PHP - Strange Results From An Array

How do I access objects inside an array of database results? I have a method that returns MySQL results as an array "$projects". Running print_r($projects) gives me this:

array(1) (
    "projects" => object Database_MySQL_Result(6) {
        protected _internal_row => integer 0
        protected _query => string(875) "SELECT [skipping remainder of long query]"
        protected _result => resource(mysql result)
        protected _total_rows => integer 53
        protected _current_row => integer 0
        protected _as_object => string(13) "Model_Project"
    }
)

If I do this: foreach($projects as $project)
{
   echo $project->PROJECT_NAME;
}

I get "Undefined property: Database_MySQL_Result::$PROJECT_NAME"

If I do this:

foreach($projects as $project)
{
   echo $project[0];
}

the browser will display projects.id for the first returned row *only*

echo $project[1] returns the project.id for the second row only. And so on.

Each returned row contains over a dozen cells.

It's almost as if I'm referring to the array incorrectly, or referring to the wrong array. I feel as though I'm making a very simple (and perhaps dumb) mistake here, but can't quite figure out what it is. Thanks.

I am writing a CRON job that will execute daily.  First it will identify from a MySql table the date in a field 'FAPforSale_repost35'  If the date is the today date it will then execute commands to delete photo images in a directory, delete the directory, and finally remove the record from the database.   I am on step one which is to build the array of records that match the days date.  When I run the code, there are no errors but I am not getting results even though the records in the test table are set for today.  Below is the select

<?php

 define( "DIR", "../zabp_employee_benefits_processor_filesSm/", true );
 
 require( '../zipconfig.php' );
 require( DIR . 'lib/db.class.php' );
 require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/configRecognition.php' );
 require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/libRecognition/MailClass.inc' );
 
 
 
$todayRepost35 = date("Y-m-d");
echo $todayRepost35;



function repostEndSelect()
 
      {
        
        global $db;
            
    $this->db = $db;
        
            
     $data = $this->db->searchQuery( "SELECT `FAPforSale_IDnumber`,  `FAPforSale_image1`,  `FAPforSale_image2`,  `FAPforSale_image3`,  `FAPforSale_repost35`  FROM  `FAP_forSaleTest` Where `FAPforSale_repost35` = '$todayRepost35' ");
    
    
         $this->FAPforSale_IDnumber = $data[0]['FAPforSale_IDnumber'];
        $this->FAPforSale_image1 = $data[0]['FAPforSale_image1'];
        $this->FAPforSale_image2 = $data[0]['FAPforSale_image2'];
        $this->FAPforSale_image3 = $data[0]['FAPforSale_image3'];
        $this->FAPforSale_repost35 = $data[0]['FAPforSale_repost35'];
        
        echo $this->FAPforSale_IDnumber;
        echo $this->FAPforSale_image1;
        echo $this->FAPforSale_image2;
        echo $this->FAPforSale_image3;
        echo $this->FAPforSale_repost35;
        
    } // ends function...
        
echo( ' Finished...' );

?>

Thanks in advance for any suggestions or direction.  Chapter two will be when I start testing the commands to delete.

I'm having trouble getting all of the results out of a query array, as it is I only get the first result and nothing else.
Here's the basic code I'm working with:
Code: [Select]
$query = "SELECT data_txt FROM jos_servicedirectory_fields_data WHERE fieldid = 19 AND itemid = $item->itemid";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
$listtags = $row['data_txt'];
$tags = "$listtags, ";

$title = "<div class='servicedirectoryItemTitle'><table class=\"sdlistingitemtitle\" cellspacing=\"5\"><tbody><tr>$listingimage<td style=\"vertical-align:top;width:690px;\">$listingbasicicon<span class=\"$listingtitleclass\">$listinglogo<a href=\"{$href}\" {$onClick} title=\"{$item->title}\">{$item->title}</a></span><br /><span class=\"listingdescription\">$listingdescription</span></td><td style=\"vertical-align:top;\">$featuredribbon$moreinfobasic</td></tr></tbody></table><div class=\"listingbottom\">Tags:<span class=\"listingtags\">  $tags </span></div></div>"; }

I've also tried using a foreach loop thinking that would pop all of the results but I end up not getting anything at all then. I'm guessing I'm setting the foreach loop up wrong. Here's how I'm trying to do it:
Code: [Select]
foreach($listtags as $value) {
$tags = $value;
}

Does it have something to do with sticking the $tags variable in the $title variable? I wouldn't think that would matter, but the strange thing is I use this exact same query in a different part of this component and just echo it directly and it works fine.

