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How can I put MySQL fields into an array

I'm trying to create and array that will be passed into a drop down ComboBox in Flash. It works fine If I put the text into the array but below where "field 1", "field 2", "field 3", "field 4", "field 5", "field 6" is needs to come from fields in MySQL. I can't seem to figure this out do I create a while or for loop to go inside the $NameArr or what?

Please help.
@mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
@mysql_select_db("$db_name") or die ("no database");
 
$result = mysql_query("SELECT * FROM job_listings WHERE employer_id='$id'");
 
$NameArr = array("field 1", "field 2", "field 3", "field 4", "field 5", "field 6");
 
 
// below is the code that is working
$NameArr = array("field 1", "field 2", "field 3", "field 4", "field 5", "field 6");
 
$numReturn = count($NameArr);
$i = 0;
print "&";
 
	while ($i < $numReturn) {
		$Name 		= $NameArr[$i];
 
		print "Name$i=$Name&";
		$i++;
	}
print "&NumItems=$numReturn&Go=Yes&";

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lallen30
Asked:
lallen30
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4 Solutions
 
shobinsunCommented:
0
 
lallen30Author Commented:
Thanks for such a quick reply I've added some code based on the code from that link but I have a problem. When I test the php page it just outputs this:

 &Name0=Array&Name1=Array&Name2=Array&Name3=Array&Name4=Array&Name5=Array&Name6=Array&Name7=Array&Name8=Array&&NumItems=9&Go=Yes&

I need it to output this:
&Name0=field 1&Name1=field 2&Name2=field 3&Name3=field 4&Name4=field 5&Name5=field 6&&NumItems=6&Go=Yes&


Any ideas on how I can fix this?
<?php
include_once "connect_to_mysql.php";
include_once "includes/id.php";
 
$result = mysql_query("SELECT * FROM job_listings WHERE employer_id='$id'");
 
 
// define the $feedList array
$feedList = array();
 
      while($row1 = mysql_fetch_array($result)){
     
            $feedid = $row1['employer_id'];
           
            $getfeedinfo = "SELECT * FROM job_listings WHERE employer_id='$id'";
            $feedinfo = mysql_query($getfeedinfo) or die(mysql_error());
           
                  while($feedinfocontent = mysql_fetch_array($feedinfo)){
 
                        $feedList[] = array(
                          'job_title' => $feedinfocontent['job_title'],
						  
                         );
                  }
      }
	  
	$numReturn = count($feedList);
	$i = 0;
	print "&";
	
		while ($i < $numReturn) {
			$Name 		= $feedList[$i];
	
			print "Name$i=$Name&";
			$i++;
		}
		
print "&NumItems=$numReturn&Go=Yes&";
?>

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alexandremattarCommented:
Try:
while($feedinfocontent = mysql_fetch_array($feedinfo)){
 
                        $feedList[] = $feedinfocontent['job_title'];
                  }

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lallen30Author Commented:
Thanks for your reply alexandremattar.
Unfortunately I'm getting this error now that I changed it:

Parse error: parse error in C:\wamp\www\collegejobs\read_job_post.php on line 25


Any other ideas would be much appreciated.
<?php
include_once "connect_to_mysql.php";
include_once "includes/id.php";
 
$result = mysql_query("SELECT * FROM job_listings WHERE employer_id='$id'");
 
 
 
 
// define the $feedList array
$feedList = array();
 
      while($row1 = mysql_fetch_array($result)){
     
            $feedid = $row1['employer_id'];
           
            $getfeedinfo = "SELECT * FROM job_listings WHERE employer_id='$id'";
            $feedinfo = mysql_query($getfeedinfo) or die(mysql_error());
           
                while($feedinfocontent = mysql_fetch_array($feedinfo)){
 
                        $feedList[] = $feedinfocontent['job_title'];
                  }
            }
      }
	  
	$numReturn = count($feedList);
	$i = 0;
	print "&";
	
		while ($i < $numReturn) {
			$Name 		= $feedList[$i];
	
			print "Name$i=$Name&";
			$i++;
		}
		
print "&NumItems=$numReturn&Go=Yes&";
?>

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shobinsunCommented:
Hello,

You have an extra '}' in your code. Please remove it. Use the original code as:

Regards
<?php
include_once "connect_to_mysql.php";
include_once "includes/id.php";
 
