus which im using for sort by

     

    Sown below the jump menus:

     

    <select name="select" onChange="MM_jumpMenu('parent',this,1)" class="basic-combo" >
              <option value="#" selected>All Locations</option>
              <option value="/index.php?view=jobs&location=1">Lincoln</option>
              <option value="/index.php?view=jobs&location=2">Mansfield</option>
      <option value="/index.php?view=jobs&location=3">Grantham</option>
            </select>

     

    and

     

    <select name="select" onChange="MM_jumpMenu('parent',this,1)" class="basic-combo" >
              <option value="#" selected>Sort By. . .</option>
              <option value="/index.php?view=jobs">All Job Types</option>
              <option value="/index.php?view=jobs&sort=1">Temporary</option>
              <option value="/index.php?view=jobs&sort=2">Permanent</option>
      <option value="/index.php?view=jobs&sort=3">Perm to Temp</option>
            </select>

     

    Here is my php

     

    <?php
$sortbylocation = intval($_GET['location']);

      switch ($sortbylocation)
      {
      case 1:
      {
      $sortlocation = "WHERE branch = 'Lincoln'";
      };break;

      case 2:
      {
      $sortlocation = "WHERE branch = 'Mansfield'";
      };break;
  
  case 3:
      {
      $sortlocation = "WHERE branch = 'Grantham'";
      };break;

      //etc with the rest of the attributes

      }




      $sortby = intval($_GET['sort']);

      switch ($sortby)
      {
      case 1:
      {
      $sqlExtra = "WHERE jobtype = 'Temporary'";
      };break;

      case 2:
      {
      $sqlExtra = "WHERE jobtype = 'Permanent'";
      };break;
  
  case 3:
      {
      $sqlExtra = "WHERE jobtype = 'Temp to Perm'";
      };break;

      //etc with the rest of the attributes

      }
?>

     

    and where its sorting:

     

    <?php
$sql = "SELECT id, agent, branch, title, location, salary_from, salary_to, frequency, jobtype, description, specifications, benefits, email, created_at, categories FROM $tbl_name ".$sortlocation." ".$sqlExtra." LIMIT $start, $limit";
$result = mysql_query($sql);
?>

     

    question is how do i make the second menu dependant on the first menu option?

    displaying a category name

    in PHP Coding Help

    Posted

    if this is more help here is the form:

    <?
<li class="even"><label class="field-title">Categories: </label> <label>
						<?  $categories = find_categories(); ?>
						<?php foreach( $categories as $category): ?>
			<input type="checkbox" name="job[cats][]" value="<?php echo $category['id']; ?>" > <?php echo $category['name']; ?> <br/>
         <?php endforeach; ?></label> 
						<span class="clearFix"> </span></li>
?>

     

    and here is the find_categories(); function

    <?
function find_categories()
{
   db_connect();

	$query =  "SELECT 
                  categories.name, 
								categories.id,
								COUNT(jobs2categories.job_id) as numjobs
						  FROM
							  categories
						  LEFT JOIN
							   jobs2categories ON categories.id = jobs2categories.category_id
							GROUP BY
							    categories.id
							 ORDER BY
							    categories.name ASC
								";

	$result = mysql_query($query);

	$result = db_result_to_array($result);

	return $result;

}
?>

    displaying a category name

    in PHP Coding Help

    Posted

    I am trying to display a category name related to a 'job' post.

     

    The function i create the job with is shown below:

     

    <?php
function create_job($params)
{
   db_connect();

	$query =  sprintf("INSERT INTO jobs set 
						jobs.agent = '%s',
		                jobs.branch = '%s',
						jobs.title = '%s',
						jobs.location = '%s',  
						jobs.salary_from = '%s',
						jobs.salary_to = '%s',
						jobs.frequency = '%s',
						jobs.jobtype = '%s',  
                            jobs.description = '%s',
					    jobs.specifications = '%s',
					    jobs.benefits = '%s',
					    jobs.email = '%s',
											  created_at = NOW() 
								", mysql_real_escape_string($params['agent']),
									 mysql_real_escape_string($params['branch']),
								   mysql_real_escape_string($params['title']),
									 mysql_real_escape_string($params['location']),
									 mysql_real_escape_string($params['salary_from']),
									 mysql_real_escape_string($params['salary_to']),
									 mysql_real_escape_string($params['frequency']),
									 mysql_real_escape_string($params['jobtype']),
									 mysql_real_escape_string($params['description']),
									 mysql_real_escape_string($params['specifications']),
									 mysql_real_escape_string($params['benefits']),
									 mysql_real_escape_string($params['email'])

								);

	$result = mysql_query($query);
	if(!$result)
	{
		 return false;
	}

	$job_id = mysql_insert_id();

	if(!empty($params['cats'])) 
	{
       foreach($params['cats'] as $cat_id) 
		 {
			 $query = sprintf("INSERT 
				                     INTO jobs2categories
				                          set
																	 category_id = '%s',
																	 job_id = '%s'",
																	 mysql_real_escape_string($cat_id),
																	 mysql_real_escape_string($job_id)
																	 );

			 $result = mysql_query($query);

			 if(!$result)
			 {
				 return false; 
			 }

		 }

   	}

	 return true;

}	?>

     

    Here is the code i use to pull the job data:

     

    <?php $query = "SELECT id, agent, branch, title, location, salary_from, salary_to, frequency, jobtype, description, specifications, benefits, email, created_at FROM jobs WHERE branch = '$ulevel' ORDER BY id DESC ";
$result = mysql_query($query) or die('Error : ' . mysql_error());

if(mysql_num_rows($result)==0) {
    echo('
<span><strong>No Jobs from '.$ulevel.' Listed</strong></span>
');
}

while(list($id, $agent, $branch, $title, $location, $salary_from, $salary_to, $frequency, $jobtype, $description, $specifications, $benefits, $email, $created_at) = mysql_fetch_array($result, MYSQL_NUM))
{
?>
      				<tr>
      					
                        <td class="col-first"><?php echo date('d.m.Y',strtotime($created_at));?></td>
      					<td class="col-second">### CATEGORY NAME ###</td>
      					<td class="col-third"><?php echo safe_output($title); ?></td>
                        <td class="col-fourth"><?php echo safe_output($location); ?></td>
                        <td class="col-five"><?php echo safe_output($jobtype); ?></td>
                        <td class="col-sixth"><?php echo safe_output($agent); ?></td>
                        <td class="col-seven">3</td>
      					<td class="row-nav"><a href="#" class="table-edit-link">Edit</a> <span class="hidden"> | </span> <a href="javascript:delArticle('<?php echo safe_output($id); ?>');" class="table-delete-link">Delete</a></td>
      				</tr>
				<?php
}



?>

     

    How would i display the related category name in the table above?

     

    Any help is appreciated

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