//First I'm assigning a $variable ($emailzipmatch) to query a database table called(repzipcodes) and having it pull and display 1 to 3 records based on matching up a customer's zip code (RepZipCode = $CustomerZipMatch) with 1 to 3 other people (GROUP BY RepId HAVING COUNT(1) <= 3") that want that customer's information from that particular zip code. <This code works fine>

// CODE WORKS BELOW
Code: [Select]
<?php
    $emailzipmatch = mysql_query("SELECT * FROM repzipcodes WHERE RepZipCode = $CustomerZipMatch GROUP BY RepId HAVING COUNT(1) <= 3") or die(mysql_error());

    $recipients = array();
    while($row = mysql_fetch_array($emailzipmatch)) 
    { 
    $recipients[] = $row['RepEmail'];

    echo "Agent's Email Address: ";
 
    echo 'font color="#FF7600"',$row['RepEmail'], '/font';
 
    echo '<br />'; 

    echo "Rep's ID: "; 
    echo '<br />';
    echo 'font color="#FF7600"',$row['RepId'], '/font';
    echo '<br />';
    echo 'hr align="left" width="50%" size="2" /';
    }


//MY PROBLEM BELOW

// For the NEXT step of the process above I would take $row['RepEmail'] and $row['RepId'] which can have 1 to 3 results and assign the 1 to 3 results a new $variable so it can be inserted into a different db table so I can track the results of the query ($emailzipmatch = ) from the top of the page: ie..

    <New Variable>   <Listed from above>
    $SentRepId 0   = RepId (results from above echo area)
    $SentRepId 1   = RepId (results from above echo area)
    $SentRepId 2   = RepId (results from above echo area)


// Below I'd like to insert the above results into a new database

    $?Variable??? = mysql_query("INSERT INTO sentemail 
   (SentRepId0, SentRepId1, SentRepId2,SentDateTime
   ) VALUES (
   '$_SESSION[RepId]', // ?????
   '$_SESSION[RepId]', // ?????
   '$_SESSION[RepId]', // ?????
    NOW()
     )") or die(mysql_error());

Thank ahead of time for any help you guys can give me. Please respond with ANY question if my coding or request isn't clear or if I've been confusing due to my lack of experience with PHP and MySQL.


MOD EDIT: code tags added.

I am running a script to retrieve numerical value from a webpage, the script echoes the numbers. However I need to add all the numbers to an array or txt file and then have them sorted from highest value to lowest value. Any hints or tips to how I would achieve this?

Say I have a table called artists with these fields (artistid, rank).   Then say I have an array called $rankadjustments with a value for each artist (like 1, 7, 3,-3, 5, 9, etc). I am doing a MYSQL query that gets the info from artists table and sorts according to the rank field like this...

$sql = "SELECT * FROM artists ORDER BY rank";

Easy enough. But what if I would like to bring that data back sorted by (rank + rankadjustment). For example, if the artist ranked 1st had a rank of "1" from mysql, but had a rankadjustment of "5" from the rankadjustment array, his "true" rank would be "6".   Again, rankadjustment is NOT a field in my table, otherwise it would obviously be simple. It's calculated on the fly and stored in an array.  FYI, the array has an index with their "artistid". For example, $rankadjustment[341][8] would be the artist with an artistid of "341" has a rank adjustment of 8.    Is there a way to do this IN the query itself? It doesn't seem possible but wanted to find out for sure. If not, what is best way to do? I assume... Get the data sorted JUST by rank and put it all into an array, then create a new array that adds rank to rankadjustment and reorder by the "new" rank amount?   I guess that wouldn't be too bad but again, wanted to make sure I'm not missing something that would allow me to do it all in the query (or just more efficiently).   Thanks!
Edited by galvin, 21 December 2014 - 09:13 PM.