$result = mysql_query("SELECT * FROM job_listings WHERE employer_id='$id'");
 
 
 
 
// define the $feedList array
$feedList = array();
 
      while($row1 = mysql_fetch_array($result)){
     
            $feedid = $row1['employer_id'];
           
            $getfeedinfo = "SELECT * FROM job_listings WHERE employer_id='$id'";
            $feedinfo = mysql_query($getfeedinfo) or die(mysql_error());
           
                while($feedinfocontent = mysql_fetch_array($feedinfo)){
 
                        $feedList[] = $feedinfocontent['job_title'];
                  }
            }
      
          
        $numReturn = count($feedList);
        $i = 0;
        print "&";
        
                while ($i < $numReturn) {
                        $Name           = $feedList[$i];
        
                        print "Name$i=$Name&";
                        $i++;
                }
                
print "&NumItems=$numReturn&Go=Yes&";
?>

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lallen30Author Commented:
You have been very helpful and thanks so much for catching that.
It's outputting the right info now but its repeating it serveral times.

Heres what it's outputing:
&Name0=Floor Supervisor&Name1=Janitor&Name2=Plumber&Name3=Floor Supervisor&Name4=Janitor&Name5=Plumber&Name6=Floor Supervisor&Name7=Janitor&Name8=Plumber&&NumItems=9&Go=Yes&

Even though there is only 3 rows it's repeating them and counting up to 9.
It's so close to being there. Is there anyway to fix this?

0
 
shobinsunCommented:
Hello,

Use this    "$numReturn = mysql_num_rows($feedinfo);"

  instead of  

  "$numReturn = count($feedList);"

Regards
0
 
lallen30Author Commented:
Again Thanks so much. It works great.

For anyone else trying to do this below is the final code. First is the PHP and below that is the Actionscript 2.0.

// php file is ComboBox_PHP_Link.php - futher down is the actionscript
<?php
include_once "connect_to_mysql.php";
include_once "includes/id.php";
 
$result = mysql_query("SELECT * FROM job_listings WHERE employer_id='$id'");
 
// define the $feedList array
$feedList = array();
 
      while($row1 = mysql_fetch_array($result)){
     
            $feedid = $row1['employer_id'];
           
            $getfeedinfo = "SELECT * FROM job_listings WHERE employer_id='$id'";
            $feedinfo = mysql_query($getfeedinfo) or die(mysql_error());
           
                while($feedinfocontent = mysql_fetch_array($feedinfo)){
 
                        $feedList[] = $feedinfocontent['job_title'];
                  }
            }
      
	  
	$numReturn = mysql_num_rows($feedinfo);
	$i = 0;
	print "&";
	
		while ($i < $numReturn) {
			$Name 		= $feedList[$i];
	
			print "Name$i=$Name&";
			$i++;
		}
		
print "&NumItems=$numReturn&Go=Yes&";
?>
 
 
 
 
//--------------------------------- actionscript is:
 
// Main Functions.. This controls your Drop Down / Combo Box.
 
// The AddItems function assigns values to the drop down box from the external source
function AddItems() {
	for (i=0; i<NumItems; i++) {
		var Name 		= eval("Name"+i);
		var DataRow 	= eval("DataRow"+i);
   		dropDown.addItem(Name, DataRow); 
	}
	//Set ChangeHandler (this is the most important line).
	dropDown.setChangeHandler("SelectItem");
}
 
// The SelectItem function tells the movie what to do when a user
// Clicks on a certain Item in the drop down movie - this is your OnChange Handler.
 
// In this example we use it to open up a new URL.
function SelectItem(){
   URLName	= dropDown.getSelectedItem().label; 
   URL		= dropDown.getSelectedItem().data;
   
   getURL(URL, "_blank");
}
 
 
// Makes sure the initial value of Go is equal to zero
_root.Go = "";
 
// Tells the thinking MC to gotoAndPlay(2) - and wait.
_root.Thinking.gotoAndPlay(2);
 
// Specify the URL to either the Text file or PHP script that loads data from DB.
// Make sure to change the path to your Data Source.
loadVariables("http://localhost/collegejobs/test/phparray/ComboBox_PHP_Link/ComboBox_PHP_Link.php", this);

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lallen30Author Commented:
Thanks so much for your help you have been a life saver.
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