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jarv

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Posts posted by jarv

    NULL is output in array

    in PHP Coding Help

    Posted

    Thanks, I'm getting there now, I used the second one:

     

    while( $rs_location = $rs_locations->fetch_row() ) {
        $markers[] = $rs_location;
    }

     

    My markers are now showing, GREAT! .... but in the middle of the Indian Ocean

     

    map.png

    NULL is output in array

    in PHP Coding Help

    Posted · Edited by jarv

    ok, I don't get it, sorry. I am just trying to follow the example using the php array they commented out
     
    here is where I currently am:
     
     

    $mysqli = mysqli_connect("localhost", "root", "", "plods");
    $sql_locations = "SELECT CONCAT(address1, ', ', address2) as address, Longitude, Latitude FROM southdowns";
    $rs_locations = mysqli_query($mysqli, $sql_locations);
 
    while( $rs_location = $rs_locations->fetch_assoc() ) {
      foreach( $rs_locations->fetch_assoc() as $rs_location => $value){
        $markers[] = $value;
      }
    }
    //while( $rs_location = $rs_locations->fetch_assoc() ) {
    //    $markers[] = $rs_location;
    //}
 
    echo "<pre>".print_r($markers)."</pre>";

    and it comes out like this:
     

    Array ( [0] => Nyetimber, Broughton House, 6-8 Sackville Street [1] => -0.137792 [2] => 51.509285 [3] => Baker Tilly, The Portland Building, 25 High Street.....

    NULL is output in array

    in PHP Coding Help

    Posted

    no no, if you look at the working example on that link

     

     

    when you call json_encode() in the javascript it should then look like this:

     

    array(
'Westminster Abbey, London',
51.4992453,
-0.1272561
),
array(
'QEII Centre, London',
51.4997296,
-0.128683
)

    NULL is output in array

    in PHP Coding Help

    Posted · Edited by jarv

    it needs to be out put like this: 

     

    array(
'Palace of Westminster, London',
51.499633,
-0.124755
),
array(
'Westminster Abbey, London',
51.4992453,
-0.1272561
)

     

    but I am getting: 

    Array ( [0] => Array ( [address] => Quintessence Fragrances Ltd, Unit 2A, Hawthorne Industrial Estate [Longitude] => 0.058305 [Latitude] => 50.798006 ) [1] => Array ( [address] => Nyetimber, Broughton House, 6-8 Sackville Street [Longitude] => -0.137792 [Latitude] => 51.509285 )

    NULL is output in array

    in PHP Coding Help

    Posted · Edited by jarv

    Please help, I am trying to get data out of my database as an array but it doesn't seem to be working?!

     

    I just get: ArrayArrayArrayArrayArray

     

    here is my code

    <?php
    
    $mysqli = mysqli_connect("localhost", "root", "", "plods");
    
    $sql_locations = "SELECT * FROM southdowns";
    
    $rs_locations = mysqli_query($mysqli, $sql_locations);
    
    foreach( $rs_locations as $rs_location ) {
    
      $markers[] = array(
    
          "{$rs_location['address1']}, {$rs_location['address2']}",
    
          $rs_location['Longitude'],
    
          $rs_location['Latitude']
    
      );
    
      echo $markers;
    
    }
    
    ?>

    adding an extra day to a date

    in PHP Coding Help

    Posted

    Yes it's all a bit mashed together

     

     

    $time variable is a date set in the database

     

     

    $time = $context->data->field_event_date[0]['value'];
 
var deadline = '". date('Y-m-d', $time) ."';
 
initializeClock('clockdiv', deadline);", 'inline');
 

    adding an extra day to a date

    in PHP Coding Help

    Posted

    Whichever... PHP?

     

    I just tried this but it came back as -16449 days 3 months -8 hours

    var deadline = '". date('Y-m-d', strtotime($time. ' + 1 days')) ."';

     

    at the moment the countdown says:

     

    0 days 13 hours 46 minutes 30 seconds

     

    I would like it to say:

     

    1 days 13 hours 46 minutes 30 seconds

    help exporting table as SQL file?

    in PHP Coding Help

    Posted

    I am getting an error to check my values  but I am using the right username, password, table name and host name.