Hello,

Having a bit of a tough time understanding what is going on in my brain! 

I'm basically pulling some data from an API using CuRL, which is working fine. The structure of the data is:

[data][tickets] - tickets being an array of support tickets in the system, which then contains ticketNumber, title, content. I want to display these values in a list, I have this code currently:

  if ($err) {
    echo "cURL Error #:" . $err;
  } else {
    $return = json_decode($response, true);
    foreach($return as $key) {
      echo $key["tickets"][1]['ticketNumber']." - ";
      echo $key["tickets"][1]['title']." - ";
      echo $key["tickets"][1]['content'];
    }
  }

Which obviously returns the ticketNumber, title & content of the first support ticket in the array, but I want to loop through and display that information for each ticket! 

Can anyone advise as to where I am going wrong? I think I need to utilise the ["tickets"][x] but not sure how!

 

Thanks in advance!

Steve

hi, i am trying to visualize a table by replacing the fixed values with images.
For example, i have a visitor statistics table on mysql. Values on the config_os table are always in three letters. WI7 for Windows7, MAC for Macintosh. I hope I can tell this correctly.
I have a display page which retrieves the values from database and displays them website statistics.
But in order to make it more cool, i am telling to PHP that When you echo to divtable WI7, replace them with specified Windows7 icon image which has identical name as WI7.gif. Here is the code:

Code: [Select]
        include("conn.php");
    $sql="SELECT *, INET_NTOA(location_ip) AS location_ip FROM xx_log_visit AS location_ip ORDER BY idvisit DESC LIMIT 10";
// I am selecting all so you can't see config_os but it is there . I cropped unnecessary code for keeping it easy to read.

    $result=mysql_query($sql);
        $numofrows = @mysql_num_rows($result);
   
   
    while($row = mysql_fetch_array($result))
    {

$system = $row['config_os'];
                $SysName = array ("AND", "IPH", "LIN", "MAC", "UNK", "W2K", "W95", "W98", "WI7", "WME", "WNT", "WOS", "WS3", "WVI", "WXP");
                $SysRename = array ("<img src='images/os/AND.gif'>","<img src='images/os/IPH.gif'>","<img src='images/os/LIN.gif'>","<img src='images/os/MAC.gif'>","<img src='images/os/UNK.gif'>","<img src='images/os/W2K.gif'>","<img src='images/os/W95.gif'>","<img src='images/os/W98.gif'>","<img src='images/os/wi7.gif'>","<img src='images/os/WME.gif'>","<img src='images/os/WNT.gif'>","<img src='images/os/WOS.gif'>","<img src='images/os/WS3.gif'>","<img src='images/os/WVI.gif'>","<img src='images/os/WXP.gif'>");
                $showsystem= str_replace($SysName, $SysRename, $system);
What I do here is simply use strreplace function.

$showsystem = str_replace($SysName, $SysRename, $system);

and in the display table, continue from code:

Code: [Select]
            echo ("
                <div class='sub15'>$showsystem</div>
            " );       
}
So for example instead of writing WI7 as visitors os, it shows WI7.gif, cool hmm

How can i automate this instead of writing each of them separately?

Thanks for any help..

I can't get the COUNT of the matching results? Please help.


