CODEWITHSUNDEEP
phpmysql
Josh
Josh

Reputation: 133

passing multiple values in a url

have a database with a list of business leads. They are all given a status of New, Hot, Cold, etc. I have links on the page to display only the New or only Hot that work fine, but I can't seem to get one working to display All. The default view is New. Here is what I'm working with, thanks in advance for your help.

$query = "SELECT * FROM contacts WHERE contactstatus = 'New' ORDER BY date DESC";

if(isset($_GET['contactstatus'])
&& in_array($_GET['contactstatus'], array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed')))
{      
$status = $_GET['contactstatus'];   
$query = "SELECT * FROM contacts WHERE contactstatus ORDER BY date DESC";  
}

if(isset($_GET['contactstatus'])
&& in_array($_GET['contactstatus'], array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed')))
{      
$status = $_GET['contactstatus'];   
$query = "SELECT * FROM contacts WHERE contactstatus = '".$status."' ORDER BY date DESC";  
}

The url that I'm using to get all is:

www.mydomain.com/leads.php?contactstatus=New&contactstatus=Hot&contactstatus=Cold&contactstatus=Rejected&contactstatus=Closed

I've also tried:

www.mydomain.com/leads.php?contactstatus=New&Hot&Cold&Rejected&Closed

Upvotes: 1

Views: 1086

Answers (3)

plod
plod

Reputation: 517

As you are trying to pass the same variable through the address have you tried using an array otherwise you are overwriting the get variable.

Like so www.mydomain.com/leads.php?contactstatus[]=New&contactstatus[]=Hot&contactstatus[]=Cold&contactstatus[]=Rejected&contactstatus[]=Closed

$status_types = Array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed');

$statuses = Array();
$query  = 'SELECT * FROM contacts ';
for($i=0, $j=count($_GET['contactstatus']); $i<$j; $i++){
    if(in_array($_GET['contactstatus'][$i], $status_types)){
        if($i==0)
            $query.= " WHERE contactstatus IN (";

        $statuses[] = $_GET['contactstatus'][$i];
        $query .= "'".$_GET['contactstatus'][$i]."'";
        if($i==($j-1))
            $query .= ")";

    }
}
$query .= ' ORDER BY contacts.date DESC';

I have set this up to automaticly select all if no contactstatus is passed.

Upvotes: 2

Don
Don

Reputation: 839

Couldn't you just set the default behavior of your php page to display all? That is, if no $_GET parameter is set (in the URL), your page simply selects all leads from database.

$query = " SELECT * FROM `contacts` ORDER BY date DESC";
$result = $mysqli->query($query);

if($result && $result->num_rows > 0){
    // do something with all leads
}

To deal with specific statuses, we have 

if( isset($_GET['contactstatus']) ){
// select leads based on status 
}

If you want to pass multiple values to a single GET parameter, see Passing multiple values for same GET variable in URL

Upvotes: 0

aynber
aynber

Reputation: 23001

The first if looks the same as the second. You're also missing the WHERE qualifier on the first query.

If you want to get all of the contacts, change this

$query = "SELECT * FROM contacts WHERE contactstatus ORDER BY contacts.date DESC";

to this

$query = "SELECT * FROM contacts ORDER BY contacts.date DESC";

Upvotes: 0

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