     

    Can anyone help me here?

    $mysqlDatabaseName ='drupal';
$mysqlUserName ='drupal';
$mysqlPassword ='xxx';
$mysqlHostName ='localhost';
$mysqlExportPath ='content_type_vacancy.sql';

$command='mysqldump --opt -h' .$mysqlHostName .' -u' .$mysqlUserName .' -p' .$mysqlPassword .' ' .$mysqlDatabaseName .' content_type_vacancy > ~/' .$mysqlExportPath;

exec($command,$output=array(),$worked);
switch($worked){
case 0:
echo 'Database <b>' .$mysqlDatabaseName .'</b> successfully exported to <b>~/' .$mysqlExportPath .'</b>';
break;
case 1:
echo 'There was a warning during the export of <b>' .$mysqlDatabaseName .'</b> to <b>~/' .$mysqlExportPath .'</b>';
break;
case 2:
echo 'There was an error during export. Please check your values:<br/><br/><table><tr><td>MySQL Database Name:</td><td><b>' .$mysqlDatabaseName .'</b></td></tr><tr><td>MySQL User Name:</td><td><b>' .$mysqlUserName .'</b></td></tr><tr><td>MySQL Password:</td><td><b>NOTSHOWN</b></td></tr><tr><td>MySQL Host Name:</td><td><b>' .$mysqlHostName .'</b></td></tr></table>';
break;
}

    adding values in a while loop

    in PHP Coding Help

    Posted

    Hi,

     

    I am trying to add values together in my while loop, so far I have:

    $q->query($DB, $SQL);
while($avrating=$q->getrow()):
$averageratings = $avrating['sum(rating)'] / $noofreviews;
$overallaverage = round($averageratings, 2) + round($averageratings, 2);
echo '<p>'.$avrating['rating_name'].': '.round($averageratings, 2).'</p>';
endwhile;
echo 'x'.$overallaverage;


$overallaverage += $overallaverage / 10;
echo 'The Average: '.$overallaverage;

     

    from this, I get:

    x7.34The Average: 8.074

     

    the total shold be: 42.34

    then dived by 10

     

    should be 4.23

    get avarage value of records in table - MySQL

    in PHP Coding Help

    Posted

    here is the code in my page:

     

    $SQL = "
SELECT sr.review_id,
sr.review_date,
sr.review,
sr.reviewer_name,
sr.reviewer_surname,

r.rating_name,
r.rating
FROM
school_reviews sr
LEFT JOIN
ratings r USING(review_id)
WHERE
sr.active = 1
AND sr.is_deleted = 0
AND sr.school_id = '" . $school_id . "'
ORDER BY
sr.review_date DESC";
$q->query($DB,$SQL);

// store results of query in array
$reviews = array();

while($row = $q->getrow())
{
// get the review id
$id = $row['review_id'];
// group reviews by review id
if(!isset($reviews[$id]))
{
$reviews[$id] = array(
'message' => $row['review'],
'date' => $row['review_date'],
'name' => $row['reviewer_name'],
'surname' => $row['reviewer_surname'],
'ratings' => array()
);
}

// group ratings by review id
$reviews[$id]['ratings'][] = array(
'name' => $row['rating_name'],
'value' => '<img src="/en/images_v2/ratings/star'.$row['rating'].'.png" />'
);
}

// loop over the reviews
foreach($reviews as $review_id => $review)
{
// output review
echo "<div class=\"schoolreview\">";
echo "<p><span class=\"right\">Posted On: " . $review['date'] . "</span>";
echo "Review by: ".$review['name']."<br />\n"
. "</p>\n";

// output ratings list for each review
echo "<ul class=\"ratings\">\n";
foreach($review['ratings'] as $rating)
{
echo "\t<li>" . $rating['name'] . ': ' . $rating['value'] . "</li>\n";
}
echo "</ul>\n";
echo "<p>Review: " . nl2br($review['message']) . "</p>";
echo "</div>";
}

    get avarage value of records in table - MySQL

    in PHP Coding Help

    Posted

    Thanks but no, I need to get the average by adding each rating for each review

     

    Review1 = Rating: Location = 4

    Review2 = Rating: Location = 4

    Review3 = Rating: Location = 5

     