$SomeVar = myname;
$queryU =  "SELECT * FROM adxone WHERE username = '".$SomeVar."'";
$resultU = mysql_query($queryU);
while($scoreP = mysql_fetch_array($resultU))
{
$scoreP['roundzAa'];
$scoreP['roundzAb'];
$scoreP['roundzAc'];
$scoreP['roundzAd'];
$scoreP['roundzAe'];
$scoreP['roundzAf'];
$scoreP['roundzAg'];
$scoreP['roundzAh'];
}


$WinV = 'resultA';
$query =  "SELECT * FROM acs WHERE resultx = '".$WinV."'";
$result = mysql_query($query);
while($scoreUsr = mysql_fetch_array($result))
{
$scoreUsr = $scoreM1['winxa'];  
$scoreUsr = $scoreM1['winxb']; 
$scoreUsr = $scoreM1['winxc']; 
$scoreUsr = $scoreM1['winxd']; 
$scoreUsr = $scoreM1['winxe']; 
$scoreUsr = $scoreM1['winxf']; 
$scoreUsr = $scoreM1['winxg']; 
$scoreUsr = $scoreM1['winxh']; 
}

$count1 = count( array_intersect($resultU, $result) );
echo($count1);

mysql_query("UPDATE comp SET id=$count1 WHERE username = '".$SomeVar."'");
?>

If I have a db with these fields: displayname, title

and I run a successful query on the db, will this code:


if (mysql_num_rows($results) == 0) {
         return "!fail";
      } else {
         while ($row = mysql_fetch_assoc($results)) {
             extract($row);
         }
         
         return $row;
      }


 create an associative array of $var = array(array('displayname' => value, 'title' => value), array('displayname' => value, 'title' => value)...)

If not, is there a way to do so, especially when I will not know the names or number of fields in a db table?

I've got my code working OK in that the correct results are drawn from the database. What I have here in the code is an array consisting of an image, accompanied by its title and thirdly a link to activate a quiz associated with the image. Everything works fine, except because the page is dynamic, I always have an unknown number of sets of data (image, title and link) which I want to display so that each set of data takes up a new row.
The code I have here, while displaying the correct data, places all of the results into the same row.
Any suggestions are most welcome.
Code: [Select]
<?php
// Query the database
$query1 = mysql_query("SELECT title FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC"); 
$query2 = mysql_query("SELECT url_small FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC"); 
$query3 = mysql_query("SELECT title FROM topics WHERE managerId='".$managerId."' AND egroup1='"."1"."' ORDER BY title ASC");
while($row1 = mysql_fetch_array($query3))
{
$linkname .= $row1['title']."<br />\n";
}
?>
                  <table>
                    <tr>
                      <td>
                      <?php
                      while ($row2 = mysql_fetch_array($query2)) 
                      { 
                      $thumbnail.= $row2['url_small'];
                      echo "<img src='wood_tool_images/{$row2['url_small']}' alt='' /><br />\n";
                      }
                      ?>
                      </td>
                      <td height="200">
                      <?php
                      echo $linkname 
                      ?>
                      </td>
                      <td>
                      <?php
                      while ($row1 = mysql_fetch_array($query1)) 
                      { 
                      $quizname.= $row1['title'];
                      echo "<a href='../{$row1['title']} Safety Quiz/{$row1['title']} Safety Quiz.php'>Take This Quiz</a><br />\n";
                      }
                      ?>
                      </td>
                     </tr>
                  </table>

Hi,

I've used this code before, but have had to make some modifications and am now getting a mysql fetch array error:

 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tesolcla/public_html/test/results10k201010.php on line 134

If anyone can help, it would be much appreciated.

Code: [Select]

<?php
$dbcnx = @mysql_connect('localhost', 'MYUSERNAME', 'MYPASSWORD');
//$dbcnx = @mysql_connect('localhost', 'root', 'mysql');
if (!$dbcnx) {
  exit('<p>Unable to connect to the ' .
      'database server at this time.</p>');
}

if (!@mysql_select_db('MYDATABASE')) {
//if (!@mysql_select_db('rac')) {
  exit('<p>Unable to locate the results ' .
      'database at this time.</p>');
}
$asc_on = '<img src="images/results_sorting/Asc.gif" border="0" />';
$asc_off= '<img src="images/results_sorting/AscOff.gif" border="0" />';
$desc_on = '<img src="images/results_sorting/Desc.gif" border="0" />';
$desc_off= '<img src="images/results_sorting/DescOff.gif" border="0" />';

