Array format

[timmah1]

I have my array like so

$inaque = array("'$dir'"=>"'$company'");

 

When I run

print_r($inaque);

 

It shows like

Array ( ['share_what_you_want'] => 'Share What You Want' )

 

How can I get this array to print out like

array ( 'share_what_you_want' => 'Share What You Want' )

 

Without the [ ]

 

This is the format that is needed for this code that has already been written by somebody else.

 

Thanks in advance

[Muddy_Funster]

that's just what print_r does, I can't fathom why anyone would write code to check what print_r outputs rather than just checking against the original array, that's just crazy.  You're either going to have to pass the print_r result to a sting and then do a regular expression replace on it, or can the code that was written by a crazy person and find something sane to do the job.

[seanlim]

var_export seems to fit your requirements quite well

[timmah1]

Thank you seanlim, it's what we needed.

 

Now, I am still learning about arrays, so I know some of my questions may be minor.

 

This is only pulling one from the database, when in fact, there is 2 right now, but the array is only showing one.

 

How can I put all of the values into the array?

include($_SERVER['DOCUMENT_ROOT']."/inc/config.db.php");
            $path = $_SERVER['DOCUMENT_ROOT']."/panel/testing/connectors/sample/projects/";
                            
                    foreach(glob($path . '*', GLOB_ONLYDIR) as $dir) {
                        $dir = basename($dir);
                        //echo $dir."<br />";
                                    
                        $companyName = "SELECT * FROM app_queue WHERE company LIKE '%$dir%' AND status = '1'";
                        $compResult = mysql_query($companyName);
                              
                            while($getComp = mysql_fetch_array($compResult)){
                                $status = $getComp['status'];
                                $company = $getComp['company'];
                            
                                    $inaque = array($dir=>$company);                                    
                            }
                                                    
                    };
                    return $inaque;

[PFMaBiSmAd]

This is the format that is needed for this code that has already been written by somebody else.

 

I can just about bet that the code is expecting an actual php array variable with specific key/value pairs. Not the print-out/print_r/var_dump/var_export (which was already suggested in your previous thread on this) of an array and not a string that looks like an array.

 

What is this code you are trying to supply data to?

[timmah1]

This is the format that is needed for this code that has already been written by somebody else.

What is this code you are trying to supply data to?

 

It's for building apps dynamically. Right now, we build them manually, what this code does is

 

Scans a directory, checks the db for a company name that matches the $dir variable, then puts items into the array.

 

That part works fine now, the problem I'm having now is that the array is only populated with 1 item, and the array should show all items.

Right now, there are 2 in thh db with a status of 1, but the array is only pulling 1 of them

[PFMaBiSmAd]

I bet you are getting the last data?

 

$inaque = array($dir=>$company);

 

^^^ That line of code overwrites the $inaque variable on each iteration of the loop. You only end up with the last set of data after the end of the loop. If your goal (which you haven't actually shown) is to build an array of arrays, you would use -

 

$inaque[] = array($dir=>$company);

[timmah1]

ok, I'm obviously lost.

 

I need t query the db with each $dir variable, and then return the result.

I can query the db and pull exactly what I need with one $dir, but I have no idea how to query the db foreach $dir variable.

 

This is the code, can somebody please help me out here?

<?php

include($_SERVER['DOCUMENT_ROOT']."/inc/config.db.php");
            $path = $_SERVER['DOCUMENT_ROOT']."/panel/testing/connectors/sample/projects/";
                            
                    foreach(glob($path . '*', GLOB_ONLYDIR) as $dir) {
                        $dir = basename($dir);
                        //echo $dir."<br />";
                                    
                        $companyName = "SELECT * FROM app_queue WHERE company LIKE '%$dir%' AND status = '1'";
                        $compResult = mysql_query($companyName);
                              
                            while($getComp = mysql_fetch_array($compResult)){
                                $status = $getComp['status'];
                                $company = $getComp['company'];
                            
                                    $inaque = array($dir=>$company);                                    
                            }
                                                    
                    };
                    return $inaque;

?>