    4+4+5 / 3reviews = Overall Location = 4.3

     

    Review1 = Rating: Accommodation = 2

    Review2 = Rating: Accommodation = 5

    Review3 = Rating: Accommodation = 1

     

    2+5+1 / 3reviews = Overall Accommodation = 2.6

    get avarage value of records in table - MySQL

    in PHP Coding Help

    Posted · Edited by jarv

    Hi,

     

    I have a table RATINGS

     

    here is an example:

     
<table border="1" cellpadding="2" cellspacing="2">
<tr>
<td>rating_id</td>
<td>review_id</td>
<td>rating_name</td>
<td>rating</td>
</tr>
 
<tr>
<td>1</td>
<td>46</td>
<td>location</td>
<td>4</td>
</tr>
 
<tr>
<td>2</td>
<td>46</td>
<td>accommodation</td>
<td>3</td>
</tr>
<tr>
<td>3</td>
<td>46</td>
<td>travel time</td>
<td>5</td>
</tr>
</table>
     

    I would like to get the average from the RATING datafield for each of the rows in my database

    Could anyone please help point me to a tutorial or be so kind to post some example code?

     

    thanks

    Loop inside of a loop?

    in PHP Coding Help

    Posted

    Hi,

    I have 2 tables of data REVIEWS and RATINGS

    I would like to show a list of my reviews and a list of rating for each of my reviews

    I have tried putting my second WHILE loop inside teh first one but that doesn't work, can anyone please help?

    here is my code:

    $SQL = "SELECT
                        *
                    FROM
                        school_reviews
                    WHERE
                        active = 1
                        AND is_deleted = 0
                        AND school_id = '" . $school_id . "'
                        ORDER BY
                        review_date DESC";
    $q->query($DB,$SQL);


    while($review = $q->getrow()):
        $reviewid = $review['review_id'];
        echo $reviewid;
        echo '<h3>'.$review['review'].'</h3>';
        

    endwhile;
    
    echo '<ul>';
    $SQL = "SELECT
                    *
                FROM
                    ratings
                WHERE
                    review_id = '".$reviewid."'";
    $q->query($DB,$SQL);
    while($rating = $q->getrow()):
        echo '<li>'.$rating['rating_name'].': '.$rating['rating'].'</li>';
    endwhile;
    echo '</ul>';

    here is an example of the review I get:

    ReviewID: 46
    Review: testing the reviews
    ReviewID: 43
    Review: Because I know John is sooo handsome and he works at Cactus
    ReviewID: 40
    Review: iuhi

    Ratings:
        Course / Course content: 5
        Teacher: 5
        Social Programme: 5
        School Facilities: 5
        Atmosphere: 5
        Location: 5
        Accommodation: 4
        Service from Cactus: 2
        Value for money: 3
        Overall Experience: 5

    Lat and Long not being returned from geocoded postcode?!

    in PHP Coding Help

    Posted

    I just got this message so hopefully I can geocode more postcodes tomorrow?!

    var_dump() results jason response string(137) "{ "error_message" : "You have exceeded your daily request quota for this API.", "results" : [], "status" : "OVER_QUERY_LIMIT" } " jason decoded object(stdClass)#1 (3) { ["error_message"]=> string(56) "You have exceeded your daily request quota for this API." ["results"]=> array(0) { } ["status"]=> string(16) "OVER_QUERY_LIMIT" } lat NULL long NULL // $qr = "SELECT * FROM pubs WHERE rsTown = 'Worthing'"; // // $res= mysql_query($qr); // $i=0; // while($row = mysql_fetch_array($res)) // { // $postcode = $row["rsPostCode"]; // $response = file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?address=' . urlencode($postcode) . '&sensor=true'); // $response = json_decode($response); // // $lat = $response->results[0]->geometry->location->lat; // $long = $response->results[0]->geometry->location->lng; // // $qr = "UPDATE pubs SET rsLat='$lat',rsLong='$long' WHERE rsPostCode = '$postcode'"; // // //echo $qr; // $res= mysql_query($qr); // echo $qr; // // $i++; // }

    Lat and Long not being returned from geocoded postcode?!

    in PHP Coding Help

    Posted

    Hi, My codes does not show any Latitude or Longitude but it did last night when I started writing this code?!