$sortfield = isset($_GET['sort']) ? $_GET['sort'] : '4';
$sorttype = isset($_GET['type']) ? $_GET['type'] : '1';

for($i=1; $i<5; $i++)
{
if($i==$sortfield)
{
if ($sorttype==1)
$srt[$i] = $asc_on.'<a href="?sort='.$i.'&type=2">'.$desc_off.'</a>';
else
$srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a>'.$desc_on;
}
else
{
$srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a><a href="?sort='.$i.'&type=2">'.$desc_off.'</a>';
}
}

$fields = array("firstname", "lastname", "time", "position");
$sorts = array("ASC", "DESC");

$field = $fields[$sortfield-1];
$sort = $sorts[$sorttype-1];

$field = $field=="" ? $fields[4] : $field;
$sort = $sort=="" ? $sorts[0] : $sort;


$sql = mysql_query("SELECT firstname, lastname, time, position FROM 10k_results ORDER BY $field $sort");


echo "<table border='1'  align='center' bordercolor='#000000' CELLPADDING=5 cellspacing='0' STYLE='font-size:13px'>";
echo "<tr bgcolor='#008000' STYLE='color:white'> <td>*</td><td><H3>First name $srt[1]</h3></td> <td><H3>Lastname $srt[2]</H3></td>  <td><H3>Time $srt[3]</H3></td>  <td><H3>Position $srt[4]</H3></td></tr>";
// keeps getting the next row until there are no more to get

 $row_counter = 1; //create row counter with default value 0

// Print out the contents of each row into a table


 while ($row = mysql_fetch_array($sql))

{   
 // Print out the contents of each row into a table   
echo "<tr>\n";    


echo "</td><td>";
echo $row_counter++; 
echo "</td>";
echo "<td>{$row['firstname']}</td>\n";    
echo "<td>{$row['lastname']}</td>\n";   
echo "<td>{$row['time']}</td>\n";    
echo "<td>{$row['position']}</td>\n";   
echo "</tr>\n";
}
echo "</table>";


?>

Hello, im new here, and i have little experience to php and mysql as i started for 2 weeks ago.
I started out with some tutorials and feeling im getting the hang of it.

Enough of me, lets get to the point:
<?php

$con = mysql_connect('localhost',$user,$pass)or die(mysql_error());
$selectdb = mysql_select_db($selectdb)or die(mysql_error());

$sql = "SELECT * From table";
$result = mysql_query($sql);
$num = mysql_num_rows($result);
$myarray = array($result);
$i =0;

while ($i < $num){

echo $myarray[$i];
$i++;
}

?>

Here i have written a dummyscript that does what the original script does, it tries to fetch the keys from the table and then trying to loop it and echo out the results.
The output in the browser is this: Resource id #3

I know this probably is a simple fix but i cant seem to get it sorted out. Hope some of you could help me get this baby work, or maybe have another way of doing it more "simple".


Thanks in advance!
Dan-Levi

I wish to order my results by its occurrence inside the $status array.

Basically all of the results with the status 'New' are to be listed first, then the results with the status Today, Tomorrow, Soon and so on.

Is this possible, or is there alternatively a different approach to ordering my results how I desire?

Code: [Select]
$status = array('New', 'Today', 'Tomorrow', 'Soon', 'Future', 'Complete');

$display->get_results("SELECT * FROM todo ORDER by status DESC");

the form results are being sent to the selected email address in the array ... but I need a copy sent to the admin email address as well.

before adding the array addresses ... the forms results were being sent to the admin, but after adding the array ... the admin address is out of the loop.

How do I get the admin@home.com address to receive a copy of the forms results?

I'm not sure if it needs to be a Bcc or what?

Thanks in advance.

_________________________________________________ _______________________________________

$to = $_POST["to"];
if(!in_array($to,array("ian@elsewhere.com","adam@elsewhere.com","keith@elsewhere.com","sue@elsewhere.com","mark@elsewhere.com","scott@elsewhere.com","joyce@elsewhere.com","joe@elsewhere.com","ken@elsewhere.com","info@elsewhere.com")))
$to = "admin@home.com";
$subject = "Order";