     

    $postcode = 'BN113ED';
$response = file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?address=' . urlencode($postcode) . '&sensor=true');
$response = json_decode($response);
            
$lat = $response->results[0]->geometry->location->lat;
$long = $response->results[0]->geometry->location->lng;

     

    here is the URL that shows the JSON

    http://maps.googleapis.com/maps/api/geocode/json?address=bn113ed&sensor=true

    Submit form to same page and then update - not working

    in PHP Coding Help

    Posted

    I am trying to submit a form and then run an update but its not working, please help

     

    here is my page:

     

    <?php
session_start();

mysql_connect("xxx", "xxx", "xxx");
mysql_select_db("xxx");

$id = $_REQUEST['id'];

$query1 = "SELECT * FROM app1_blog";
$result1 = mysql_query($query1);

$qry="SELECT id,title,date,article,image1,image2,image3 FROM app1_blog WHERE id = '$id'";
$cur=mysql_query($qry);
while($i=mysql_fetch_row($cur))
{$id=$i[0];
$title=$i[1];
$date=$i[2];
$article=$i[3];
$image1=$i[4];
$image2=$i[5];
$image3=$i[6];
}





if(isset($_POST['submit'])){
echo 'hello';
$new_title=mysql_real_escape_string(stripslashes($_POST["title"]));
$new_article=mysql_real_escape_string(stripslashes($_POST["article"]));
echo $new_title;
}
//$sql = "UPDATE app1_blog SET title='$new_title',article='$new_article' WHERE id = '$id'";

//$msg = 'Blog edited';
//$result = mysql_query($sql,$link) or die('Error: ' . mysql_error() . '<br>SQL: ' . $sql);

//header("Location: edit.php?msg=$msg&id=$id");

?>
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="initial-scale=1.0, user-scalable=no">
  <meta name="apple-mobile-web-app-capable" content="yes">
  <meta name="apple-mobile-web-app-status-bar-style" content="black">
  <title></title>
 
 
 
  <link rel="stylesheet" href="https://d10ajoocuyu32n.cloudfront.net/mobile/1.3.1/jquery.mobile-1.3.1.min.css">
 
  <!-- Extra Codiqa features -->
  <link rel="stylesheet" href="codiqa.ext.css">
 
  <!-- jQuery and jQuery Mobile -->
  <script src="https://d10ajoocuyu32n.cloudfront.net/jquery-1.9.1.min.js"></script>
  <script src="https://d10ajoocuyu32n.cloudfront.net/mobile/1.3.1/jquery.mobile-1.3.1.min.js"></script>

  <!-- Extra Codiqa features -->
  <script src="https://d10ajoocuyu32n.cloudfront.net/codiqa.ext.js"></script>
   
</head>

<body>
<!-- Home -->
<div data-role="page" id="page<?php echo $id;?>">
    <div data-theme="a" data-role="header">
        <h3>
            Edit blog post
        </h3>
    </div>
    <div data-role="content">
        <h2>
            Blog
        </h2>
        <form method="post" action="<?php print $_SERVER["PHP_SELF"]; ?>" name="form">
        <div data-role="fieldcontain">
            <label for="textinput1">
                Title
            </label>
            <input name="title" id="textinput1" placeholder="" value="<?php echo $title;?>" type="text">
            <input name="id" placeholder="" value="<?php echo $id;?>" type="hidden">
        </div>
        <div data-role="fieldcontain">
            <label for="textinput2">
                Date
            </label>
            <input name="date" id="textinput2" placeholder="" value="<?php echo $date;?>" type="text">
        </div>
        <div data-role="fieldcontain">
            <label for="textinput3">
                Article
            </label>
            <textarea name="article" id="textarea1" placeholder=""><?php echo $article;?></textarea>
        </div>
            <img src="<?php echo $image1;?>" alt="image" width="288" >
            <img src="<?php echo $image2;?>" alt="image" width="288" >
            <img src="<?php echo $image3;?>" alt="image" width="288" >
        <input data-theme="b" value="Submit" type="submit">
        </form>
    </div>
</div>
</body>
</html>
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